Solve Derivative Problem: Plane Descent from h to (0,0)

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The discussion revolves around finding a smooth landing path for a plane descending from height y=h at x=-L to (0,0) using a cubic function y=ax^3 + bx^2 + cx + d. Key conditions include ensuring the first derivative y' is negative and zero at x=0, leading to the conclusion that c=0 and d must also be 0. The second derivative y'' must be greater than 0 to maintain a smooth descent. Participants clarify that dx/dt can be constant by comparing it to a driver maintaining speed on a hillside, and they suggest using the chain rule to determine dy/dt. The conversation emphasizes the mathematical relationships necessary for the plane's descent.
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Homework Statement


A plane starts its descent from height y=h at x =-L to land at (0,0). Choose a, b, c, d so its landing path y=ax^3 + bx^2 + cx + d is smooth.With dx/dt=V=constant, find dy/dt and d2y/dt2 at x=0 and x =-L. (To keep d2y/dt2 small, a coast-to-coast plane starts down L>100 miles from the airport.)

The Attempt at a Solution


For a smooth landing the derivative must be negative and 0 at x=0.
y'=3ax^2+2bx+c < 0
y'(0)=0
c=0
d must be 0 because y(0)=0.
second derivative must be > 0.
y''=6ax+2b >0

And that's as far as I get. I'd appreciate any tips and can someone explain to me how dx/dt can be constant? And how do I find dy/dt?
 
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kristo said:
… can someone explain to me how dx/dt can be constant? And how do I find dy/dt?

Hi kristo! :smile:

Think of the curve as a hillside … the driver can drive at whatever speed he likes.

And to find dy/dt, use the chain rule. :wink:
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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