Calculus: Early Transcendentals 8th Edition

The distance between the cars is increasing at a rate of $65~mi/h$
Let $z$ be the distance between the two cars. We can find $z$ after 2 hours: $z^2 = x^2+y^2$ $z = \sqrt{x^2+y^2}$ $z = \sqrt{(50~mi)^2+(120~mi)^2}$ $z = 130~mi$ We can differentiate both sides of the equation with respect to $t$: $z^2 = x^2+y^2$ $2z~\frac{dz}{dt} = 2x~\frac{dx}{dt} + 2y~\frac{dy}{dt}$ $\frac{dz}{dt} = \frac{1}{z}~(x~\frac{dx}{dt} + y~\frac{dy}{dt})$ $\frac{dz}{dt} = \frac{1}{130}~[(50)(25) + (120)(60)]$ $\frac{dz}{dt} = 65~mi/h$ The distance between the cars is increasing at a rate of $65~mi/h$.