# Normal plane and tangent line

• Mar 31st 2011, 01:29 PM
Connected
Normal plane and tangent line
Given $f(t)=(\cos t,\sin t,t)$ then find the tangent line and normal plane if $f\left( \dfrac{\pi }{4} \right)=\left( \dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},\dfrac{\pi }{4} \right).$

Let $x(t)=\cos t,y(t)=\sin t,z(t)=t$ so normal plane is given by

$\left\langle \left( x'\left( {{t}_{0}} \right),y'\left( {{t}_{0}} \right),z'\left( {{t}_{0}} \right) \right),\left( x-x\left( {{t}_{0}} \right),y-y\left( {{t}_{0}} \right),z-z\left( {{t}_{0}} \right) \right) \right\rangle =0.$ Is that correct?

How to find the tangent line?
• Mar 31st 2011, 01:43 PM
Plato
Quote:

Originally Posted by Connected
Given $f(t)=(\cos t,\sin t,t)$ then find the tangent line and normal plane if $f\left( \dfrac{\pi }{4} \right)=\left( \dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},\dfrac{\pi }{4} \right).$

You have the point.
Now find the vector $\nabla f\left( {\frac{\pi }{4}} \right)$.
That vector is the normal of the plane and the direction vector of the line.
• Mar 31st 2011, 05:13 PM
Connected
Haven't seen gradient yet on my class, is there another way to do it?

By the way, is my procedure correct?
• Mar 31st 2011, 05:54 PM
Plato
Quote:

Originally Posted by Connected
Haven't seen gradient yet on my class, is there another way to do it?

Because this in a truly international forum, I have no idea where you are located. Nonetheless, you are being cheated if the above quote is in fact true. If I were you then I would complain to the educational authority. There is absolutely no reason to ask questions that are not based in standard practices.