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# 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 speed is $600 ft/s$ when it has risen $3000 ft.$(a) How fast is the distance from the television camera to the rocket changing at that moment?(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?

## a) 360 $\mathrm{ft} / \mathrm{s}$b) 0.096 $\mathrm{rad} / \mathrm{s}$

Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

So for these related rates problems it's always a good idea to draw yourself a picture. So I have a rocket and don't laugh at my horrible drawing. But there's my rocket ship And 400 ft away is my tv camera. And obviously that's not a tv camera, that's an old fashioned television. So I've got my rocket ship and the elevation of the camera will change. So we're going to call that feta. And I'm gonna connect this because I'm going to end up with a right triangle. I call this X. Y. And Z. It should be a right triangle where X squared plus Y squared equals the square. So that will help us with the first part of the question. Now, the distance from the camera Is changing at a speed of 600 ft/s. So that's D X. D. T. That's what how the rocket speed is changing and that's 600 feet for a second when it's risen 300 ft or 3000 ft. So when X is 3000, it's changing at this rate and this is going to be a constant. And then I'm asked, how far is the distance from the television to the camera. So I'm asked to find Z. DZ DT, what does DZ DT equal under these conditions? All right. So at that moment It's not 400, sorry, it's 4000 ft left zero. So at that moment X squared three 1000 squared plus Y squared equals E squared. So at that moment for that instant Z will be fixed. So you need to put in your calculator 3000 squared Or recognize that you have a 345 triangle. You recognize that you have a 345 triangle. And you'll be able to recognize that Z will be 5000 ft at that moment. 345. Okay. So now that I have that information, I have to find disease E D T. Which means I'm gonna have to take the derivatives. So going to another screen X squared plus 4000 because remember that's a constant equals Z squared. So taking the derivative, bring down the 22 x times the derivative effects which should be dX DT zero because the derivative of constant is zero to Z. DZ DT I just need to plug in the values that I'm given. So X 3000 dX DT is 600 And Z was 5000. And then to solve for DZ D T or if you wanted to, you could rewrite this formula and divide both sides by two Z. And he would end up with X divided by z times DX GT equals DZ DT. I'll pause. So whether you solve for DZ DT and then substitute in or substitute and then solve DZ D T. Well equal. I'm going to use this up here 1000 Divided by 5000 times 600. Yeah. And we get 360 feet per second. So at that moment, this distance is changing by a rate Of, what did I say? 360 ft her second. And then you're asked how fast is the distance? No. Oh you're asked to find d theta DT how fast is the camera angle changing? Okay. So I still have X, Y and Z. But I'm not going to use easy anymore. And I know that this is 400 or 4000. And this is data. So what's the relationship between these legs of the triangle and data? That would be tangent? So the tangent of theta is X over y. Now 400 is going to be 4000 will be a constant. So I'm going to multiply that by both sides and this is going to be the formula that I am going to take the derivative of. So when I take the derivative, this is a constant. So 4000 times the derivative of tangent seeking squared times the derivative of theta with respect to t. Now I go back to the other page and remember this was 600. This is what I'm trying to find. So I do need one more piece of information here. I need to know data. So I'm gonna come back to this formula and remember X at that moment is three 1000 pete. So I'm gonna solve this for tangent or for to so tha tha will be the inverse tangent of three Force. And I don't know if you're supposed to have it in radiance or in degrees. So because I'm going to leave mine and radiant. So when I do Inverse Tangent of three Force I get approximately 0.64 35 radiance. Okay. Now that I know that I'm gonna plug that in for third up so seek it Squared of .6435. Oops. I didn't leave myself room D. TheTA DT. Which means D theTA DT Will be 600 Divided by 4000. Seek it squared .643 five and plugging that into my calculator. I get approximately .096 Radiance. Uh huh. Second that it's changing.

Mount Vernon Nazarene University

#### Topics

Derivatives

Differentiation

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp