ski jump derivative problem

JrBunny

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Oct 25, 2006
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I have been having trouble with this calculus problem for a couple of hours now, hopefully you can help me...ok here it goes.

-In the sport of ski jumping, the ski jumper glides down a straight path called the inrun that usually makes an angle of about 38 degrees with the horizontal and onto another straight path called the takeoff that usually makes an angle of about 11 degrees with the horizontal. The inrun and the takeoff are usually connected by a curved path called the transition. Find a quadratic function that models the transition assuming the overall path should be differentiable. (Use the following guidelines for your model: Place the start of the takeoff at the origin so that your quadratic function can have the form ax^2 + bx. Assume the inrun ends at a horizontal distance 5 meters from where the takeoff starts. Use the approximations tan(38) = 0.8 and tan(11) = 0.2. Slopes must be equal at the points where the inrun joins the transition and where the transition joins the takeoff. A diagram would be helpful).

PLEASE! HELP ME! The only thing I have been able to do so far is draw a rough sketch of the graph. Also I derived the equation ax^2 + bx to 2ax + b. But after that I am lost again please help me!
 
hint for the ski jump problem

HINT: when a line makes an angle of theta with the horizontal then its slope is given by tan(theta). I know this is a tough problem. I hope that there is someone who is smart enough to help me solve it.
 
Try using the definition of slope to find your parabola.

equation of parabola is given by \(\displaystyle y=ax^{2}+bx\)

\(\displaystyle m=2ax+b\)

If I am reading the problem correctly, you have a slope of -1/5 at x=0 and -4/5 at x=-5

\(\displaystyle 2a(-5)+b=-4/5\\2a(0)+b=-1/5\)
 
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