Question

at a concert, a band is playing on a platform that extends $p$ feet from a wall behind the band, and the platform is rising from ground level, as shown in the figure above. a light source is $l$ feet from the wall, and the platform casts a lengthening shadow on the wall as the platform rises. at time $t$ seconds, the platform is $h$ feet above the ground, and the height of the shadow is $s$ feet. the quantities are related by the equation $\frac{1}{l}(h + s)=\frac{1}{p}s$, where $l$ and $p$ are constants. which of the following best expresses the rate of change of $h$ with respect to time in terms of the rate of change of $s$ with respect to time?
a $\frac{dh}{dt}=\frac{l}{p}s - s$
b $\frac{dh}{dt}=\frac{l}{p}s-\frac{ds}{dt}$
c $\frac{dh}{dt}=\frac{l}{p}\frac{ds}{dt}-s$
d $\frac{dh}{dt}=\frac{l}{p}\frac{ds}{dt}-\frac{ds}{dt}$

Explanation:

Step1: Differentiate the given equation

Differentiate $\frac{1}{L}(h + s)=\frac{1}{P}s$ with respect to $t$. Using the sum - rule and constant - multiple rule of differentiation, we get $\frac{1}{L}(\frac{dh}{dt}+\frac{ds}{dt})=\frac{1}{P}\frac{ds}{dt}$.

Step2: Solve for $\frac{dh}{dt}$

First, multiply both sides of the equation $\frac{1}{L}(\frac{dh}{dt}+\frac{ds}{dt})=\frac{1}{P}\frac{ds}{dt}$ by $L$: $\frac{dh}{dt}+\frac{ds}{dt}=\frac{L}{P}\frac{ds}{dt}$. Then, subtract $\frac{ds}{dt}$ from both sides: $\frac{dh}{dt}=\frac{L}{P}\frac{ds}{dt}-\frac{ds}{dt}$.

Answer:

D. $\frac{dh}{dt}=\frac{L}{P}\frac{ds}{dt}-\frac{ds}{dt}$