Find a equation for the line tangent to the curve

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SigepBrandon

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Hey everyone, this is my first post and I'm having trouble with determining an equation of a tangent line to a curve. Here is the exercise-

Determine an equation for the line tangent to r(t)=<t sin(t), t cos(t), t> at the point (pi/2, 0, pi/2) on the curve
Click to expand...

First I took the derivative
r'(t)= <sin(t) + t*cos(t), cos(t) - t*sin(t), 0>

Then evaluated r'(t) at the point (pi/2, 0, pi/2)
r'(pi/2,0,pi/2)= <sin(pi/2) + pi/2*cos(pi/2), cos(0)-0*sin(0), 0> => <1+0, 1-0, 0>

Which leaves me with an answer of <1, 0, 0>

Is this correct or where did I go wrong?
 
Thanks Dr. Steve.
So my answer should be <1,0,1> This does not seem like an equation of a tangent line though... I'm familiar with y-y[sub:4burcfba]0[/sub:4burcfba]=m((x-x[sub:4burcfba]0[/sub:4burcfba])+b) but not sure how that relates in three space, or if it's even needed...
 
Hello, SigepBrandon!

Welcome aboard!

Edit: I overlooked another error you made.

\(\displaystyle \text{Determine an equation for the line tangent to: }\:r(t) \:=\:\bigg\langle t\sin t,\;t\cos t,\; t \bigg\rangle\,\text{ at the point }\left(\tfrac{\pi}{2},\;0,\;\tfrac{\pi}{2}\right)\,\text{ on the curve.}\)

\(\displaystyle \text{First I took the derivative: }\;r'(t) \:=\: \bigg\langle \sin t + t\cos t,\;\cos t - t\sin t,\; 0\bigg\rangle\) . The derivative of t is 1.

\(\displaystyle \text{Then evaluated }r'(t)\text{ at the point }\left(\tfrac{\pi}{2},\:0,\:\tfrac{\pi}{2}\right)\)

. . \(\displaystyle \underbrace{r'\left(\tfrac{\pi}{2},\:0,\:\tfrac{\pi}{2}\right)}_{??} \:=\: \bigg\langle \sin\tfrac{\pi}{2} + \tfrac{\pi}{2}\cos\tfrac{\pi}{2},\;\cos 0-0\sin0,\; 0\bigg\rangle\)

. . This is all wrong . . .
Click to expand...

You have \(\displaystyle r'\) as function of three variables . . .


\(\displaystyle \text{At the point }\left(\tfrac{\pi}{2},\:0,\:\tfrac{\pi}{2}\right),\,\text{ we have: }\,t = \tfrac{\pi}{2}\)

. . \(\displaystyle \text{and }that\text{ what we substitute into }r'(t).\)


\(\displaystyle r'\left(\tfrac{\pi}{2}\right) \;=\;\bigg\langle \sin\tfrac{\pi}{2} + \tfrac{\pi}{2}\cos\tfrac{\pi}{2},\;\cos\tfrac{\pi}{2} - \tfrac{\pi}{2}\sin\tfrac{\pi}{2},\;1\bigg\rangle \;=\;\bigg\langle 1,\;\text{-}\tfrac{\pi}{2},\;1\bigg\rangle\)



It seems that you are not familiar with lines in three dimensions.
I recommend that you remedy that.

 
Thanks for the help Soroban!

I did think that r'(t) was a function of 3 variables. To write it parametrically:

x'(t) = sin(t) + t*cos(t)
y'(t) = cos(t) - t*sin(t)
z'(t) = 1

and the above evaluated at <x,y,z>=<pi/2, 0, pi/2> in order to find the slope of the line in each direction and utilize that to find an equation of the tangent line at (pi/2, 0, pi/2)

If i may ask- how did you find t to be equal to pi/2?
 
Ok. So I found the equation of a tangent line at r(t[sub:sr2bsg9y]o[/sub:sr2bsg9y]): L(t)=r(t[sub:sr2bsg9y]0[/sub:sr2bsg9y]) + tr'(t[sub:sr2bsg9y]0[/sub:sr2bsg9y]

since the problem lists (pi/2,0,pi/2) as a point on the curve, that is my r(t[sub:sr2bsg9y]0[/sub:sr2bsg9y]) and r'(t[sub:sr2bsg9y]0[/sub:sr2bsg9y]) is evaluated at pi/2;

L(t)=(pi/2, 0, pi/2)+t<1,-(pi/2),1> giving me a equation of a tangent line at the point.

If this is correct, how does someone know that pi/2 is supposed to be plugged into r'(t)?
 
It says so in the original question. If you look at the third component of the vector, it reads \(\displaystyle t=\frac{\pi}{2}\).
 
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