# Thread: Normal plane and tangent line

1. ## Normal plane and tangent line

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

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

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

How to find the tangent line?

2. Originally Posted by Connected
Given $\displaystyle f(t)=(\cos t,\sin t,t)$ then find the tangent line and normal plane if $\displaystyle f\left( \dfrac{\pi }{4} \right)=\left( \dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},\dfrac{\pi }{4} \right).$
You have the point.
Now find the vector $\displaystyle \nabla f\left( {\frac{\pi }{4}} \right)$.
That vector is the normal of the plane and the direction vector of the line.

3. Haven't seen gradient yet on my class, is there another way to do it?

By the way, is my procedure correct?

4. 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.