Example:
A particle is moving around a circle of radius 5
around the origin. As it passes through the point (3,4), its x
position is changing at a rate dxdt|x=3,y=4=2.
How fast is y
changing at that instant?
Solution: We know that the equation for the circle is x2+y2=25. To find a relationship between the rates of change of x and y with respect to time, we can implicitely differentiate the equation above with respect to t.
2xdxdt+2ydydt=0.
We know that when the particle is passing (3,4), then its velocity is dxdt|x=3,y=4=2, so we can solve for
dydt|x=3,y=4. Since
2(3)(2)+2(4)dydt|x=3,y=4=0, we have
dydt|x=3,y=4=−128=−32. |