a street light is at the top of a 10 - ft - tall pole. a 6 - ft tall woman walks away from the pole with a…
a street light is at the top of a 10 - ft - tall pole. a 6 - ft tall woman walks away from the pole with a speed of 4 ft/sec along a straight path (see figure). how fast is the tip of her shadow moving, in ft/sec, when she is 50 ft from the base of the pole?
Answer
Explanation:
Step1: Set up similar - triangles relationship
Let $x$ be the distance of the woman from the base of the pole and $y$ be the distance from the base of the pole to the tip of her shadow. Then, by similar triangles, $\frac{10}{y}=\frac{6}{y - x}$. Cross - multiplying gives $10(y - x)=6y$. Expanding, we get $10y-10x = 6y$, and then $4y = 10x$, so $y=\frac{5}{2}x$.
Step2: Differentiate with respect to time
Differentiate both sides of $y=\frac{5}{2}x$ with respect to time $t$. Using the chain rule, $\frac{dy}{dt}=\frac{5}{2}\frac{dx}{dt}$.
Step3: Substitute the given value of $\frac{dx}{dt}$
We are given that $\frac{dx}{dt}=4$ ft/sec. Substituting this into the equation $\frac{dy}{dt}=\frac{5}{2}\frac{dx}{dt}$, we have $\frac{dy}{dt}=\frac{5}{2}\times4$.
Answer:
10