1 THINKING FORWARD - Calculus

Question

The derivative along a cut edge: Let

\begin{equation}f(x,y)=x^2y^3\end{equation}

Find the rate of change of f in the (positive) y direction when the surface is cut by the plane with equation x = 2. Find the rate of change of f in the (positive) x direction when the surface is cut by the plane with equation y = 1. 

Step-by-Step Solution

Answer

The rate of change of f in the (positive) y direction when the surface is cut by the plane with equation x = 2 is

\begin{equation}12y^2\end{equation}

The rate of change of f in the (positive) x direction when the surface is cut by the plane with equation y = 1 is

\begin{equation}2x\end{equation}

Step 1. Given Infomation

The equation is \begin{equation}f(x,y)=x^2y^3\end{equation}

Step 2. Calculation

->To find the rate of change of f in the positive y direction when the surface is cut by the plane x = 2, we need to take the partial derivative of f with respect to y and evaluate it at the point (2, y) on the surface. The partial derivative of f with respect to y is 

\begin{equation}(\frac{\partial f}{\partial y})=x^2(\frac{\partial}{\partial y}y^3)\end{equation}

\begin{equation}(\frac{\partial f}{\partial y})=x^2(3*y^{3-1})\end{equation}

\begin{equation}(\frac{\partial f}{\partial y})=3x^2y^2\end{equation}

The surface is cut by the plane x=2 this implies substituting it in the partial derivative

\begin{equation}(\frac{\partial f}{\partial y})_{x=2}=3(2^2)y^2\end{equation}

\begin{equation}(\frac{\partial f}{\partial y})_{x=2}=3(4)y^2\end{equation}

\begin{equation}(\frac{\partial f}{\partial y})_{x=2}=12y^2\end{equation}

The rate of change of f in the positive y direction is 12y^2.

->To find the rate of change of f in the positive x direction when the surface is cut by the plane y = 1, we need to take the partial derivative of f with respect to x and evaluate it at the point (x, 1) on the surface. The partial derivative of f with respect to x is 

\begin{equation}(\frac{\partial f}{\partial x})=(\frac{\partial}{\partial x}x^2)y^3\end{equation}

\begin{equation}(\frac{\partial f}{\partial x})=(2*x^{2-1})y^3\end{equation}

\begin{equation}(\frac{\partial f}{\partial x})=(2x)y^3\end{equation}

The surface is cut by the plane y=1 this implies substituting it in the partial derivative

\begin{equation}(\frac{\partial f}{\partial x})_{y=1}=(2x)(1^3)\end{equation}

\begin{equation}(\frac{\partial f}{\partial x})=(2x)(1)\end{equation}

\begin{equation}(\frac{\partial f}{\partial x})=(2x)\end{equation}

The rate of change of f in the positive x direction is 2x