Find the vector equation of the tangent line to the vector function r(t)= 2sin(t) i + 2cos(t) j + 4sin^2 (t) k at (1, sqrt(3), 1).

Answer 1

The first step is to find the derivative of r(t):

r'(t) = 2cos(t) i - 2sin(t) j + 8sin(t)cos(t) k

Then evaluate r'(t) at t = π/3 (since the point of interest is when t = π/3):

r'(π/3) = 2cos(π/3) i - 2sin(π/3) j + 8sin(π/3)cos(π/3) k

= i - √3 j + 4k

This gives us the direction of the tangent line at the point (1,√3,1). Now we just need a point on the line to obtain the vector equation.

We can use the point (1,√3,1) as our point on the line. Therefore, the vector equation of the tangent line is:

r(t) = (1,√3,1) + t(i - √3 j + 4k)

Answer 2

wait how did we come with the point of interest is t= pi/3? I thought you would have to solve each component with the points. Like 2sint=1, 2cost=sqrt(3) and 4sin^2 (t) =1. I then solved for t and got pi/6

Answer 3

You are correct. My apologies for the mistake. We need to solve for t in each component to find the point of interest:

2sin(t) = 1 --> t = π/6 or 5π/6
2cos(t) = √3 --> t = π/6 or 11π/6
4sin^2(t) = 1 --> t = π/6 or 5π/6

The only value of t that satisfies all three equations is t = π/6. Therefore, the point of interest is r(π/6) = (1, √3, 1).

We can now find the derivative as before:

r'(t) = 2cos(t) i - 2sin(t) j + 8sin(t)cos(t) k
r'(π/6) = i - √3 j + 4k

And the vector equation of the tangent line is still:

r(t) = (1,√3,1) + t(i - √3 j + 4k)

Answer 4

I got r'(pi/6) = 2cos(pi/6) i - 2sin(pi/6) j + 8 cos(pi/6)sin(pi/6) which would be sqrt(3) i - j + 2sqrt(3) k ?

Answer 5

You are correct! Thank you for catching my mistake. The correct value of the derivative of `r(t)` at `t = π/6` is:

r'(t) = 2cos(t) i - 2sin(t) j + 8sin(t)cos(t) k
r'(π/6) = 2cos(π/6) i - 2sin(π/6) j + 8sin(π/6)cos(π/6) k
r'(π/6) = sqrt(3) i - j + 2sqrt(3) k

Using the point of interest `(1, √3, 1)`, the vector equation of the tangent line is:

r(t) = (1, √3, 1) + t(sqrt(3) i - j + 2sqrt(3) k)