Question

Question: Find the equation of the tangent plane to the given parametric surface at the specified point. If you have software that graphs parametric surfaces, use a computer to graph the surface and the tangent plane

\(r\left( {u,v} \right) = u\cos v\,\,i + u\sin v\,\,j + v\,\,k;u = 1,v = \frac{\pi }{3}\)

Short answer

The equation of the tangent plane is \(\frac{{\sqrt 3 }}{2}x - \frac{1}{2}y + z = \frac{\pi }{3}\)

Step-by-step solution

Solution for x, y and z

\(x,y,z\)points are done by given\(r\left( {u,v} \right)\)

\(\begin{aligned}x& = u\cos v\\y = u\sin v\\z& = v\end{aligned}\)

\(\begin{aligned}x& = (1)\cos \left( {\frac{\pi }{3}} \right)\\x& = \frac{1}{2}\end{aligned}\)

\(\begin{aligned}y& = (1)\sin \left( {\frac{\pi }{3}} \right)\\y& = \frac{{\sqrt 3 }}{2}\end{aligned}\)

\(z = \left( {\frac{\pi }{3}} \right)\)

Solution of the tangent vector

Let, consider the tangent vector\({r_u}\)and\({r_v}\)

\(\begin{aligned}{r_u}& = \frac{{\partial x}}{{\partial u}}i + \frac{{\partial y}}{{\partial u}}j + \frac{{\partial z}}{{\partial u}}k\\& = \cos vi + \sin vj + (0)k\\{r_v}& = \frac{{\partial x}}{{\partial v}}i + \frac{{\partial y}}{{\partial v}}j + \frac{{\partial z}}{{\partial v}}k\\ = - u\sin vi + u\cos vj + k\end{aligned}\)

Solution for the tangent plane

The tangent pane is given by\({r_u} \times {r_v}\)

\(\begin{aligned}{r_u} \times {r_v}& = \left| {\begin{aligned}{*{20}{c}}i&j&k\\{\cos v}&{\sin v}&0\\{ - u\sin v}&{u\cos v}&1\end{aligned}} \right|\\& = (\sin v - 0)i - (\cos v - 0)j + (u{\cos ^2}v + u{\sin ^2}v)k\\& = \sin vi - \cos vj + uk\end{aligned}\)

Substitute u as 1 and v as\(\frac{\pi }{3}\)

So, the tangent plane

\(\begin{aligned}\sin vi - \cos vj + uk& = \sin \left( {\frac{\pi }{3}} \right)i - \cos \left( {\frac{\pi }{3}} \right)j + (1)k\\& = \frac{{\sqrt 3 }}{2}i - \frac{1}{2}j + k\end{aligned}\)

The equation of the tangent plane 

Let’s substitute points\(\left( {\frac{1}{2},\frac{{\sqrt 3 }}{2},\frac{\pi }{3}} \right)\)in tangent plane

\(\begin{aligned}\frac{{\sqrt 3 }}{2}\left( {x - \frac{1}{2}} \right) - \frac{1}{2}\left( {y - \frac{{\sqrt 3 }}{2}} \right) + \left( {z - \frac{\pi }{3}} \right)& = 0\\\frac{{\sqrt 3 }}{2}x - \frac{{\sqrt 3 }}{4} - \frac{1}{2}y + \frac{{\sqrt 3 }}{4} + z - \frac{\pi }{3}& = 0\\\frac{{\sqrt 3 }}{2}x - \frac{1}{2}y + z& = \frac{\pi }{3}\end{aligned}\)

The equation of the tangent plane is \(\frac{{\sqrt 3 }}{2}x - \frac{1}{2}y + z = \frac{\pi }{3}\).