Ugh, a question I could do with my eyes closed 7 months ago is taking me three hours, and 7 attempts to do now. I eventually got it after being stupid with Arctan, but I want to do it another way, so yeah....
A television camera at ground level is filming the lift off of a rocket that is rising vertically with a postition s = 25t^2 m, where t is in seconds. Find the rate of change of the angle after 10 seconds.
They give you the diagram, noting that the distance from the camera to the rocket at launch is 1000 m
So it's a simple triangle with the base of 1000 m constant and the height of 25t^2.
So this is what I did ( let x represent the angle, since I can't find no delta symbol on here
):
tanx = s/1000
1000tan(x) = s
1000tan(x) = 25t^2
derive w/ respect to t
1000 [sec(x)]^2 dx/dt = 50t
dx/dt = 50t * 1/1000 * [tan(x)]^2
dx/dt = 1/2 * [ 2500 / 1000 ] ^2
dx/dt = 3.125
But that's wrong, the answer is 2/29. Which I eventually got using arctan, but I want to do it this way. I can do it this way right? Bah!! Final in 4 hours :S
A television camera at ground level is filming the lift off of a rocket that is rising vertically with a postition s = 25t^2 m, where t is in seconds. Find the rate of change of the angle after 10 seconds.
They give you the diagram, noting that the distance from the camera to the rocket at launch is 1000 m
So it's a simple triangle with the base of 1000 m constant and the height of 25t^2.
So this is what I did ( let x represent the angle, since I can't find no delta symbol on here
):tanx = s/1000
1000tan(x) = s
1000tan(x) = 25t^2
derive w/ respect to t
1000 [sec(x)]^2 dx/dt = 50t
dx/dt = 50t * 1/1000 * [tan(x)]^2
dx/dt = 1/2 * [ 2500 / 1000 ] ^2
dx/dt = 3.125
But that's wrong, the answer is 2/29. Which I eventually got using arctan, but I want to do it this way. I can do it this way right? Bah!! Final in 4 hours :S
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