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).
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
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.
To find the vector equation of the tangent line to the curve at a specific point, we need to determine two things: the position vector of the point on the curve and the direction vector of the tangent line at that point.
First, let's find the position vector of the point on the curve. We are given the values of t at which the curve passes through the point (1, √3, 1). So we need to find the value of t that satisfies the equation of the curve for this point.
The equation of the curve is:
r(t) = 2sin(t)i + 2cos(t)j + 4sin^2(t)k
To find the value of t, we can set the x-coordinate, y-coordinate, and z-coordinate equal to their respective values at the given point:
2sin(t) = 1
2cos(t) = √3
4sin^2(t) = 1
From the first equation, we can solve for sin(t):
sin(t) = 1/2
From the second equation, we can solve for cos(t):
cos(t) = √3/2
Now, we need to determine the value of t that satisfies these two conditions. We can use the trigonometric identity sin^2(t) + cos^2(t) = 1 and substitute the values of sin(t) and cos(t) into it:
(1/2)^2 + (√3/2)^2 = 1
1/4 + 3/4 = 1
4/4 = 1
This confirms that the values of sin(t) and cos(t) satisfy the equation, and therefore t = π/3.
Now, let's find the direction vector of the tangent line at this point. To find the direction vector, we need to differentiate the position vector with respect to t.
dr(t)/dt = d/dt (2sin(t)i + 2cos(t)j + 4sin^2(t)k)
= 2cos(t)i - 2sin(t)j + 8sin(t)cos(t)k
Substituting t = π/3, we get:
dr(π/3)/dt = 2cos(π/3)i - 2sin(π/3)j + 8sin(π/3)cos(π/3)k
= 2(1/2)i - 2(√3/2)j + 8(1/2)(√3/2)k
= i - √3j + 2√3k
The obtained vector is the direction vector of the tangent line at the point (1, √3, 1).
Finally, we can write the vector equation of the tangent line:
r(t) = (1, √3, 1) + t(i - √3j + 2√3k)
Therefore, the vector equation of the tangent line to the curve r(t) = 2sin(t)i + 2cos(t)j + 4sin^2(t)k at the point (1, √3, 1) is given by r(t) = (1, √3, 1) + t(i - √3j + 2√3k), where t is a parameter.