Question

A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let's assume the rocket rises vertically and its speed is 800 ft/s when it has risen 3000 ft. (Round your answers to three decimal places.)
(a) How fast is the distance from the television camera to the rocket changing at that moment?
ft/s

(b) If the television camera is always kept aimed at the rocket, how fast is the camera's angle of elevation changing at that same moment?
rad/s

Answer 1

Answers:

• Part (a)    480 feet per second
• Part (b)   0.128 radians per second

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Explanation for part (a)

• t = time in seconds
• x = horizontal distance from the camera to the launch pad
• y = vertical distance from the launch pad to the rocket's location
• z = distance from camera to the rocket at time t

All distances mentioned are in feet.

We'll have a right triangle which allows us to apply the pythagorean theorem. Refer to the diagram below.

a^2+b^2 = c^2

x^2+y^2 = z^2

Apply the derivative to both sides with respect to t. We'll use implicit differentiation and the chain rule.

$x^2+y^2 = z^2\\\\\frac{d}{dt}[x^2+y^2] = \frac{d}{dt}[z^2]\\\\\frac{d}{dt}[x^2]+\frac{d}{dt}[y^2] = \frac{d}{dt}[z^2]\\\\2x*\frac{dx}{dt}+2y*\frac{dy}{dt}=2z*\frac{dz}{dt}\\\\x*\frac{dx}{dt}+y*\frac{dy}{dt}=z*\frac{dz}{dt}$

Now we'll plug in (x,y,z) = (4000,3000,5000). The x and y values are given. The z value is found by use of the pythagorean theorem. Ie, you solve 4000^2+3000^2 = z^2 to get z = 5000. Or you could note that this is a scaled copy of the 3-4-5 right triangle.

We know that dx/dt = 0 because the horizontal distance, the x distance, is not changing. The rocket is only changing in the y direction. Or you could say that the horizontal speed is zero.

The vertical speed is dy/dt = 800 ft/s and it's when y = 3000. It's likely that dy/dt isn't the same value through the rocket's journey; however, all we care about is the instant when y = 3000.

Let's plug all that in and isolate dz/dt

$4000*0+3000*800=5000*\frac{dz}{dt}\\\\2,400,000=5000*\frac{dz}{dt}\\\\\frac{dz}{dt} = \frac{2,400,000}{5000}\\\\\frac{dz}{dt} = 480$

At the exact instant that the rocket is 3000 ft in the air, the distance between the camera and the rocket is changing by an instantaneous speed of 480 ft/s.

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Explanation for part (b)

Again, refer to the diagram below.

We have theta (symbol $\theta$) as the angle of elevation. As the rocket's height increases, so does the angle theta.

We can tie together the opposite side y with the adjacent side x with the tangent function of this angle.

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(\theta) = \frac{y}{x}$

Like before, we'll apply implicit differentiation. This time we'll use the quotient rule as well.

$\tan(\theta) = \frac{y}{x}\\\\\frac{d}{dt}[\tan(\theta)] = \frac{d}{dt}\left[\frac{y}{x}\right]\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{\frac{dy}{dt}*x - y*\frac{dx}{dt}}{x^2}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{800*4000 - 3000*0}{4000^2}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{3,200,000}{16,000,000}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{32}{160}\\\\\sec^2(\theta)*\frac{d\theta}{dt} = \frac{1}{5}$

Let's take a brief detour. We'll return to this later. Recall earlier that $\tan(\theta) = \frac{y}{x}$

If we plug in y = 3000 and x = 4000, then we end up with $\tan(\theta) = \frac{3}{4}$ which becomes $\tan^2(\theta) = \frac{9}{16}$

Apply this trig identity

$\sec^2(\theta) = 1 + \tan^2(\theta)$

and you should end up with $\sec^2(\theta) = 1+\frac{9}{16} = \frac{25}{16}$

So we can now return to the equation we want to solve

$\sec^2(\theta)*\frac{d\theta}{dt} = \frac{1}{5}\\\\\frac{25}{16}*\frac{d\theta}{dt} = \frac{1}{5}\\\\\frac{d\theta}{dt} = \frac{1}{5}*\frac{16}{25}\\\\\frac{d\theta}{dt} = \frac{16}{125}\\\\\frac{d\theta}{dt} = 0.128$

This means that at the instant the rocket is 3000 ft in the air, the angle of elevation theta is increasing by 0.128 radians per second.

This is approximately 7.334 degrees per second.

Answer 2

The distance from the television camera to the rocket is changing at 480 ft/s while the camera's angle of elevation is changing at 0.128 rad/s

Let x represent the distance from the camera to the rocket and let h represent the height of the rocket.

a)

$x^2=h^2+4000^2\\\\2x\frac{dx}{dt}=2h\frac{dh}{dt} \\\\x\frac{dx}{dt}=h\frac{dh}{dt} \\\\At \ h=3000\ ft, \frac{dh}{dt}=800\ ft/s;\\x^2=3000^{2} +4000^2\\x=5000\\\\\\x\frac{dx}{dt}=h\frac{dh}{dt} \\\\5000\frac{dx}{dt}=3000*800\\\\\frac{dx}{dt}=480\ ft/s$

b)

$tan(\theta)=\frac{h}{4000} \\\\h=4000tan(\theta)\\\\\frac{dh}{dt}=4000sec^2(\theta)\frac{d\theta}{dt} \\\\\\At\ h=3000\ ft;\\\\tan\theta = \frac{3000}{4000}=\frac{3}{4} \\\\sec^2(\theta)=1+tan^2(\theta)=1+(\frac{3}{4})^2=\frac{25}{16} \\\\\\\frac{dh}{dt}=4000sec^2(\theta)\frac{d\theta}{dt}\\\\800=4000*\frac{25}{16}* \frac{d\theta}{dt}\\\\\frac{d\theta}{dt}=0.128\ rad/s$

Hence, The distance from the television camera to the rocket is changing at 480 ft/s while the camera's angle of elevation is changing at 0.128 rad/s

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