Parametric angle of descent

https://imgur.com/a/kkvXCPq

Could anyone help me with part ci please?
I know I need to measure the angle from the bottom and that the height it reaches is 80m when t is 2 giving x as 80-40 root 2 and use tan but I get 88 point something and the answer is 82.9 3dp. Could someone help me understand where I’ve gone wrong?

Reply 1

mqb2766

Original post by username79352

https://imgur.com/a/kkvXCPq
Could anyone help me with part ci please?
I know I need to measure the angle from the bottom and that the height it reaches is 80m when t is 2 giving x as 80-40 root 2 and use tan but I get 88 point something and the answer is 82.9 3dp. Could someone help me understand where I’ve gone wrong?

Not really sure what youve done but you want something like the gradient/tangent so
dy/dx = tan(phi)
when it returns to the ground at the end of the path.

(edited 1 month ago)

Reply 2

username79352

Original post by mqb2766

Not really sure what youve done but you want something like the gradient/tangent so
dy/dx = tan(phi)
when it returns to the ground at the end of the path.

Could you please explain why I would do that?

This is what I did: https://imgur.com/a/7rYScQf

Reply 3

mqb2766

Original post by username79352

Could you please explain why I would do that?
This is what I did: https://imgur.com/a/7rYScQf

Again, get your pencil out and run it over the curve with your pencil acting as the tangent at a point. They want the angle of descent, so roughly the gradient of the tangent/your pencil, at the end of the curve/path. Its got nothing to do with forming a triangle at the max height.

So the gradient of the tangent at a point is dy/dx at that point and that corresponds to tan() of the angle. You can get dy/dx either eliminating t and forming the quartic y(x) or by doing dy/dx = dy/dt dt/dx. Id probably do the latter.

(edited 1 month ago)

Reply 4

username79352

I got -8 as the gradient at the point the rollercoaster reaches the end so the angle is -82.87…? In degrees

Would this method be the same for these type of questions? To find the gradient via dy by dx and then use tan to find the angle?

Reply 5

mqb2766

Original post by username79352

I got -8 as the gradient at the point the rollercoaster reaches the end so the angle is -82.87…? In degrees
Would this method be the same for these type of questions? To find the gradient via dy by dx and then use tan to find the angle?

I got 82.9 or whatever it was. So arctan(-8) and yes if it asks for something like this, so the angle of approach or descent or ... at a point. If you think about it in tems of finite differences so small changes in cartesian coordinates Dy and Dx, then your heading ... angle corresponds to arctan(Dy / Dx) in the limit so arctan(dy/dx).

Reply 6

username79352

Ok thank you so much for your help! I really appreciate it :smile:

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