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 z-axis (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!

1. IrishBoy123

this 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

2. IrishBoy123

actually, 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