at a concert, a band is playing on a platform that extends p feet from a wall behind the band, and the…

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 1/l(h + s)=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?

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 1/l(h + s)=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?

Answer

Explanation:

Step1: Differentiate the given equation

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

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

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

Answer:

$\frac{dh}{dt}=\left(\frac{L - P}{P}\right)\frac{ds}{dt}$