A particle moves along the curve y= sqrt x . When y=2 the x-component of its position is increasing at the rate of 4 units per second. a) What is the value of dy/dt when y=2?

Question

b) How fast is the distance from the particlr to the origin changing when y=2?

c)What is the value of d(angle)/dt when y=2

Answer 1

y =x^(1/2)

dy/dt = (1/2)x^(-1/2)

When y = 2, x=4

so dy/dt = (1/2)(1/2) = 1/4 = 0.25

Answer 2

Given y=2, you can find the value of x in that moment. 2=sqrt x, so x=4.

As y=sqrt{x}, you can get dy/dt= (1/2) (x^(-1/2)) (dx/dt). Hence dy/dt = (1/2) (1/2) (4) = 1 unit per second.

Note that sqrt x= x^(1/2).

For part (b).

The distance formula is r = sqrt{ x^2 + y^2}, so r^2 = x^2+ y^2 which can be written as

r^2 = x^+y^2 = x^2 + (sqrt x)^2= x^2 + x

I have written in this form, since it's an easier form to work with. If y=2, then x=4, so r= sqrt 20

Differentiating the formula r^2= x^2+ x you get the following

2r (dr/dt) = 2x (dx/dt) + (dx/dt),

By plugging the values x=4, r=sqrt 20, dx/dt=4 , we get 2\sqrt 20 dr/dt= 2(4)(4) + 4= 36.

Thus dr/dt = 72/sqrt 20.