Question

a spotlight on the ground shines on a wall 12 m away. if a man 2 m tall walks from the spotlight toward the building at a speed of 1.3 m/s, how fast (in m/s) is the length of his shadow on the building decreasing when he is 4 m from the building? (round your answer to one decimal place.)

1. -/1 points

a man starts walking north at 5 ft/s from a point p. five minutes later a woman starts walking south at 7 ft/s from a point 500 ft due east of p. at what rate (in ft/s) are the people moving apart 15 minutes after the woman starts walking? (round your answer to two decimal places.)

Explanation:

Step1: Set up similar - triangles relationship for the first problem

Let $x$ be the distance of the man from the spotlight and $y$ be the length of his shadow on the building. The distance from the spotlight to the building is $L = 12$ m. By similar triangles, $\frac{2}{y}=\frac{x}{12}$, so $y=\frac{24}{x}$. The man is walking towards the building at a speed of $\frac{dx}{dt}=1.3$ m/s. When he is 4 m from the building, $x = 12 - 4=8$ m.

Step2: Differentiate the equation with respect to time $t$

Differentiate $y=\frac{24}{x}$ with respect to $t$ using the chain - rule. We have $\frac{dy}{dt}=-\frac{24}{x^{2}}\cdot\frac{dx}{dt}$.

Step3: Substitute the known values

Substitute $x = 8$ m and $\frac{dx}{dt}=1.3$ m/s into the equation for $\frac{dy}{dt}$. Then $\frac{dy}{dt}=-\frac{24}{8^{2}}\times1.3=-\frac{24}{64}\times1.3=-\frac{3}{8}\times1.3=- 0.4875\approx - 0.5$ m/s. The negative sign indicates that the length of the shadow is decreasing.

Step4: Set up the distance formula for the second problem

Let $t$ be the time (in seconds) since the woman starts walking. The man has been walking for $t + 300$ seconds. The distance the man has walked north is $d_1=5(t + 300)$ ft, and the distance the woman has walked south is $d_2 = 7t$ ft. The horizontal distance between them is $x = 500$ ft. The distance $D$ between them is given by $D^{2}=x^{2}+(d_1 + d_2)^{2}=500^{2}+(5(t + 300)+7t)^{2}=250000+(5t+1500 + 7t)^{2}=250000+(12t + 1500)^{2}$.

Step5: Differentiate the equation with respect to time $t$

Differentiate $D^{2}=250000+(12t + 1500)^{2}$ with respect to $t$. Using the chain - rule, $2D\frac{dD}{dt}=2(12t + 1500)\times12$.

Step6: Calculate the values of $D$ and $t$

When $t = 15\times60=900$ s, $d_1=5\times(900 + 300)=5\times1200 = 6000$ ft, $d_2=7\times900 = 6300$ ft. Then $D=\sqrt{500^{2}+(6000 + 6300)^{2}}=\sqrt{250000+(12300)^{2}}=\sqrt{250000 + 151290000}=\sqrt{151540000}\approx12309.34$ ft.

Step7: Substitute the values to find $\frac{dD}{dt}$

Substitute $t = 900$ s and $D\approx12309.34$ ft into $2D\frac{dD}{dt}=2(12t + 1500)\times12$. First, when $t = 900$, $12t+1500=12\times900 + 1500=10800+1500 = 12300$. Then $2D\frac{dD}{dt}=2\times12300\times12$. So $\frac{dD}{dt}=\frac{12300\times12}{D}=\frac{12300\times12}{12309.34}\approx11.99$ ft/s.

Answer:

• For the first problem: $0.5$ m/s
• For the second problem: $11.99$ ft/s