This section covers:

Implicit Differentiation
Equation of the Tangent Line with Implicit Differentiation
Related Rates
More Practice

Introduction to Implicit Differentiation

Up to now, we’ve differentiated in explicit form, since, for example, \(y\) has been explicitly written as a function of \(x\).

But sometimes, we can’t get an equation with a “\(y\)” only on one side; we may have multiply “\(y\)’s” in the equation. In these cases, we have to differentiate “implicitly”, meaning that some “\(y\)’s” are “inside” the equation. This is called implicit differentiation, and we actually have to use the chain rule to do this. Here’s an example of an equation that we’d have to differentiate implicitly: \(y=7{{x}^{2}}y-2{{y}^{2}}-\sqrt{{xy}}\). Do you see how it’d be really difficult to get \(y\) alone on one side?

The main thing to remember is when you are differentiating with respect to “\(x\)” and what you are differentiating only has “\(\boldsymbol {x}\)’s” in it (or constants), you just get the derivative the normal way (since the \(dx\)’s cancel out). But if you are differentiating with respect to \(x\), and what you are differentiating has another variable in it, like “\(y\)”, you have to multiply by \(\displaystyle \frac{{dy}}{{dx}}\text{ (}{y}’)\):

\(\require{cancel} \displaystyle \frac{d}{{dx}}\left( {{{x}^{2}}} \right)=2x\cdot \frac{{\cancel{{dx}}}}{{\cancel{{dx}}}}=2x\): variables agree, so just use the power rule (chain rule has no effect)

\(\displaystyle \frac{d}{{dx}}\left( {{{y}^{2}}} \right)=2y\cdot \frac{{dy}}{{dx}}\): variables disagree, so use the power rule, and then the chain rule

After we do the differentiation, we want to solve for the \(\displaystyle \frac{{dy}}{{dx}}\) by getting it to one side by itself (and we may have both \(x\)’s and \(y\)’s on the other side, which is fine). Here are the steps for differentiating implicitly:

Implicit Differentiation

Differentiate both sides of equation with respect to \(\boldsymbol{x}\). When are you differentiating variables other than \(x\) (such as “\(y\)”), remember to multiply that term by \(\displaystyle \frac{{dy}}{{dx}}\text{ (}{y}’)\)!
Move to the left side of the equation and move all other terms to the right side (even if they have \(x\)’s and \(y\)’s in them).
Factor out the \({y}’\) on the left side of the equation, and solve for \({y}’\).

Example: \(\displaystyle \begin{align}{{x}^{2}}+{{y}^{2}}&=25\\2x+2y{y}’&=0\\2y{y}’&=-2x\\{y}’&=-\frac{x}{y}\end{align}\)

I know it looks a bit scary, but it’s really not that bad!

Let’s do some non-trig problems first. Do you see how we have to use the chain rule a lot more? And note how the algebra can get really complicated!

Implicit Differentiation Problem	Solution
Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation: \({{x}^{2}}+{{y}^{2}}=25\)	\(\displaystyle \begin{align}{{x}^{2}}+{{y}^{2}}&=25\\2x+2y{y}’&=0\\2y{y}’&=-2x\\{y}’&=-\frac{x}{y}\end{align}\)
Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation: \(\displaystyle \frac{{x+y}}{{{{y}^{2}}}}=xy\)	\(\displaystyle \begin{align}\frac{{x+y}}{{{{y}^{2}}}}&=xy\\\frac{{{{y}^{2}}\left( {1+{y}’} \right)-\left( {x+y} \right)\left( {2y{y}’} \right)}}{{{{y}^{4}}}}&=x{y}’+y\\{{y}^{2}}+{y}'{{y}^{2}}-2xy{y}’-2{{y}^{2}}{y}’&=\left( {x{y}’+y} \right){{y}^{4}}\\{{y}^{2}}+{y}'{{y}^{2}}-2xy{y}’-2{{y}^{2}}{y}’&=x{{y}^{4}}{y}’+{{y}^{5}}\\{y}'{{y}^{2}}-2xy{y}’-2{{y}^{2}}{y}’-x{{y}^{4}}{y}’&={{y}^{5}}-{{y}^{2}}\\{y}’\left( {{{y}^{2}}-2xy-2{{y}^{2}}-x{{y}^{4}}} \right)&={{y}^{5}}-{{y}^{2}}\\{y}’&=\frac{{{{y}^{5}}-{{y}^{2}}}}{{-{{y}^{2}}-2xy-x{{y}^{4}}}}=\frac{{{{y}^{4}}-y}}{{-y-2x-x{{y}^{3}}}}\end{align}\)
Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation: \(\sqrt{{xy}}+2x{{y}^{2}}=3\)	\(\displaystyle \begin{align}{{\left( {xy} \right)}^{{\frac{1}{2}}}}+2x{{y}^{2}}&=3\\\frac{1}{2}{{\left( {xy} \right)}^{{-\frac{1}{2}}}}\left( {x{y}’+y} \right)+2\left( {x\cdot 2y{y}’+{{y}^{2}}} \right)=0\\\cancel{{{{{\left( {xy} \right)}}^{{\frac{1}{2}}}}}}\left[ {\frac{1}{2}\cancel{{{{{\left( {xy} \right)}}^{{-\frac{1}{2}}}}}}\left( {x{y}’+y} \right)+2\left( {2xy{y}’+{{y}^{2}}} \right)} \right]&=0\cdot {{\left( {xy} \right)}^{{\frac{1}{2}}}}\,\,\,\,\text{multiply both sides by }{{\left( {xy} \right)}^{{\frac{1}{2}}}}\\\frac{1}{2}\left( {x{y}’+y} \right)+4xy{y}'{{\left( {xy} \right)}^{{\frac{1}{2}}}}+2{{y}^{2}}{{\left( {xy} \right)}^{{\frac{1}{2}}}}&=0\\x{y}’+y+8xy{y}'{{\left( {xy} \right)}^{{\frac{1}{2}}}}+4{{y}^{2}}{{\left( {xy} \right)}^{{\frac{1}{2}}}}&=0\,\,\,\,\,\text{multiply both sides by }2\\{y}’\left( {8{{x}^{{\frac{3}{2}}}}{{y}^{{\frac{3}{2}}}}+x} \right)&=-y-4{{x}^{{\frac{1}{2}}}}{{y}^{{\frac{5}{2}}}}\\{y}’&=\frac{{-y-4{{x}^{{\frac{1}{2}}}}{{y}^{{\frac{5}{2}}}}}}{{8{{x}^{{\frac{3}{2}}}}{{y}^{{\frac{3}{2}}}}+x}}=-\frac{{y+4{{x}^{{\frac{1}{2}}}}{{y}^{{\frac{5}{2}}}}}}{{8{{x}^{{\frac{3}{2}}}}{{y}^{{\frac{3}{2}}}}+x}}\end{align}\)

Let’s try some problems with trig now. Note that with trig functions, it’s hard to know where to stop simplifying, so there are several “correct” answers.

Implicit Differentiation Problem	Solution
Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation: \(\sin x+4\cos \left( {3y} \right)=1\)	\(\displaystyle \begin{align}\sin x+4\cos \left( {3y} \right)&=1\\\cos x+4\cdot -\sin \left( {3y} \right)\cdot 3\cdot {y}’&=0\\-12{y}’\sin \left( {3y} \right)&=-\cos x\\{y}’&=\frac{{-\cos x}}{{-12\sin \left( {3y} \right)}}=\frac{{\cos x}}{{12\sin \left( {3y} \right)}}=\frac{1}{{12}}\cos x\csc \left( {3y} \right)\end{align}\)
Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation: \(\cos x=x\left( {2+\csc y} \right)\)	\(\displaystyle \begin{align}\cos x&=x\left( {2+\csc y} \right)\\-\sin x&=x\left( {0+-\csc y\cot y\cdot {y}’} \right)+\left( {2+\csc y} \right)\left( 1 \right)\\-\sin x&=-x{y}’\csc y\cot y+2+\csc y\\x{y}’\csc y\cot y&=2+\csc y+\sin x\\{y}’&=\frac{{2+\csc y+\sin x}}{{x\csc y\cot y}}\\&=\frac{2}{{x\csc y\cot y}}+\frac{{\cancel{{\csc y}}}}{{x\cancel{{\csc y}}\cot y}}+\frac{{\sin x}}{{x\csc y\cot y}}\,\,\,\,\,(\text{expand)}\\&=\frac{2}{{x\frac{1}{{\sin y}}\cdot \frac{{\cos y}}{{\sin y}}}}+\frac{1}{{x\cdot \frac{{\cos y}}{{\sin y}}}}+\frac{{\sin x}}{{x\cdot \frac{1}{{\sin y}}\cdot \frac{{\cos y}}{{\sin y}}}}\,\,\,\,\,\,\text{(put in terms of sin, cos)}\\&=\frac{{2{{{\sin }}^{2}}y}}{{x\cos y}}+\frac{{\tan y}}{x}+\frac{{\sin x\cdot {{{\sin }}^{2}}y}}{{x\cos y}}\\&=\frac{{2\sin y\tan y+\tan y+\sin x\sin y\tan y}}{x}\end{align}\) (see how we can get carried away with simplifying?)
Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation: \(\displaystyle \begin{array}{c}{{\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi x} \right)} \right)}^{2}}\\=4\end{array}\)	\(\displaystyle \begin{align}{{\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}^{2}}&=4\\2{{\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}^{1}}\left( {\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi +{{{\sec }}^{2}}\left( {\pi y} \right)\cdot \pi {y}’} \right)&=0\\\frac{{\cancel{{2\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}}\left( {\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi +{{{\sec }}^{2}}\left( {\pi y} \right)\cdot \pi {y}’} \right)}}{{\cancel{{2\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}}}}&=\frac{0}{{2\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}}\\\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi +{{\sec }^{2}}\left( {\pi y} \right)\cdot \pi {y}’&=0\\{{\sec }^{2}}\left( {\pi y} \right)\cdot \pi {y}’&=-\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi \\{y}’&=\frac{{-\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \cancel{\pi }}}{{{{{\sec }}^{2}}\left( {\pi y} \right)y\cdot \cancel{\pi }}}\\=-\sec \left( {\pi x} \right)\tan &\left( {\pi x} \right){{\cos }^{2}}\left( {\pi y} \right)\end{align}\)
Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation: \(\displaystyle \cos (xy)=5x\)	\(\displaystyle \begin{align}\cos \left( {xy} \right)&=5x\\-\sin \left( {xy} \right)\left( {x{y}’+y} \right)&=5\\\sin \left( {xy} \right)\left( {x{y}’+y} \right)&=-5\\x{y}’\sin \left( {xy} \right)+y\sin \left( {xy} \right)=-5\\x{y}’\sin \left( {xy} \right)&=-5-y\sin \left( {xy} \right)\\{y}’&=-\frac{{5+y\sin \left( {xy} \right)}}{{x\sin \left( {xy} \right)}}=\frac{{-5\csc \left( {xy} \right)+y}}{x}\end{align}\)

Implicit Differentiation Problem

Solution

Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation:

\(\sin x+4\cos \left( {3y} \right)=1\)

\(\displaystyle \begin{align}\sin x+4\cos \left( {3y} \right)&=1\\\cos x+4\cdot -\sin \left( {3y} \right)\cdot 3\cdot {y}’&=0\\-12{y}’\sin \left( {3y} \right)&=-\cos x\\{y}’&=\frac{{-\cos x}}{{-12\sin \left( {3y} \right)}}=\frac{{\cos x}}{{12\sin \left( {3y} \right)}}=\frac{1}{{12}}\cos x\csc \left( {3y} \right)\end{align}\)

Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation:

\(\cos x=x\left( {2+\csc y} \right)\)

\(\displaystyle \begin{align}\cos x&=x\left( {2+\csc y} \right)\\-\sin x&=x\left( {0+-\csc y\cot y\cdot {y}’} \right)+\left( {2+\csc y} \right)\left( 1 \right)\\-\sin x&=-x{y}’\csc y\cot y+2+\csc y\\x{y}’\csc y\cot y&=2+\csc y+\sin x\\{y}’&=\frac{{2+\csc y+\sin x}}{{x\csc y\cot y}}\\&=\frac{2}{{x\csc y\cot y}}+\frac{{\cancel{{\csc y}}}}{{x\cancel{{\csc y}}\cot y}}+\frac{{\sin x}}{{x\csc y\cot y}}\,\,\,\,\,(\text{expand)}\\&=\frac{2}{{x\frac{1}{{\sin y}}\cdot \frac{{\cos y}}{{\sin y}}}}+\frac{1}{{x\cdot \frac{{\cos y}}{{\sin y}}}}+\frac{{\sin x}}{{x\cdot \frac{1}{{\sin y}}\cdot \frac{{\cos y}}{{\sin y}}}}\,\,\,\,\,\,\text{(put in terms of sin, cos)}\\&=\frac{{2{{{\sin }}^{2}}y}}{{x\cos y}}+\frac{{\tan y}}{x}+\frac{{\sin x\cdot {{{\sin }}^{2}}y}}{{x\cos y}}\\&=\frac{{2\sin y\tan y+\tan y+\sin x\sin y\tan y}}{x}\end{align}\)

(see how we can get carried away with simplifying?)

Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation:

\(\displaystyle \begin{array}{c}{{\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi x} \right)} \right)}^{2}}\\=4\end{array}\)

\(\displaystyle \begin{align}{{\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}^{2}}&=4\\2{{\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}^{1}}\left( {\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi +{{{\sec }}^{2}}\left( {\pi y} \right)\cdot \pi {y}’} \right)&=0\\\frac{{\cancel{{2\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}}\left( {\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi +{{{\sec }}^{2}}\left( {\pi y} \right)\cdot \pi {y}’} \right)}}{{\cancel{{2\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}}}}&=\frac{0}{{2\left( {\sec \left( {\pi x} \right)+\tan \left( {\pi y} \right)} \right)}}\\\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi +{{\sec }^{2}}\left( {\pi y} \right)\cdot \pi {y}’&=0\\{{\sec }^{2}}\left( {\pi y} \right)\cdot \pi {y}’&=-\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \pi \\{y}’&=\frac{{-\sec \left( {\pi x} \right)\tan \left( {\pi x} \right)\cdot \cancel{\pi }}}{{{{{\sec }}^{2}}\left( {\pi y} \right)y\cdot \cancel{\pi }}}\\=-\sec \left( {\pi x} \right)\tan &\left( {\pi x} \right){{\cos }^{2}}\left( {\pi y} \right)\end{align}\)

Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation:

\(\displaystyle \cos (xy)=5x\)

\(\displaystyle \begin{align}\cos \left( {xy} \right)&=5x\\-\sin \left( {xy} \right)\left( {x{y}’+y} \right)&=5\\\sin \left( {xy} \right)\left( {x{y}’+y} \right)&=-5\\x{y}’\sin \left( {xy} \right)+y\sin \left( {xy} \right)=-5\\x{y}’\sin \left( {xy} \right)&=-5-y\sin \left( {xy} \right)\\{y}’&=-\frac{{5+y\sin \left( {xy} \right)}}{{x\sin \left( {xy} \right)}}=\frac{{-5\csc \left( {xy} \right)+y}}{x}\end{align}\)

Equation of the Tangent Line with Implicit Differentiation

Here are some problems where you have to use implicit differentiation to find the derivative at a certain point, and the slope of the tangent line to the graph at a certain point. The last problem asks to find the equation of the tangent line and normal line (the line perpendicular to the tangent line – take the negative reciprocal of the slope) at a certain point.

Note that we learned about finding the Equation of the Tangent Line in the Equation of the Tangent Line, Tangent Line Approximation, and Rates of Change section.

Implicit Differentiation Tangent Problem	Solution
Find \(\displaystyle \frac{{dx}}{{dy}}\) by implicit differentiation and evaluate the derivative at point \(\left( {-2,4} \right)\): \({{x}^{2}}+{{y}^{3}}=68\)	\(\displaystyle \begin{align}{{x}^{2}}+{{y}^{3}}&=68\\2x+3{{y}^{2}}{y}’&=0\\3{{y}^{2}}{y}’&=-2x\\{y}’&=-\frac{{2x}}{{3{{y}^{2}}}}\end{align}\) \(\displaystyle \text{At }\left( {-2,4} \right),\,\,\,{y}’=-\frac{{2\left( {-2} \right)}}{{3{{{\left( 4 \right)}}^{2}}}}=-\frac{{-4}}{{48}}=\frac{1}{{12}}\)
Find the slope of the tangent line to the graph at point \(\left( {2,1} \right)\): \(\displaystyle {{x}^{3}}+{{y}^{3}}=\frac{9}{2}xy\)	\(\displaystyle \begin{align}{{x}^{3}}+{{y}^{3}}&=\frac{9}{2}xy\\3{{x}^{2}}+3{{y}^{2}}{y}’&=\frac{9}{2}\left( {x{y}’+y\cdot 1} \right)\\6{{x}^{2}}+6{{y}^{2}}{y}’&=9\left( {x{y}’+y} \right)\\6{{x}^{2}}+6{{y}^{2}}{y}’&=9x{y}’+9y\\6{{y}^{2}}{y}’-9x{y}’&=-6{{x}^{2}}+9y\\{y}’\left( {6{{y}^{2}}-9x} \right)&=-6{{x}^{2}}+9y\\{y}’&=\frac{{-6{{x}^{2}}+9y}}{{6{{y}^{2}}-9x}}\\{y}’&=\frac{{-2{{x}^{2}}+3y}}{{2{{y}^{2}}-3x}}\end{align}\) \(\displaystyle \text{At }\left( {2,1} \right),\,\,\,{y}’=\frac{{-2{{{\left( 2 \right)}}^{2}}+3\left( 1 \right)}}{{2{{{\left( 1 \right)}}^{2}}-3\left( 2 \right)}}=\frac{5}{4}\)
Find the equation of for the tangent line and the normal line to the circle at the point \(\left( {3,4} \right)\): \({{x}^{2}}+{{y}^{2}}=25\)	\(\displaystyle \begin{align}\\{{x}^{2}}+{{y}^{2}}&=25\\2x+2y{y}’&=0\\2y{y}’&=-2x\\{y}’&=-\frac{x}{y}\end{align}\) \(\displaystyle \begin{array}{c}\text{Tangent Line:}\\\text{At }\left( {3,4} \right),\,\,\,{y}’=-\frac{3}{4}\\y-{{y}_{1}}=-\frac{3}{4}\left( {x-{{x}_{1}}} \right)\\y-4=-\frac{3}{4}\left( {x-3} \right)\\y=-\frac{3}{4}x+\frac{{25}}{4}\end{array}\) \(\displaystyle \begin{array}{c}\text{Normal Line:}\\\text{At }\left( {3,4} \right),\,\,\,{y}’=-\frac{3}{4}\\y-{{y}_{1}}=\frac{4}{3}\left( {x-{{x}_{1}}} \right)\\y-4=\frac{4}{3}\left( {x-3} \right)\\y=\frac{4}{3}x\end{array}\)

Here’s a problem where we have to use implicit differentiation to find the points at which the graph of the equation has a horizontal and vertical tangent line. Remember that for the horizontal tangent line, we set the numerator to 0, and for the vertical tangent line, we set the denominator to 0.

Implicit Differentiation and Horizontal/Vertical Tangent Lines
Use implicit differentiation to find the points at which the graph of the equation has a horizontal and vertical tangent line: \(2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\)
Use Implicit Differentiation to get \({y}’\): \(\displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\4x+2y{y}’-4+2{y}’+0=0\\2y{y}’+2{y}’=-4x+4\\{y}’\left( {2y+2} \right)=-4x+4\end{array}\) \(\displaystyle {y}’=\frac{{-4x+4}}{{2y+2}}\) \(\displaystyle {y}’=\frac{{2-2x}}{{y+1}}\)	Points at Horizontal Tangent (set numerator to 0): \(\displaystyle \begin{array}{c}2-2x=0\\x=1\end{array}\) Now find \(y\) (use original): \(\displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\2{{\left( 1 \right)}^{2}}+{{y}^{2}}-4\left( 1 \right)+2y+2=0\\{{y}^{2}}+2y=0\\y\left( {y+2} \right)=0\\y=0,\,\,-2\\\left( {1,0} \right)\,\,\text{and}\,\,\left( {1,-2} \right)\end{array}\)	Points at Vertical Tangent (set denominator to 0): \(\displaystyle \begin{array}{c}y+1=0\\y=-1\end{array}\) Now find \(x\) (use original): \(\displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\2{{x}^{2}}+{{\left( {-1} \right)}^{2}}-4x+2\left( {-1} \right)+2=0\\2{{x}^{2}}-4x+1=0\end{array}\) \(\displaystyle x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\) \(\displaystyle =\frac{{-\left( {-4} \right)\pm \sqrt{{{{{\left( {-4} \right)}}^{2}}-4\cdot 2\cdot 1}}}}{{2\cdot 2}}\) \(\displaystyle x=\frac{{4\pm \sqrt{8}}}{4}=\frac{{4\pm 2\sqrt{2}}}{4}=\frac{{2\pm \sqrt{2}}}{2}\) \(\displaystyle \left( {\frac{{2-\sqrt{2}}}{2},\,\,-1} \right)\,\,\text{and}\,\,\left( {\frac{{2+\sqrt{2}}}{2},\,\,-1} \right)\)

Implicit Differentiation and Horizontal/Vertical Tangent Lines

Use implicit differentiation to find the points at which the graph of the equation has a horizontal and vertical tangent line:

\(2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\)

Use Implicit Differentiation to get \({y}’\):

\(\displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\4x+2y{y}’-4+2{y}’+0=0\\2y{y}’+2{y}’=-4x+4\\{y}’\left( {2y+2} \right)=-4x+4\end{array}\)

\(\displaystyle {y}’=\frac{{-4x+4}}{{2y+2}}\)

\(\displaystyle {y}’=\frac{{2-2x}}{{y+1}}\)

Points at Horizontal Tangent (set numerator to 0):

\(\displaystyle \begin{array}{c}2-2x=0\\x=1\end{array}\)

Now find \(y\) (use original):

\(\displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\2{{\left( 1 \right)}^{2}}+{{y}^{2}}-4\left( 1 \right)+2y+2=0\\{{y}^{2}}+2y=0\\y\left( {y+2} \right)=0\\y=0,\,\,-2\\\left( {1,0} \right)\,\,\text{and}\,\,\left( {1,-2} \right)\end{array}\)

Points at Vertical Tangent (set denominator to 0):

\(\displaystyle \begin{array}{c}y+1=0\\y=-1\end{array}\)

Now find \(x\) (use original):

\(\displaystyle \begin{array}{c}2{{x}^{2}}+{{y}^{2}}-4x+2y+2=0\\2{{x}^{2}}+{{\left( {-1} \right)}^{2}}-4x+2\left( {-1} \right)+2=0\\2{{x}^{2}}-4x+1=0\end{array}\)

\(\displaystyle x=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}\)

\(\displaystyle =\frac{{-\left( {-4} \right)\pm \sqrt{{{{{\left( {-4} \right)}}^{2}}-4\cdot 2\cdot 1}}}}{{2\cdot 2}}\)

\(\displaystyle x=\frac{{4\pm \sqrt{8}}}{4}=\frac{{4\pm 2\sqrt{2}}}{4}=\frac{{2\pm \sqrt{2}}}{2}\)

\(\displaystyle \left( {\frac{{2-\sqrt{2}}}{2},\,\,-1} \right)\,\,\text{and}\,\,\left( {\frac{{2+\sqrt{2}}}{2},\,\,-1} \right)\)

Using Implicit Differentiation to Find Higher Order Derivatives

Here’s a problem where we have to use implicit differentiation twice to find the second derivative \(\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}\,\,\,or\,\,\,{y}”\). This one is really tricky, since we need to substitute what we got for \(\displaystyle \frac{{dy}}{{dx}}\,\,or\,\,{y}’\) to simplify.

Note that in this example, we also substituted the original function back in to simplify further. (The reason I substituted at the end is because the resulting answer was one of the choices on a multiple choice test question. Your answer could be in a number of different forms.)

Problem	Solution
Use implicit differentiation to find \(\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}\): \({{y}^{5}}={{x}^{{10}}}\) (Note that this is the same as \(\displaystyle \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=1\))	\(\displaystyle \begin{align}{{y}^{5}}&={{x}^{{10}}}\\5{{y}^{4}}{y}’&=10{{x}^{9}}\\{y}’&=\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\end{align}\) \(\displaystyle {y}’=\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\times \frac{{xy}}{{xy}}\,=\frac{{10{{x}^{{10}}}y}}{{5x{{y}^{5}}}}=\frac{{10y}}{{5x}}\times \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=\frac{{10y}}{{5x}}\times 1\,=\frac{{2y}}{x}\) Notice the cool trick to simplify before taking the second derivative. Since we ended up with \(\displaystyle \frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\) after differentiating once, and we know from the original equation that \(\displaystyle \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=1\), we can multiply by \(\displaystyle \frac{{xy}}{{xy}}\) to substitute and simplify!
Now take the second derivative using implicit differentiation and the quotient rule: \(\require{cancel} \displaystyle \begin{align}{y}’&=\frac{{2y}}{x}\\{y}^{\prime \prime}&=\frac{{x\cdot 2{y}’-2y\left( 1 \right)}}{{{{x}^{2}}}}\\&=\frac{{\cancel{x}\cdot 2\left( {\frac{{2y}}{{\cancel{x}}}} \right)-2y}}{{{{x}^{2}}}}\,\,\,\,\,\,\,\,\,\,\,\,(\text{plug}\,\text{ }{y}’\,\,\text{back in)}\\&=\frac{{4y-2y}}{{{{x}^{2}}}}=\frac{{2y}}{{{{x}^{2}}}}\end{align}\)

Problem

Solution

Use implicit differentiation to find \(\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}\):

\({{y}^{5}}={{x}^{{10}}}\)

(Note that this is the same as \(\displaystyle \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=1\))

\(\displaystyle \begin{align}{{y}^{5}}&={{x}^{{10}}}\\5{{y}^{4}}{y}’&=10{{x}^{9}}\\{y}’&=\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\end{align}\) \(\displaystyle {y}’=\frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\times \frac{{xy}}{{xy}}\,=\frac{{10{{x}^{{10}}}y}}{{5x{{y}^{5}}}}=\frac{{10y}}{{5x}}\times \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=\frac{{10y}}{{5x}}\times 1\,=\frac{{2y}}{x}\)

Notice the cool trick to simplify before taking the second derivative. Since we ended up with \(\displaystyle \frac{{10{{x}^{9}}}}{{5{{y}^{4}}}}\) after differentiating once, and we know from the original equation that \(\displaystyle \frac{{{{x}^{{10}}}}}{{{{y}^{5}}}}=1\), we can multiply by \(\displaystyle \frac{{xy}}{{xy}}\) to substitute and simplify!

Now take the second derivative using implicit differentiation and the quotient rule:

\(\require{cancel} \displaystyle \begin{align}{y}’&=\frac{{2y}}{x}\\{y}^{\prime \prime}&=\frac{{x\cdot 2{y}’-2y\left( 1 \right)}}{{{{x}^{2}}}}\\&=\frac{{\cancel{x}\cdot 2\left( {\frac{{2y}}{{\cancel{x}}}} \right)-2y}}{{{{x}^{2}}}}\,\,\,\,\,\,\,\,\,\,\,\,(\text{plug}\,\text{ }{y}’\,\,\text{back in)}\\&=\frac{{4y-2y}}{{{{x}^{2}}}}=\frac{{2y}}{{{{x}^{2}}}}\end{align}\)

Related Rates

I used to have such a problem with related rates problems, until I began writing down the steps to do them. And it helps to remember that the rates in these problems typically are differentiated with respect to time, or \(\displaystyle \frac{{d\left( {\text{something}} \right)}}{{dt}}\).

Hope these steps help:

Draw a picture and label any amounts that could be changing. These amounts will be variables, even if the problem gives you an amount at a certain time for these parts (the “when”). In other words, understand what is changing and what isn’t. I find it helpful to draw an arrow to show where the picture is “moving”.
On the picture you’ve drawn, label places that are changing (you’ll need variables), and places that aren’t changing (constants). For constants, you can put numbers on the picture, if they are given.
Write down exactly what the problem gives you, and what you need to solve for. For example, you may write down “Find \(\displaystyle \frac{{dA}}{{dt}}\) when r = 6”. Remember again that the rates (things that are changing) have “dt” (with respect to time) in them – pretty much always. Also, determine what snapshot in time (or in another variable) the question is asking for (the “when”).
For the rates, remember that rates with words like “filling up” means a positive volume rate, “emptying out” means a negative volume rate.
Now we have to figure out a way to relate all the variables together. Write an equation that relates all of the given information and variables that you’ve written down. A lot of times this is a geometry equation for volume, area, or perimeter. Also, you’ll see the Pythagorean Equation or the trigonometric functions, such as SOH-CAH-TOA. Remember that shapes stay in proportion, so we may have to use geometrically similar figures and set up proportions.
Simplify by trying to put everything in as few variables as possible before differentiating (you may have to substitute some variables by solving in terms of other variables). For example, you may need to relate length and width together if you have a perimeter. I find this the hardest part – how to know what to plug in before I differentiate, and what to plug in after. Remember that every variable you have before differentiating will eventually have a “dt” denominator after differentiating, so get rid of any variables where rates of that variable aren’t given or needed in the problem!
The rule of thumb is that when you have values that say “when” something happens, these are put in after differentiating.
Use implicit differentiation to differentiate with respect to time. Hopefully the rates (\(\displaystyle \frac{{d\left( {\text{something}} \right)}}{{dt}}\)) you have left at this point are rates that either you have values for, or need values for.
Fill in the equation with what you know (for example, the “when”s). Solve the equation for what you need (what you don’t know).
Make sure you are answering the question that is being asked, and the units are correct. You’d hate to do all this work, and then answer the wrong question. Remember when dealing with angles, make sure the calculator is in the correct mode (radians or degrees)!

Here are some examples:

Related Rates Problem	Steps and Solution
A snowball is melting at a rate of 1 cubic inch per minute. At what rate is the radius changing when the snowball has a radius of 2 inches?	Draw a picture. Note that since volume and radius are changing, we have to use variables for them. Let \(r=\) radius, \(V=\) volume. Define what we have and what we want when. We have: \(\displaystyle \frac{{dV}}{{dt}}=-1\) (since volume is decreasing). We want: find \(\displaystyle \frac{{dr}}{{dt}}\) when \(r=2\). Determine what equation we need: we need to relate the radius of a circle to its volume (sphere): \(\displaystyle {{V}_{{sphere}}}=\frac{4}{3}\pi {{r}^{3}}\). Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time: \(\displaystyle \require{cancel} \frac{{dV}}{{dt}}=\cancel{3}\cdot \frac{4}{{\cancel{3}}}\pi \,{{r}^{2}}\frac{{dr}}{{dt}}=4\pi \,{{r}^{2}}\frac{{dr}}{{dt}}\) . Plug in what we know at the time we know it, and solve for what we need: \(\displaystyle 4\pi {{r}^{2}}\frac{{dr}}{{dt}};\,\,\,\,\,-1=4\pi {{\left( 2 \right)}^{2}}\cdot \frac{{dr}}{{dt}};\,\,\,\,\,\frac{{dr}}{{dt}}=\frac{{-1}}{{16\pi }}\,\,\approx \,\,-.0199\,\,\text{in/min}\). Make sure we’re answering the question: yes, –.0199 inches per minute is the rate the radius is changing. It makes sense that it is negative, since radius is decreasing along with the volume.
A 40-foot ladder is resting against a wall, and its base is slipping on the floor (away from the wall) at a rate of 2 feet per minute. Find the rate of change of the height of the ladder at the time when the base is 20 feet from the base of the wall.	Draw a picture. Note that since height (\(y\)) and distance from base (\(x\)) are changing, we have to use variables for them. The ladder (slanted) will be the hypotenuse of a right triangle formed by the wall, the floor, and the ladder. Define what we have and what we want when. We have: \(\displaystyle \frac{{dx}}{{dt}}=2\). We want: find \(\displaystyle \frac{{dy}}{{dt}}\) when \(x=20\). Determine what equation we need: we need to relate the variables together; let’s use the Pythagorean Theorem: \({{x}^{2}}+{{y}^{2}}={{40}^{2}}\). Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time: \(\displaystyle 2x\frac{{dx}}{{dt}}+2y\frac{{dy}}{{dt}}=0\). Plug in what we know at the time we know it, and solve for what we need. The tricky part here is that we need to also use the Pythagorean Theorem again to get what \(y\) (height) will be when \(x=20\) (distance from base of wall): \({{20}^{2}}+{{y}^{2}}={{40}^{2}};\,\,\,\,y=\sqrt{{1200}}\approx 34.641\). Now plug everything in: \(\displaystyle 2x\frac{{dx}}{{dt}}+2y\frac{{dy}}{{dt}}=0;\,\,\,\,2\left( {20} \right)\left( 2 \right)+2\left( {34.641} \right)\frac{{dy}}{{dt}}=0\) \(\displaystyle \frac{{dy}}{{dt}}=-\frac{{2\left( {20} \right)\left( 2 \right)}}{{2\left( {34.641} \right)}}\approx -1.155\,\,\text{ft/min}\). Make sure we’re answering the question: yes, –1.155 feet per minute is the rate the height of the ladder is changing. It makes sense that it is negative, since the ladder is slipping down the wall.

Related Rates Problem

Steps and Solution

A snowball is melting at a rate of 1 cubic inch per minute.

At what rate is the radius changing when the snowball has a radius of 2 inches?

Draw a picture. Note that since volume and radius are changing, we have to use variables for them. Let \(r=\) radius, \(V=\) volume.
Define what we have and what we want when. We have: \(\displaystyle \frac{{dV}}{{dt}}=-1\) (since volume is decreasing). We want: find \(\displaystyle \frac{{dr}}{{dt}}\) when \(r=2\).
Determine what equation we need: we need to relate the radius of a circle to its volume (sphere): \(\displaystyle {{V}_{{sphere}}}=\frac{4}{3}\pi {{r}^{3}}\). Make sure that all variables will differentiate to a rate we either have or need: yes!
Differentiate with respect to time: \(\displaystyle \require{cancel} \frac{{dV}}{{dt}}=\cancel{3}\cdot \frac{4}{{\cancel{3}}}\pi \,{{r}^{2}}\frac{{dr}}{{dt}}=4\pi \,{{r}^{2}}\frac{{dr}}{{dt}}\) .
Plug in what we know at the time we know it, and solve for what we need: \(\displaystyle 4\pi {{r}^{2}}\frac{{dr}}{{dt}};\,\,\,\,\,-1=4\pi {{\left( 2 \right)}^{2}}\cdot \frac{{dr}}{{dt}};\,\,\,\,\,\frac{{dr}}{{dt}}=\frac{{-1}}{{16\pi }}\,\,\approx \,\,-.0199\,\,\text{in/min}\).
Make sure we’re answering the question: yes, –.0199 inches per minute is the rate the radius is changing. It makes sense that it is negative, since radius is decreasing along with the volume.

A 40-foot ladder is resting against a wall, and its base is slipping on the floor (away from the wall) at a rate of 2 feet per minute.

Find the rate of change of the height of the ladder at the time when the base is 20 feet from the base of the wall.

Draw a picture. Note that since height (\(y\)) and distance from base (\(x\)) are changing, we have to use variables for them. The ladder (slanted) will be the hypotenuse of a right triangle formed by the wall, the floor, and the ladder.
Define what we have and what we want when. We have: \(\displaystyle \frac{{dx}}{{dt}}=2\). We want: find \(\displaystyle \frac{{dy}}{{dt}}\) when \(x=20\).
Determine what equation we need: we need to relate the variables together; let’s use the Pythagorean Theorem: \({{x}^{2}}+{{y}^{2}}={{40}^{2}}\). Make sure that all variables will differentiate to a rate we either have or need: yes!
Differentiate with respect to time: \(\displaystyle 2x\frac{{dx}}{{dt}}+2y\frac{{dy}}{{dt}}=0\).
Plug in what we know at the time we know it, and solve for what we need. The tricky part here is that we need to also use the Pythagorean Theorem again to get what \(y\) (height) will be when \(x=20\) (distance from base of wall): \({{20}^{2}}+{{y}^{2}}={{40}^{2}};\,\,\,\,y=\sqrt{{1200}}\approx 34.641\). Now plug everything in: \(\displaystyle 2x\frac{{dx}}{{dt}}+2y\frac{{dy}}{{dt}}=0;\,\,\,\,2\left( {20} \right)\left( 2 \right)+2\left( {34.641} \right)\frac{{dy}}{{dt}}=0\) \(\displaystyle \frac{{dy}}{{dt}}=-\frac{{2\left( {20} \right)\left( 2 \right)}}{{2\left( {34.641} \right)}}\approx -1.155\,\,\text{ft/min}\).
Make sure we’re answering the question: yes, –1.155 feet per minute is the rate the height of the ladder is changing. It makes sense that it is negative, since the ladder is slipping down the wall.

Here are more problems:

Related Rates Problem	Steps and Solution
Water is being poured into a cylindrical can that is 20 inches in height and has a radius of 8 inches. The water is being poured at a rate of 3 cubic inches per second. How fast is the height of the water in the can changing when the height is 8 feet deep?	Draw a picture. Note that since the radius and height of the actual can is not changing, we can use constants for them. Since the height (and volume) of the water inside is changing, we have to use variables: Let \(h=\) height of water, \(V=\) volume of water. Define what we have and what we want when. We have: height of can \(=20\), radius of can (and water) \(=8\), \(\displaystyle \frac{{dV}}{{dt}}=3\) (rate of volume of water in can). We want: find \(\displaystyle \frac{{dh}}{{dt}}\) (rate of height of water) when \(h=8\) (height of water). Determine what equation we need: We need to relate the radius and height of a cylinder to its volume: \({{V}_{{cylinder}}}=\pi {{r}^{2}}h\). Differentiate with respect to time: \(\displaystyle \frac{{dV}}{{dt}}=2\pi r\frac{{dr}}{{dt}}\). Uh oh; we have a value for \(h\) and not \(r\), and we want to solve for \(\displaystyle \frac{{dh}}{{dt}}\), not \(\displaystyle \frac{{dr}}{{dt}}\), so we’d better backtrack and put \(r\) in terms of \(h\), before differentiating. To get \(r\) in terms of \(h\), we have to use proportions of similar figures (figures with the same shape; we can do this since we know the shape of the water is similar to the shape of the can): \(\displaystyle \frac{r}{h}=\frac{8}{{20}};\,\,\,\,\,r=\frac{{8h}}{{20}}=.4h\). Now plug in and differentiate again, and we’re on the right track: \(\displaystyle {{V}_{{cylinder}}}=\pi {{r}^{2}}h=\,\pi {{\left( {.4h} \right)}^{2}}h=.16\pi {{h}^{3}};\,\,\,\,\frac{{dV}}{{dt}}=.48\pi {{h}^{2}}\frac{{dh}}{{dt}}\). We now have everything we need! Plug in what we know at the time we know it, and solve for what we need: \(\displaystyle \frac{{dV}}{{dt}}=.48\pi {{h}^{2}}\frac{{dh}}{{dt}};\,\,\,\,\,3=.48\pi {{\left( 8 \right)}^{2}}\frac{{dh}}{{dt}};\,\,\,\,\frac{{dh}}{{dt}}\approx .031\,\,\text{ft/sec}\). Make sure we’re answering the question: Yes, .031 feet per second is the rate the height of the water is changing. It makes sense that it is positive, since the can is filling up.

Related Rates Problem

Steps and Solution

Water is being poured into a cylindrical can that is 20 inches in height and has a radius of 8 inches.

The water is being poured at a rate of 3 cubic inches per second.

How fast is the height of the water in the can changing when the height is 8 feet deep?

Draw a picture. Note that since the radius and height of the actual can is not changing, we can use constants for them. Since the height (and volume) of the water inside is changing, we have to use variables: Let \(h=\) height of water, \(V=\) volume of water.
Define what we have and what we want when. We have: height of can \(=20\), radius of can (and water) \(=8\), \(\displaystyle \frac{{dV}}{{dt}}=3\) (rate of volume of water in can). We want: find \(\displaystyle \frac{{dh}}{{dt}}\) (rate of height of water) when \(h=8\) (height of water).
Determine what equation we need: We need to relate the radius and height of a cylinder to its volume: \({{V}_{{cylinder}}}=\pi {{r}^{2}}h\).
Differentiate with respect to time: \(\displaystyle \frac{{dV}}{{dt}}=2\pi r\frac{{dr}}{{dt}}\). Uh oh; we have a value for \(h\) and not \(r\), and we want to solve for \(\displaystyle \frac{{dh}}{{dt}}\), not \(\displaystyle \frac{{dr}}{{dt}}\), so we’d better backtrack and put \(r\) in terms of \(h\), before differentiating. To get \(r\) in terms of \(h\), we have to use proportions of similar figures (figures with the same shape; we can do this since we know the shape of the water is similar to the shape of the can): \(\displaystyle \frac{r}{h}=\frac{8}{{20}};\,\,\,\,\,r=\frac{{8h}}{{20}}=.4h\). Now plug in and differentiate again, and we’re on the right track: \(\displaystyle {{V}_{{cylinder}}}=\pi {{r}^{2}}h=\,\pi {{\left( {.4h} \right)}^{2}}h=.16\pi {{h}^{3}};\,\,\,\,\frac{{dV}}{{dt}}=.48\pi {{h}^{2}}\frac{{dh}}{{dt}}\). We now have everything we need!
Plug in what we know at the time we know it, and solve for what we need: \(\displaystyle \frac{{dV}}{{dt}}=.48\pi {{h}^{2}}\frac{{dh}}{{dt}};\,\,\,\,\,3=.48\pi {{\left( 8 \right)}^{2}}\frac{{dh}}{{dt}};\,\,\,\,\frac{{dh}}{{dt}}\approx .031\,\,\text{ft/sec}\).
Make sure we’re answering the question: Yes, .031 feet per second is the rate the height of the water is changing. It makes sense that it is positive, since the can is filling up.

Related Rates Problem	Steps and Solution
A woman 5 feet tall walks at a rate of 6 feet per second away from a lamppost. When the woman is 6 feet from the lamppost, her shadow is 8 feet tall. At what rate is the length of her shadow changing when she is 12 feet from the lamppost? At what rate is the tip of her shadow changing when she is 12 feet from the lamppost?	Draw a picture. Remember that the woman’s shadow is on the ground in front of her; we’ll use \(y\) for this distance. The distance from the base of the lamppost to the woman is \(x\). We need variables for both of these distances, since they are changing. Define what we have and what we want when. We have the rate the woman walks from the lamppost, \(\displaystyle \frac{{dx}}{{dt}}=6\). Now this is tricky: to get the rate of the length of her shadow increasing, we want ,\(\displaystyle \frac{{dy}}{{dt}}\) but if we want the rate of the tip of her shadow increasing, we want \(\displaystyle \frac{{d\left( {x+y} \right)}}{{dt}}\). (This is because we need to measure the tip of the shadow in reference to the base of the lamppost). We also have the height of the woman (5 feet), and we know that when the woman is 6 feet from the lamppost \((x=6),\) and her shadow is 8 feet (\(y=8\)). It turns out that we will need the height \(h\) of the lamppost to get the equation we need to differentiate, so from similar triangles, we can get this: (see last triangle to the left): \(\displaystyle \frac{{\text{side of large }\Delta }}{{\text{side of small }\Delta }}=\frac{{\text{bottom of large }\Delta }}{{\text{bottom of small }\Delta }}:\,\,\,\,\,\,\frac{h}{5}=\frac{{14}}{8};\,\,\,h=8.75\) Determine what equation we need: We need to relate \(x\) and \(y\) to what we know (\(h\) and woman’s height), which we can do using similar triangles again: \(\displaystyle \frac{{\text{8}\text{.75}}}{5}=\frac{{x+y}}{y},\) or (simplified), \(3.75y=5x\). Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time: \(\displaystyle 3.75\frac{{dy}}{{dt}}=5\frac{{dx}}{{dt}}\). Plug in what we know at the time we know it, and solve for what we need: Now plug in: \(\displaystyle 3.75\frac{{dy}}{{dt}}=5\frac{{dx}}{{dt}};\,\,\,\,3.75\frac{{dy}}{{dt}}=5\left( 6 \right);\,\,\,\frac{{dy}}{{dt}}=8\) feet per second. This is the rate of the length of the shadow. To get the rate of the tip of the shadow, we need \(\displaystyle \frac{{d\left( {x+y} \right)}}{{dt}}\), which is actually \(\displaystyle \frac{{dx}}{{dt}}+\frac{{dy}}{{dt}}\) (note that could have also defined a variable to represent \(x + y\), and do some subtraction in setting it up). The rate of the tip of the shadow is \(6+8=14\) feet per second.

Related Rates Problem

Steps and Solution

A woman 5 feet tall walks at a rate of 6 feet per second away from a lamppost. When the woman is 6 feet from the lamppost, her shadow is 8 feet tall.

At what rate is the length of her shadow changing when she is 12 feet from the lamppost?

At what rate is the tip of her shadow changing when she is 12 feet from the lamppost?

Draw a picture. Remember that the woman’s shadow is on the ground in front of her; we’ll use \(y\) for this distance. The distance from the base of the lamppost to the woman is \(x\). We need variables for both of these distances, since they are changing.
Define what we have and what we want when. We have the rate the woman walks from the lamppost, \(\displaystyle \frac{{dx}}{{dt}}=6\). Now this is tricky: to get the rate of the length of her shadow increasing, we want ,\(\displaystyle \frac{{dy}}{{dt}}\) but if we want the rate of the tip of her shadow increasing, we want \(\displaystyle \frac{{d\left( {x+y} \right)}}{{dt}}\). (This is because we need to measure the tip of the shadow in reference to the base of the lamppost). We also have the height of the woman (5 feet), and we know that when the woman is 6 feet from the lamppost \((x=6),\) and her shadow is 8 feet (\(y=8\)). It turns out that we will need the height \(h\) of the lamppost to get the equation we need to differentiate, so from similar triangles, we can get this: (see last triangle to the left): \(\displaystyle \frac{{\text{side of large }\Delta }}{{\text{side of small }\Delta }}=\frac{{\text{bottom of large }\Delta }}{{\text{bottom of small }\Delta }}:\,\,\,\,\,\,\frac{h}{5}=\frac{{14}}{8};\,\,\,h=8.75\)
Determine what equation we need: We need to relate \(x\) and \(y\) to what we know (\(h\) and woman’s height), which we can do using similar triangles again: \(\displaystyle \frac{{\text{8}\text{.75}}}{5}=\frac{{x+y}}{y},\) or (simplified), \(3.75y=5x\). Make sure that all variables will differentiate to a rate we either have or need: yes!
Differentiate with respect to time: \(\displaystyle 3.75\frac{{dy}}{{dt}}=5\frac{{dx}}{{dt}}\).
Plug in what we know at the time we know it, and solve for what we need: Now plug in: \(\displaystyle 3.75\frac{{dy}}{{dt}}=5\frac{{dx}}{{dt}};\,\,\,\,3.75\frac{{dy}}{{dt}}=5\left( 6 \right);\,\,\,\frac{{dy}}{{dt}}=8\) feet per second. This is the rate of the length of the shadow.
To get the rate of the tip of the shadow, we need \(\displaystyle \frac{{d\left( {x+y} \right)}}{{dt}}\), which is actually \(\displaystyle \frac{{dx}}{{dt}}+\frac{{dy}}{{dt}}\) (note that could have also defined a variable to represent \(x + y\), and do some subtraction in setting it up). The rate of the tip of the shadow is \(6+8=14\) feet per second.

Here’s one that involves right triangle trigonometry:

Related Rates Problem	Steps and Solution
A hot air balloon is rising straight up and is tracked by a range finder 400 feet from point of lift-off. When the range finder’s angle of elevation is \(\displaystyle \frac{\pi }{3}\), the angle is increasing at a rate of .15 radians per minute. How fast is the balloon rising at that time?	Draw a picture. Note that since height and the angle of elevation are changing, we have to use variables for them. Let \(h=\) height, \(\theta =\) angle of elevation. Define what we have and what we want when. We have: \(\displaystyle \frac{{d\theta }}{{dt}}=.15\) (in radians). We want: find \(\displaystyle \frac{{dh}}{{dt}}\) when \(\displaystyle \theta =\frac{\pi }{3}\). We also know the distance on the ground from the range finder to the balloon is 400. Determine what equation we need: We need to relate the height and base of the right triangle to the angle of elevation; we can use trigonometry: \(\displaystyle \tan \theta =\frac{h}{{400}}\). Make sure that all variables will differentiate to a rate we either have or need: yes! Differentiate with respect to time: \(\displaystyle \frac{{d\left( {\tan \theta } \right)}}{{dt}}=\frac{1}{{400}}\frac{{dh}}{{dt}};\,\,\,\,\,{{\sec }^{2}}\theta \frac{{d\theta }}{{dt}}=\frac{1}{{400}}\frac{{dh}}{{dt}}\). Plug in what we know at the time we know it, and solve for what we need (calculator in radians): \(\displaystyle \begin{align}{{\sec }^{2}}\theta \frac{{d\theta }}{{dt}}&=\frac{1}{{400}}\frac{{dh}}{{dt}};\,\,\,\,\,\left( {{{{\sec }}^{2}}\left( {\frac{\pi }{3}} \right)} \right)\left( {.15} \right)=\frac{1}{{400}}\frac{{dh}}{{dt}}\\{{2}^{2}}\left( {.15} \right)&=\frac{1}{{400}}\frac{{dh}}{{dt}};\,\,\,\,\,\frac{{dh}}{{dt}}=240\,\,\text{rad/min}\end{align}\) \(\displaystyle (\text{note that }\sec \left( {\frac{\pi }{3}} \right)=\frac{1}{{\cos \left( {\frac{\pi }{3}} \right)}}=\frac{1}{{\frac{1}{2}}}=2)\) Make sure we’re answering the question: Yes, 240 radians per minute is the rate the angle of elevation is changing. It makes sense that it is positive since the angle is increasing along with the height.

Related Rates Problem

Steps and Solution

A hot air balloon is rising straight up and is tracked by a range finder 400 feet from point of lift-off.

When the range finder’s angle of elevation is \(\displaystyle \frac{\pi }{3}\), the angle is increasing at a rate of .15 radians per minute.

How fast is the balloon rising at that time?

Draw a picture. Note that since height and the angle of elevation are changing, we have to use variables for them. Let \(h=\) height, \(\theta =\) angle of elevation.
Define what we have and what we want when. We have: \(\displaystyle \frac{{d\theta }}{{dt}}=.15\) (in radians). We want: find \(\displaystyle \frac{{dh}}{{dt}}\) when \(\displaystyle \theta =\frac{\pi }{3}\). We also know the distance on the ground from the range finder to the balloon is 400.
Determine what equation we need: We need to relate the height and base of the right triangle to the angle of elevation; we can use trigonometry: \(\displaystyle \tan \theta =\frac{h}{{400}}\). Make sure that all variables will differentiate to a rate we either have or need: yes!
Differentiate with respect to time: \(\displaystyle \frac{{d\left( {\tan \theta } \right)}}{{dt}}=\frac{1}{{400}}\frac{{dh}}{{dt}};\,\,\,\,\,{{\sec }^{2}}\theta \frac{{d\theta }}{{dt}}=\frac{1}{{400}}\frac{{dh}}{{dt}}\).
Plug in what we know at the time we know it, and solve for what we need (calculator in radians): \(\displaystyle \begin{align}{{\sec }^{2}}\theta \frac{{d\theta }}{{dt}}&=\frac{1}{{400}}\frac{{dh}}{{dt}};\,\,\,\,\,\left( {{{{\sec }}^{2}}\left( {\frac{\pi }{3}} \right)} \right)\left( {.15} \right)=\frac{1}{{400}}\frac{{dh}}{{dt}}\\{{2}^{2}}\left( {.15} \right)&=\frac{1}{{400}}\frac{{dh}}{{dt}};\,\,\,\,\,\frac{{dh}}{{dt}}=240\,\,\text{rad/min}\end{align}\) \(\displaystyle (\text{note that }\sec \left( {\frac{\pi }{3}} \right)=\frac{1}{{\cos \left( {\frac{\pi }{3}} \right)}}=\frac{1}{{\frac{1}{2}}}=2)\)
Make sure we’re answering the question: Yes, 240 radians per minute is the rate the angle of elevation is changing. It makes sense that it is positive since the angle is increasing along with the height.

Note: See Jake’s Math Lessons here for more information and problems on related rates.

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