Find a relation between dx/dt and dy/dt

  • #1

Homework Statement


A particle moves counterclockwise around the ellipse with equation 9x^2 + 16y^2 = 25.

a). In which of the four quadrants in dx/dt > 0? Explain.
b). Find a relation between dx/dt and dy/dt.
c). At what rate is the x-coordinate changing when the particle passes the point (-1,1) if its y-coordinate is increasing at a rate of 6 m/s?
d). Find dy/dt when the particle is at the top and bottom of the ellipse.


Homework Equations


None


The Attempt at a Solution



a). I don't see how I could solve this with differentiation so I drew a picture of the ellipse. If the particle is travelling counter-clockwise, x will be increasing over time in quadrants 3 and 4.

b). Implicitly differentiating for x and y both as functions of t I get
dx/dt = (-32y*dy/dt)/18x

c). Plugging in the values for the above formula...
dx/dt = -32*6/18 = -32/3

d). If the particle is at the top and bottom of the ellipse, then x is zero, and thus dy/dt is zero because

dy/dt = (-18x*dx/dt)/32y = 0 @ x = 0


Just checking if I did this correctly particularly a) as I don't see how I can "explicitly" show that dx/dt > 0 in quadrants 3 and 4 outside of explaining why in English.
 

Answers and Replies

  • #2
33,741
5,432

Homework Statement


A particle moves counterclockwise around the ellipse with equation 9x^2 + 16y^2 = 25.

a). In which of the four quadrants in dx/dt > 0? Explain.
b). Find a relation between dx/dt and dy/dt.
c). At what rate is the x-coordinate changing when the particle passes the point (-1,1) if its y-coordinate is increasing at a rate of 6 m/s?
d). Find dy/dt when the particle is at the top and bottom of the ellipse.


Homework Equations


None


The Attempt at a Solution



a). I don't see how I could solve this with differentiation so I drew a picture of the ellipse. If the particle is travelling counter-clockwise, x will be increasing over time in quadrants 3 and 4.

b). Implicitly differentiating for x and y both as functions of t I get
dx/dt = (-32y*dy/dt)/18x

c). Plugging in the values for the above formula...
dx/dt = -32*6/18 = -32/3

d). If the particle is at the top and bottom of the ellipse, then x is zero, and thus dy/dt is zero because

dy/dt = (-18x*dx/dt)/32y = 0 @ x = 0


Just checking if I did this correctly particularly a) as I don't see how I can "explicitly" show that dx/dt > 0 in quadrants 3 and 4 outside of explaining why in English.
Since the ellipse was not given parametrically (i.e., as x = f(t) and y = g(t) for some functions f and g), I believe that you did what you were supposed to do for part a. IOW, look at the graph and determine visually that dx/dt > 0 where x is increasing.

The only thing you should add are some units in part c. They're telling you the units at which y is changing, so you should report the same units when you say how x is changing. Also, it's probably a good idea to simplify the fraction.
 
Last edited:
  • #3
Since the ellipse was not given parametrically (i.e., as x = f(t) and y = g(t) for some functions f and g), I believe that you did what you were supposed to do for part a. IOW, look at the graph and determine visually that dx/dt > 0 where x is increasing.

The only thing you should add are some units in part c. They're telling you the units at which y is changing, so you should report the same units when you say how x is changing. Also, it's probably a good idea to simplify the fraction.
Will do, thanks.
 
Last edited by a moderator:
  • #4
33,741
5,432
What you can say for part a, is by direct observation, x is increasing in the 3rd and 4th quadrants.
 

Related Threads on Find a relation between dx/dt and dy/dt

Replies
4
Views
3K
  • Last Post
Replies
5
Views
5K
Replies
2
Views
2K
Replies
5
Views
2K
Replies
5
Views
1K
  • Last Post
Replies
2
Views
3K
Replies
43
Views
4K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
17K
Top