To find the tangent vector, we need to take the derivative of r(t):
r'(t) = 2cos(t) i - 2sin(t) j + 8sin(t)cos(t) k
At the point (1, sqrt(3), 1), we have:
r'(π/3) = 2cos(π/3) i - 2sin(π/3) j + 8sin(π/3)cos(π/3) k = i - 2sqrt(3) j + 4k
So the tangent line at the point (1, sqrt(3), 1) is given by:
r(t) = <1, sqrt(3), 1> + t<i, -2sqrt(3), 4>
or in component form:
x = 1 + t
y = sqrt(3) - 2sqrt(3)t
z = 1 + 4t
find the vector equation of the tangent line to r(t) = 2sin(t) i + 2cos(t) j +4 sin^(2) t k at ( 1, sqrt(3) , 1).
3 answers
Where did you get π/3 from?
We need to find the value of `t` when `r(t)` passes through the point `(1, sqrt(3), 1)`.
Let's first write out the coordinates of `r(t)`:
```
r(t) = 2sin(t) i + 2cos(t) j + 4sin^2(t) k
```
For `r(t)` to pass through the point `(1, sqrt(3), 1)`, we must have:
```
2sin(t) = 1
2cos(t) = sqrt(3)
4sin^2(t) = 1
```
Solving for `t`, we get `t = π/3` (since `sin(π/3) = 1/2` and `cos(π/3) = sqrt(3)/2`).
Therefore, we use `t = π/3` to get the tangent vector and equation.
Let's first write out the coordinates of `r(t)`:
```
r(t) = 2sin(t) i + 2cos(t) j + 4sin^2(t) k
```
For `r(t)` to pass through the point `(1, sqrt(3), 1)`, we must have:
```
2sin(t) = 1
2cos(t) = sqrt(3)
4sin^2(t) = 1
```
Solving for `t`, we get `t = π/3` (since `sin(π/3) = 1/2` and `cos(π/3) = sqrt(3)/2`).
Therefore, we use `t = π/3` to get the tangent vector and equation.