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rsst123
 one year ago
*WILL MEDAL*
A curve C has the parametrization x = a sint cosα , y = bsint sinα , z = c cost, t ≥ 0 where a, b, c, α are all positive constants
a)Show that C lies on the ellipsoid x^2/a^2 +y^2/b^2+z^2/c^2=1
(b) Show that C also lies on a plane that contains the zaxis
(c) Describe the curve C. Give its equation
I understand how to get a , by replacing the values for x,y,and z, but I am having trouble how to explain b and c. can someone please help thank you!
rsst123
 one year ago
*WILL MEDAL* A curve C has the parametrization x = a sint cosα , y = bsint sinα , z = c cost, t ≥ 0 where a, b, c, α are all positive constants a)Show that C lies on the ellipsoid x^2/a^2 +y^2/b^2+z^2/c^2=1 (b) Show that C also lies on a plane that contains the zaxis (c) Describe the curve C. Give its equation I understand how to get a , by replacing the values for x,y,and z, but I am having trouble how to explain b and c. can someone please help thank you!

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1this is real ugly but it will help you nail b) as \( \vec r = <a \ sint \ cosα , \ b \ sint \ sinα , c \ cost>\), then the tangent vector \(\vec t\) is \(<a \ cost \ cosα , \ b \ cost \ sinα ,  c \ sint>\) and you can [ugly bit coming up] fix tangent vectors at t = 0 and \(t = \pi /2\) as \(t_1 = <a cos \alpha,b sin \alpha,0>\) and \(t_2 = <0,0,c>\) and then cross product them to give a normal  \(\vec n\)  to the plane of this curve so \(\vec n = \vec t_1 \times \vec t_2 = \) \[\left\begin{matrix}\hat i & \hat j & \hat k\\ a \cos \alpha & b \sin \alpha & 0 \\ 0 & 0 & c\end{matrix}\right\] \( = <cbsin\alpha, \ acos\alpha, \ 0>\) you can already see that \(\hat k\) is in the plane  but dot product \(\vec n\) with \(\hat k\) and you are done there's surely a more elegant way. the gradient \(\nabla\) operator usually knocks these right out of the ground and frankly that would be easiest way. if i had a bit more time i think i'd **unparameterise** this curve, \(\nabla\) it for the normal, and use this info to answer c) . it must just be an ellipse transformed by \(\alpha\) out of the xy plane. maybe you have access to some software that will show that :p hope this helps....:p

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1actually, that answer was not great. mea culpa! if you just do \(\vec r \times \ \vec t\) and you get the normal vector to the curve for all value of the parameter t, not just the 0 and \(\pi/2\) \(\vec r \times \ \vec t\) looks something like \(<cb \ sin\alpha, ac \ cos\alpha, 0>\), ie no \(\hat z\) component meaning that z axis is in the plane itself..... :p
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