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Related rates,...again
I also have another question (I'm not doing too good one these realted rates). A camera, located 2 km from the launch pad, is tracking the rocket that is fired straight up. When the height of the rocket is 20 km, the camera is
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A missile rises vertically from a point on the ground 75,000 feet from a radar station. If the missile is rising at the rate of 16,500 feet per minute at the instant when it is 38,000 feet high, what is the rate of change(in
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Solve the integral: (1x^2)^(3/2)/x^2 This is what I have so far. I used trig substitution. x=sin theta dx=cos theta dtheta [1(sin^2 theta)^(3/2)]*[cos theta dtheta]/(sin^2 theta) [(cos^3 theta)]*[cos theta dtheta]/(sin^2 theta)
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asked by gita on September 10, 2013 
More related rates.
Heyy, its me again. So last night I asked a related rate question, and I think I still don't get it. I tried it a few differnt ways and I think I'm just missing something...I don't think a speed of a rocket can be 0.018km/hr:S. So
asked by Tracy on November 3, 2006 
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Finding critical numbers g(theta) = 16(theta)  4tan(theta) I can only get the derivative of g to be g'(theta) = 16  4sec^2(theta). Am I supposed to move on to get sec^2(theta) = 4, then sec(theta) = 2?
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If the hands on a clocktower's clock are 2ft and 3ft long, how fast is the distance between the tips of the hands of the clock changing at 9:20? I know that the derivative of the minute hand is pi/30 rad/min and derivative of the
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What is a simplified form of the expression (sec^3 theta)  (sec theta/cot^2 theta)? a. 0 b. sec thetatan theta c. cos theta d. sec theta***** Every time I attempt to work out this problem I end up with one. I know the answer is
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