Question

a street light is mounted at the top of a 15 - ft pole. a man 6 feet tall walks away from the pole with a speed of 4 ft/s along a straight path. how fast (in ft/s) is the tip of his shadow moving when he is 30 feet from the pole?

1. -/1 points

two cars start moving from the same point. one travels south at 24 mi/h and the other travels west at 18 mi/h. at what rate (in mi/h) is the distance between the cars increasing three hours later?

Explanation:

Step1: Set up similar - triangles for the first problem

Let $x$ be the distance of the man from the pole and $y$ be the distance from the pole to the tip of his shadow. By similar triangles, $\frac{6}{15}=\frac{y - x}{y}$, which simplifies to $6y=15y - 15x$, then $9y = 15x$, or $y=\frac{5}{3}x$.

Step2: Differentiate with respect to time

Differentiating both sides of $y=\frac{5}{3}x$ with respect to time $t$, we get $\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}$. Given $\frac{dx}{dt}=4$ ft/s, so $\frac{dy}{dt}=\frac{5}{3}\times4=\frac{20}{3}$ ft/s.

Step3: Set up the Pythagorean - theorem for the second problem

Let $x$ be the distance of the car moving west and $y$ be the distance of the car moving south, and $z$ be the distance between the two cars. Then $z^{2}=x^{2}+y^{2}$.

Step4: Find $x$, $y$, and $z$ after 3 hours

After 3 hours, $x = 18\times3=54$ miles and $y = 24\times3 = 72$ miles. Then $z=\sqrt{54^{2}+72^{2}}=\sqrt{(18\times3)^{2}+(18\times4)^{2}}=18\sqrt{3^{2}+4^{2}}=90$ miles.

Step5: Differentiate the Pythagorean equation

Differentiating $z^{2}=x^{2}+y^{2}$ with respect to time $t$ gives $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$.

Step6: Substitute the known values

We know that $\frac{dx}{dt}=18$ mi/h and $\frac{dy}{dt}=24$ mi/h, $x = 54$ miles, $y = 72$ miles, and $z = 90$ miles. Substituting these values into $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$, we have $90\frac{dz}{dt}=54\times18 + 72\times24$.
$90\frac{dz}{dt}=972+1728=2700$. Then $\frac{dz}{dt}=30$ mi/h.

Answer:

1. $\frac{20}{3}$ ft/s
2. $30$ mi/h