Calc I for Engineers

[sofay]

QUESTION: A particle is moving along the curve

. As the particle passes through the point

, its

-coordinate increases at a rate of

units per second. Find the rate of change of the distance from the particle to the origin at this instant.

I don't even know how to begin this problem...
so It would be nice if someone could explain it.

Thanks,

sofay

 

[MarkFL]

I would begin by stating the relationship between the distance of the particle from the origin and its coordinates:

\(\displaystyle d^2(t)=x^2(t)+y^2(t)\)

What do you get when you implicitly differentiate with respect to t?

 

[sofay]

I would begin by stating the relationship between the distance of the particle from the origin and its coordinates:

\(\displaystyle d^2(t)=x^2(t)+y^2(t)\)

What do you get when you implicitly differentiate with respect to t?

Well don't I have to multiply both x and y by their respective dx/dt or dy/dt since they're differentiable according to time?

So then it would be
= 2x*(dx/dt) + 2y*(dy/dt)

 

[MarkFL]

Yes, you have the right idea with the chain rule, but you want to include the left side as well:

\(\displaystyle 2d(t)\dfrac{dd}{dt}=2x(t)\dfrac{dx}{dt}+2y(t) \dfrac{dy}{dt}\)

Divide through by \(\displaystyle 2d(t)\):

\(\displaystyle \dfrac{dd}{dt}=\dfrac{x(t)\dfrac{dx}{dt}+y(t) \dfrac{dy}{dt}}{d(t)}\)

We are given:

\(\displaystyle x(t)=3,\,y(t)=12,\,\dfrac{dx}{dt}=5\dfrac{\text{u}}{\text{s}}\).

How can we find \(\displaystyle d(t)\) and \(\displaystyle \dfrac{dy}{dt}\)?