Answer to Question #172685 in Calculus for Todd Phillips

Question

While walking on one bank of a 25 m wide canal you see a child 50m down the canal on the far bank slip, hit his head, and fall into the water. If you can swim 1.5m/s through the water and can run 5.0m/s along the side of the pool, what distance should you swim through the water to minimize the time to get to the child?

Answer 1

The length is 50 m and the width is 25 m.

Distance AB

$$AB^2=50^2+25^2$$

Let I run $x\ m$ along the side of the pool. Then I swim

$$BD=\sqrt{(50-x)^2+25^2}$$

the time to get to the child is

$$t=t(x)=\dfrac{x}{5}+\dfrac{\sqrt{(50-x)^2+25^2}}{1.5}, 0\leq x\leq 50$$

Find the first derivative with respect to $x$

$$t'(x)=\dfrac{1}{5}-\dfrac{50-x}{1.5\sqrt{(50-x)^2+25^2}}$$

Find the critical number(s)

$$t'(x)=0=>\dfrac{1}{5}-\dfrac{50-x}{1.5\sqrt{(50-x)^2+25^2}}=0$$

$$10(50-x)=3\sqrt{(50-x)^2+25^2}$$

$$100(50-x)^2=9(50-x)^2+9(625)$$

$$91(50-x)^2=9(625)$$

$$x=50-\dfrac{75}{\sqrt{91}}$$

First Derivative Test

If $0<x<50-\dfrac{75}{\sqrt{91}}, t'(x)<0, t(x)$ decreases

If $50-\dfrac{75}{\sqrt{91}}<x<50, t'(x)>0,t(x)$ increases.

Then function $t(x)$ has a local minimum at $x=50-\dfrac{75}{\sqrt{91}}.$

Sincd the function $t(x)$ has the only extremum, then the function $x(t)$ has the absolute minimum at $x=50-\dfrac{75}{\sqrt{91}}.$

$$BD=\sqrt{(50-50+\dfrac{75}{\sqrt{91}})^2+25^2}=\dfrac{250}{\sqrt{91}}$$

$$\approx26.207(m)$$

I should swim 26.207 m through the water to minimize the time to get to the child.