SOLUTION 16: Car B starts 30 miles directly east of car A.

Here is a diagram showing the positions of cars A and B after $t$ hours. Assume that distances $x, y,$ and $z$ are all functions of time $t$.

GIVEN: $\ \ \ \displaystyle{ dx \over dt } = -90 \ mph$ (It is negative since $x$ is decreasing.) and $\displaystyle{ dy \over dt } = 60 \ mph$.

FIND: $\ \ \ \displaystyle{ dz \over dt }$ when $\ \$ a.) $\ t=1/5 \ hr.$ $\ \$ b.) $\ t=1/3 \ hr.$

Using the Pythagorean Theorem we get the following distance equation:

$$x^2+y^2=z^2$$

Differentiating the distance equation, we get

$$D \{ x^2+y^2 \} = \{ z^2 \} \ \ \ \ \longrightarrow$$ $$2x \displaystyle{ dx \over dt } + 2y \displaystyle{ dy \over dt } = 2 z \displaystyle{ dz \over dt } \ \ \ \ \longrightarrow$$

$\Big($ Multiply both sides of the equation by $\displaystyle{ 1 \over 2 }. \Big)$

$$(**) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \displaystyle{ dx \over dt } + y \displaystyle{ dy \over dt } = z \displaystyle{ dz \over dt } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$

Recall that: $$(Distance) = (Constant \ Rate)(Time)$$

$\ \ \ \ \$ a.) If $t=1/5 \ hr.$, then car B has traveled west $(90 \ mph)(1/5 \ hr.) = 18 \ miles,$ so that $x=30-18=12 \ miles$; and car A has traveled north $(60 \ mph)(1/5 \ hr.) = 12 \ miles,$ so that $y=12 \ miles.$ Using the distance equation we get that $\ (12)^2+(12)^2= z^2 \ \ \rightarrow \ z^2= 288 \ \ \rightarrow \ z = \sqrt{288} \approx 16.97 \ miles.$

Now let $\displaystyle{ dx \over dt } = -90, \displaystyle{ dy \over dt } = 60, x=12, y=12$ and $\ z=\sqrt{288}$ in equation (**), getting

$$(12)(-90) + (12)(60) = (\sqrt{288}) \displaystyle{ dz \over dt } \ \ \ \ \longrightarrow$$ $$-1080 + 720 = \sqrt{288} \displaystyle{ dz \over dt } \ \ \ \ \longrightarrow$$ $$\displaystyle{ dz \over dt } = { -360 \over \sqrt{288} } \approx -21.21 \ mph$$

$\ \ \ \ \$ b.) If $t=1/3 \ hr.$, then car B has traveled west $(90 \ mph)(1/3 \ hr.) = 30 \ miles,$ so that $x=30-30=0 \ miles$; and car A has traveled north $(60 \ mph)(1/3 \ hr.) = 20 \ miles,$ so that $y=20 \ miles.$ Using the distance equation we get that $\ (0)^2+(20)^2= z^2 \ \ \ \rightarrow \ z^2= 400 \ \ \ \rightarrow \ z = 20 \ miles.$

Now let $\displaystyle{ dx \over dt } = -90, \displaystyle{ dy \over dt } = 45, x=0, y=20$ and $\ z=20$ in equation (**), getting

$$(0)(-90) + (20)(60) = (20) \displaystyle{ dz \over dt } \ \ \ \ \longrightarrow$$ $$1200 = 20 \displaystyle{ dz \over dt } \ \ \ \ \longrightarrow$$ $$\displaystyle{ dz \over dt } = 60 \ mph$$