Question

5 two people start walking from the same point. one person travels west at a rate of 3 miles per hour, and the other walks north at a rate of 4 miles per hour. at what rate is the distance between the two people changing 1 hour after they start walking?

Explanation:

Step1: Define variables

Let $x$ be the distance the west - walking person travels, $y$ be the distance the north - walking person travels, and $z$ be the distance between them. By the Pythagorean theorem, $z^{2}=x^{2}+y^{2}$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of $z^{2}=x^{2}+y^{2}$ with respect to $t$ gives $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$, which simplifies to $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$.

Step3: Find $x$, $y$, and $z$ at $t = 1$

We know that $\frac{dx}{dt}=3$ miles per hour and $\frac{dy}{dt}=4$ miles per hour. At $t = 1$ hour, $x=\frac{dx}{dt}\times t=3\times1 = 3$ miles and $y=\frac{dy}{dt}\times t=4\times1 = 4$ miles. Using the Pythagorean theorem, $z=\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+4^{2}} = 5$ miles.

Step4: Solve for $\frac{dz}{dt}$

Substitute $x = 3$, $y = 4$, $z = 5$, $\frac{dx}{dt}=3$, and $\frac{dy}{dt}=4$ into $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$. We get $5\frac{dz}{dt}=3\times3 + 4\times4$. So $5\frac{dz}{dt}=9 + 16=25$. Then $\frac{dz}{dt}=5$ miles per hour.

Answer:

5 miles per hour