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This is the question: Two cars leave from the same point at the same time. One car travels north at a rate of 60 miles per hour and the other travels east at a speed of 80 miles per hour. How fast is the distance between the cars changing when 2 hours has elapsed?

When I first do these problems I try to think of what formula I can use to help me, for this one I chose the Pythagorean theorem. I created a right triangle and set the x (or east car) to 160 because that's how far this car would have traveled after two hours, and the y (north car) I set to 120 for the same reason. $160^2+120^2=r^2$ and I found $r=200$. This is how I set it up:

$$x^2+y^2=r^2$$ $${d\over dv}(x^2+y^2)={d\over dv}(200)$$ $$2x{dx\over dv}+2y{dy\over dv}=0$$ $$2x{dx\over dv}=-2y{dy\over dv}$$ I then set ${dx\over dv}=80mi/hr$ and${dy\over dv}=60mi/hr$

$$2x(80mi/hr)=-2y(60mi/hr)$$ $$-{x\over y}={60\over 80}$$ $$-{x\over y}=0.75$$

Now I'm stuck, what have I done wrong?

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When you take the derivative of $x^2+y^2=r^2$, you don't plug in $r=200$ first. So it would be as follows: $$x^2+y^2=r^2\\ 2x\frac{dx}{dv}+2y\frac{dy}{dv}=2r\frac{dr}{dv}$$ Now plug in $x=160$, $y=120$, $r=200$, $\frac{dx}{dv}=80$, $\frac{dy}{dv}=60$, and solve for $\frac{dr}{dv}$.

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You've correctly differentiated the left hand side of $x^2 + y^2 = r^2$, but remember that the right hand side is not constant; it is changing with time. The right side differentiated should be $2r\frac{dr}{dv}$. From there you should be able to solve the problem.

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