Question
\( 1 \leftarrow \) The density of a thin circular plate of radius 4 is given by \( \rho(x, y)=4+x y \). The edge of the plate is described by the parametric equations \( x=4 \cos t, y=4 \sin t \), for \( 0 \leq t \leq 2 \pi \). a. Find the rate of change of the density with respect to \( t \) on the edge of the plate. b. At what point(s) on the edge of the plate is the density a maximum? First, find some intermediate derivatives. \( \frac{\partial \rho}{\partial x}=y \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d x}{d t}=-4 \) sin \( t \) (Type an expression using \( t \) as the variable.) \( \frac{\partial \rho}{\partial y}=\square \) (Type an expression using \( x \) and \( y \) as the variables.)
Ask by Huff John. in the United States
Mar 17,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\[
\frac{\partial \rho}{\partial y} = x
\]
Solution
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The Deep Dive
To find the intermediate derivative, we have: \(\frac{\partial \rho}{\partial y} = 4 + x\) Now let's summarize the expressions you've found so far: For the density function, we have: \[ \rho(x, y) = 4 + x y \] The partial derivatives are: \[ \frac{\partial \rho}{\partial x} = y \] \[ \frac{\partial \rho}{\partial y} = 4 + x \] And you've correctly identified the expression for \( \frac{dx}{dt} \): \[ \frac{dx}{dt} = -4 \sin t \] Now for \( \frac{dy}{dt} \): \[ \frac{dy}{dt} = 4 \cos t \] Now you have all the pieces you need to proceed with parts (a) and (b)!
