Question

bc unit 4 quiz

1. 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?

(a) dh/dt = l/p s - s
(b) dh/dt = l/p s - ds/dt
(c) dh/dt = l/p ds/dt - s
(d) dh/dt = l/p ds/dt - ds/dt

1. a 10 - foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at the rate of 1 foot per second. which of the following is true about the distance between the top of the ladder and the ground when the base of the ladder is 9 feet from the wall?

(a) the distance is increasing at a rate of 9/√19 feet per second.
(b) the distance is decreasing at a rate of 9/√19 feet per second.
(c) the distance is increasing at a rate of √19/9 feet per second.
(d) the distance is decreasing at a rate of √19/9 feet per second.

Explanation:

Step1: Differentiate given equation

Given $\frac{1}{L}(h + s)=\frac{1}{P}s$. Multiply both sides by $LP$ to get $P(h + s)=Ls$. Differentiate with respect to time $t$ using the sum - rule and chain - rule. The derivative of $P(h + s)$ with respect to $t$ is $P(\frac{dh}{dt}+\frac{ds}{dt})$, and the derivative of $Ls$ with respect to $t$ is $L\frac{ds}{dt}$. So, $P(\frac{dh}{dt}+\frac{ds}{dt})=L\frac{ds}{dt}$.

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

Expand the left - hand side: $P\frac{dh}{dt}+P\frac{ds}{dt}=L\frac{ds}{dt}$. Then, isolate $\frac{dh}{dt}$: $P\frac{dh}{dt}=L\frac{ds}{dt}-P\frac{ds}{dt}$. Factor out $\frac{ds}{dt}$ on the right - hand side: $P\frac{dh}{dt}=(L - P)\frac{ds}{dt}$. Divide both sides by $P$ to get $\frac{dh}{dt}=\frac{L}{P}\frac{ds}{dt}-\frac{ds}{dt}$.

for question 4:

Step1: Establish the relationship

Let $x$ be the distance of the base of the ladder from the wall and $y$ be the distance of the top of the ladder from the ground. By the Pythagorean theorem, $x^{2}+y^{2}=100$ (since the length of the ladder $L = 10$).

Step2: Differentiate with respect to time

Differentiate both sides of the equation $x^{2}+y^{2}=100$ with respect to time $t$. Using the chain - rule, we get $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. Then, $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$.

Step3: Find $y$ when $x = 9$

When $x = 9$, we can find $y$ using $x^{2}+y^{2}=100$. So, $y=\sqrt{100 - 81}=\sqrt{19}$. We know that $\frac{dx}{dt}=1$.

Step4: Calculate $\frac{dy}{dt}$

Substitute $x = 9$, $y=\sqrt{19}$, and $\frac{dx}{dt}=1$ into $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$. We get $\frac{dy}{dt}=-\frac{9}{\sqrt{19}}$. The negative sign indicates that the distance $y$ (between the top of the ladder and the ground) is decreasing.

Answer:

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