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Related Rates Elipse Question

  • Thread starter tmlrlz
  • Start date
  • #1
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0

Homework Statement


A particle is moving along the ellipse x2/16 + y2/4 = 1. At each time t its x and y coordinates are given by x = 4cost, y = 2sint. At what rate is the particle's distance from the origin changing at time t? At what rate is the distance from the origin changing when t = pi/4?


Homework Equations


x2/16 + y2/4 = 1
x = 4cost
y = 2sint
dx/dt
dy/dt

The Attempt at a Solution


I am fairly sure i can do this question, the only problem is that i am not so sure what it is asking for. I'm assuming that i am to find dy/dx.
dy/dx = (dy/dt)(dt/dx)
dy/dt = 2cost
dx/dt = -4sint

x2/16 + y2/4 = 1
2x(dx/dt) (1/16) + 2y(dy/dt)(1/4) = 0
(x/8) (-4sint) + (y/2)(2cost) = 0
(-xsint)/2 + ycost = 0
At t = pi/4
2ycos(pi/4) = xsin(pi/4)
2y(1/√2) = x(1/√2)
x = 2y

dy/dx = (dy/dt) (dt/dx)
= (2cost)(-1/4sint)
= (-1/2) (cost/sint)
= (-1/2)(cott)
at t = pi/4
dy/dx = (-1/2)(1/1)
= -1/2

Does this mean that the answer is -1/2 because then i'm not sure why they gave us the equation for the ellipse because it doesnt play any effect in finding dy/dx. Thank you.
 
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Answers and Replies

  • #2
33,273
4,981

Homework Statement


A particle is moving along the ellipse x2/16 + y2/4 = 1. At each time t its x and y coordinates are given by x = 4cost, y = 2sint. At what rate is the particle's distance from the origin changing at time t? At what rate is the distance from the origin changing when t = pi/4?


Homework Equations


x2/16 + y2/4 = 1
x = 4cost
y = 2sint
dx/dt
dy/dt

The Attempt at a Solution


I am fairly sure i can do this question, the only problem is that i am not so sure what it is asking for. I'm assuming that i am to find dy/dx.
No, they're not asking for dy/dx. You wrote what the problem is asking for in your problem statement:
At what rate is the particle's distance from the origin changing at time t? At what rate is the distance from the origin changing when t = pi/4?
What expression represents the particle's distance from the origin?
dy/dx = (dy/dt)(dt/dx)
dy/dt = 2cost
dx/dt = -4sint

x2/16 + y2/4 = 1
2x(dx/dt) (1/16) + 2y(dy/dt)(1/4) = 0
(x/8) (-4sint) + (y/2)(2cost) = 0
(-xsint)/2 + ycost = 0
At t = pi/4
2ycos(pi/4) = xsin(pi/4)
2y(1/√2) = x(1/√2)
x = 2y

dy/dx = (dy/dt) (dt/dx)
= (2cost)(-1/4sint)
= (-1/2) (cost/sint)
= (-1/2)(cott)
at t = pi/4
dy/dx = (-1/2)(1/1)
= -1/2

Does this mean that the answer is -1/2 because then i'm not sure why they gave us the equation for the ellipse because it doesnt play any effect in finding dy/dx. Thank you.
 
  • #3
29
0
No, they're not asking for dy/dx. You wrote what the problem is asking for in your problem statement:


What expression represents the particle's distance from the origin?
the rate of change of x with respect to t plus the rate of change of y with respect to t:
dx/dt + dy/dt ?
 

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