Question
upstudy study bank question image url

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? \( \frac{\partial \rho}{\partial x}=y \) (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.) \( \frac{\partial \rho}{\partial y}=x \) (Type an expression using \( x \) and \( y \) as the variables.) \( \frac{d y}{d t}=4 \) cos \( t \) (Type an expression using \( t \) as the variable.) a. What is the rate of change of the density with respect to \( t \) on the edge of the plate? \( \rho^{\prime}(t)=\square \) (Type an expression using \( t \) as the variable.)

Ask by Lawson Spencer. in the United States
Mar 17,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

a. The rate of change of the density with respect to \( t \) on the edge of the plate is \( \rho'(t) = 16\cos 2t \). b. The density is a maximum at the points \( (2\sqrt{2}, 2\sqrt{2}) \) and \( (-2\sqrt{2}, -2\sqrt{2}) \).

Solution

Sign in to Unlock Answers for Free!

A Learning Platform Trusted by Millions of Real Students and Teachers.

star-icon Unlock

Answered by UpStudy AI and reviewed by a Professional Tutor

Mind Expander

To find the rate of change of the density with respect to \( t \) on the edge of the plate, we can use the chain rule. The density function is given as \( \rho(x, y) = 4 + xy \), and we need to express this in terms of \( t \) by substituting \( x(t) = 4 \cos t \) and \( y(t) = 4 \sin t \). Using the chain rule, we find: \[ \frac{d\rho}{dt} = \frac{\partial \rho}{\partial x} \frac{dx}{dt} + \frac{\partial \rho}{\partial y} \frac{dy}{dt} \] Given: \[ \frac{\partial \rho}{\partial x} = y \quad \text{and} \quad \frac{\partial \rho}{\partial y} = x \] Also, we have: \[ \frac{dx}{dt} = -4 \sin t \quad \text{and} \quad \frac{dy}{dt} = 4 \cos t \] Now substituting \( y \) and \( x \): \[ \rho^{\prime}(t) = y(-4 \sin t) + x(4 \cos t) \] Substituting \( y = 4 \sin t \) and \( x = 4 \cos t \): \[ \rho^{\prime}(t) = (4 \sin t)(-4 \sin t) + (4 \cos t)(4 \cos t) \] \[ = -16 \sin^2 t + 16 \cos^2 t \] \[ = 16(\cos^2 t - \sin^2 t) \] So the final expression is: \[ \rho^{\prime}(t) = 16(\cos^2 t - \sin^2 t) \]

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy