Find a relation between dx/dt and dy/dt

[physicsernaw]

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 traveling 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 = 0Just 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.

 

[Mark44]

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 traveling 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 = 0Just 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.

 

[physicsernaw]

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.

 

[Mark44]

What you can say for part a, is by direct observation, x is increasing in the 3rd and 4th quadrants.