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