Question

I have a question about parametrizing a tangent line.

So I found that the parametrized tangent line has a vector function of R(u)=(2i+5k)+u(8i-j+8k)

The book has the vector R(u)=(2i+5k)+u(4i-j+4k). Is this essentially the same answer except with a 2 factored out? Would it still be technically correct?

Answer 1

Lets say the first answer you got is denoted \(a_1\) and the answer in the back of the book we denote as \(a_2\).
We can see the difference by calculating:
\[a_2-a_1\]
If they are the same answer, then \(a_1=a_2\) and therefore,
\[a_2-a_1=a_2-a_2=a_1-a_1=0\]
Basically, subtract both functions from each other. If the final result is zero, then they are the same.

Answer 2

Yes if the original vector by the scalar u was <8,-2,8> instead of <8,-1,8>. But I think that's probably just a typo. Since u is a scalar (a constant) it can absorb the 2 they factored out from the vector. This happens a lot with things like constants of integration.

Answer 3

Ok, I kind of figured it was a typo. My book is full of them.

Just to be sure, the original equation is: r(t)=2t^2i+(1-t)j+(3+2t^2)k at P(2,0,5)

So r'(t)=4ti-j+4tk
r'(2)=8i-j+8k
R(u)=(2i+5k)+u(8i-j+8k)

Answer 4

Never mind. I figured out the problem.

Instead of plugging in the x value for t, you must first solve for t by setting each part of the equation to x, y, and z.

2=2t^2
0=1-t
5=3+2t^2

t=1.

Therefore, R(u)=(2i+5k)+u(4i-j+4k).