Related Rates with shadows

Question

On a morning of a day when the sun will pass directly​ overhead, the shadow of a 72​-ft building on level ground is 30 ft​ long, as shown in the figure. At the moment in​ question, the angle theta the sun makes with the ground is increasing at the rate of 0.24degrees​/min. At what rate is the shadow​ decreasing? Express your answer in​ in./min. (Remember to use​ radians.)

Answer 1

Another way. No figure was shown, so we use the setup of the legs being the height of the 72 ft building and x representing distance of shadow length from the building. Also, theta is the angle of elevation along the hypotenuse extending from the end of the shadow to the top of the building.

cot theta = x/72

dx/dt = (dx/d theta) X (d theta/dt)

d theta/dt = 0.24 pi/180 = 0.004189 rad/min

The right triangle is actually a 5-12-13 rt triangle, multiplied by 6

dx/d theta = -72 csc² theta = -72(13/12)² = - 169/2 = -84.5

So -84.5 X 0.004189 = -0.35395, approximately -0.354 ft/min

Answer 2

The building, shadow and sun rays form a right triangle. The building is the vertical side. The shadow is the horizontal side. The sun rays are the hypotenuse.

The vertical side is constant at 72 since the building height doesn't change.

Since the length of the shadow is changing, we need to use a variable for that. Let's call it x.

Since the angle that the sun rays are making with the ground is changing, we need a variable for that too. Let's call it A.

Therefore, we have a right triangle with angle A, opposite side of 72 and adjacent side of x.

Therefore, we can write an equation with those variables using trigonometry:

tan A = 72/x

They give us the rate at which the angle A is changing: 0.24 deg/min.

First, we need it in radians. Multiply by pi/180 deg = 0.0042 rad/min.

This means that dA/dt = 0.0042

Obviously, this is a derivative, so we need to take the derivative of our tan equation (with respect to t).

d/dt (tan A) = d/dt (72/x)

Note we need to implicit differentiation since A and x are both function of time t.

sec²A * dA/dt = (-72/x²) * dx/dt

We know dA/dt = 0.0042.

At this moment, the length of the shadow is 30 so x = 30.

We are looking for dx/dt so leave that alone.

We still need A. We can actually solve for sec A.

At this moment, x = 30 and the vertical side is 72. Since sec A = hypotenuse / adjacent, if we find hypotenuse, we can find sec A.

30² + 72² = h²

6084 = h²

78 = h

So sec A = 78/30 = 2.6

sec²A = (2.6)² = 6.76

So in our derivative equation, we have:

(6.76)*(0.0042) = (-72/30²)*dx/dt

0.028392 = (-0.08)*dx/dt

-0.3549 = dx/dt

The length of the shadow is decreasing (the derivative is negative) at a rate of 0.3549 ft / min.