Solve the problem. Round your answer, if appropriate.
-A man $6 \mathrm { ft }$ tall walks at a rate of $3 \mathrm { ft } / \mathrm { sec }$ away from a lamppost that is $18 \mathrm { ft }$ high. At what rate is the length of his shadow changing when he is $40 \mathrm { ft }$ away from the lamppost? (Do not round your answer)
A) $\frac { 3 } { 2 } \mathrm { ft } / \mathrm { sec }$
B) $20 \mathrm { ft } / \mathrm { sec }$
C) $\frac { 3 } { 4 } \mathrm { ft } / \mathrm { sec }$
D) $\frac { 3 } { 8 } \mathrm { ft } / \mathrm { sec }$

Question

Quizplus · Accepted Answer

The rate at which the length of his shadow is changing can be found using similar triangles. Let $x$ be the distance of the man from the lamppost and $y$ be the length of his shadow. The relationship between the heights and distances is $\frac{6}{y} = \frac{18}{x+y}$. Differentiating both sides with respect to time $t$ gives $\frac{dy}{dt}$ when $x = 40$. Given $dx/dt = 3 \, \text{ft/sec}$, solving the equation yields $\frac{dy}{dt} = \frac{3}{2} \, \text{ft/sec}$.

Solve the Problem 6ft6 \mathrm { ft }6ft Tall Walks at a Rate Of

Solve the Problem $6 \mathrm { ft }$ Tall Walks at a Rate Of