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Joko123's question at Yahoo! Answers (Sketching a cone)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Here is the question:

Consider the Surface: r = ucos(v)i + usin(v)j + (h*u/r)k 0<u<r and 0<v<2*pi

(should be less then equal to for constraints)

Show that this represents the curved surface of a circular cone height h and base radius r. Sketch this cone?


I have tried multiple times to solve this and cant seem to grasp a solution. PLEASE HELP
Here is a link to the question:

Please Help, How to Sketch a circular cone, height h and base r? - Yahoo!7 Answers

I have posted a link there to this topic so the OP can find my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello Joko123,

Consider the surface: $$S:\left \{ \begin{matrix}x=u\cos v\\y=u\sin v\\z=\dfrac{h}{r}u\end{matrix}\right. \qquad (u\in\mathbb{R},\;v\in\mathbb{R})$$ We have $x^2+y^2=u^2\cos^2v+u^2\sin^2v=u^2(\cos^2v+\sin^2v)=u^2$ and $u=rz/h$, so: $$S:z^2=\frac{h^2}{r^2}(x^2+y^2)$$ and we know that a equation of this form is the equation of an unbounded conical surface. Now, consider the contraints $0\leq u\leq r,\;0\leq v\leq 2\pi$. For $u\in [0,r]$ we have $$\left \{ \begin{matrix}x=u\cos v\\y=u\sin v\\z=\dfrac{h}{r}u\end{matrix}\right. \qquad (v\in [0,2\pi])$$ that is, a circle on the plane $z=hu/r$.

If $u=0$, we get the point $(0,0,0)$ (circle with radius $0$).

If $u=r$, we get $$\left \{ \begin{matrix}x=r\cos v\\y=r\sin v\\z=h\end{matrix}\right. \qquad (v\in [0,2\pi])$$ that is, a circle on the plane $z=h$ with center at $(0,0,h)$ and radius $r$. Now, is easy to sketch the cone (choose only the part above):

DoubleCone.png