Normal plane and tangent line

**Connected** · March 31st 2011, 12:29 PM

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?

**Plato** · March 31st 2011, 12:43 PM

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.

**Connected** · March 31st 2011, 04:13 PM

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

By the way, is my procedure correct?

**Plato** · March 31st 2011, 04:54 PM

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.

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