suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 45…

Explanation:

Step1: Define variables

Let $y$ be the position of the police - officer relative to the intersection (initially $y =-\frac{1}{2}$ mile and $\frac{dy}{dt}=45$ mph), $x$ be the position of the other car relative to the intersection (initially $x=\frac{1}{2}$ mile), and $z$ be the straight - line distance between the officer and the other car. By the Pythagorean theorem, $z^{2}=x^{2}+y^{2}$.

Step2: Differentiate the equation with respect to time $t$

Differentiating $z^{2}=x^{2}+y^{2}$ with respect to $t$ gives $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$, which simplifies to $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$. When $x = \frac{1}{2}$ mile, $y=-\frac{1}{2}$ mile, then $z=\sqrt{(\frac{1}{2})^{2}+(-\frac{1}{2})^{2}}=\frac{\sqrt{2}}{2}$ mile.

Step3: Solve for $\frac{dx}{dt}$ in the first case

We know that $\frac{dz}{dt}=15$ mph, $x = \frac{1}{2}$ mile, $y =-\frac{1}{2}$ mile, $\frac{dy}{dt}=45$ mph. Substitute these values into $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$: $\frac{\sqrt{2}}{2}\times15=\frac{1}{2}\frac{dx}{dt}+(-\frac{1}{2})\times45$. Multiply through by 2 to get $\sqrt{2}\times15=\frac{dx}{dt}-45$. $\frac{dx}{dt}=45 + 15\sqrt{2}\approx45+15\times1.414=45 + 21.21=66.21$ mph.

Step4: Solve for $\frac{dx}{dt}$ in the second case

We know that $\frac{dz}{dt}=- 15$ mph, $x=\frac{1}{2}$ mile, $y =-\frac{1}{2}$ mile, $\frac{dy}{dt}=45$ mph. Substitute into $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$: $\frac{\sqrt{2}}{2}\times(-15)=\frac{1}{2}\frac{dx}{dt}+(-\frac{1}{2})\times45$. Multiply through by 2: $-\sqrt{2}\times15=\frac{dx}{dt}-45$. $\frac{dx}{dt}=45-15\sqrt{2}\approx45 - 21.21 = 23.79$ mph.

Answer:

1. $45 + 15\sqrt{2}\approx66.21$
2. $45-15\sqrt{2}\approx23.79$