Calculus I - Related Rates

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Section 3.11 : Related Rates

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1. In the following assume that \(x\) and \(y\) are both functions of \(t\). Given \(x = - 2\), \(y = 1\) and \(x' = - 4\) determine \(y'\) for the following equation.

\[6{y^2} + {x^2} = 2 - {x^3}{{\bf{e}}^{4 - 4y}}\]

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Start Solution

The first thing that we need to do here is use implicit differentiation to differentiate the equation with respect to \(t\).

\[12y\,y' + 2x\,x' = - 3{x^2}x'{{\bf{e}}^{4 - 4y}} + 4{x^3}{{\bf{e}}^{4 - 4y}}y'\] Show Step 2

All we need to do now is plug in the given information and solve for \(y'\).

\[12y' + 16 = 48 - 32y'\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{y' = {\textstyle{8 \over {11}}}}}\]