Find a unit tangent vector to r(t)=e^(2t)i+e^(-1)j+(t^2+4)k at (1,1,4).
To get the point (1,1,4) it appears there's an error. t=0 produces (1,1/e,4) as written. Is the coefficient of j e^(-t)?
oh yes it is.
r(t)=e^(2t)i+e^(-t)j+(t^2+4)k at (1,1,4)
Cool. Now just differentiate term by term:
r'(t) = 2e^(2t)i - e^(-t)j + 2tk
r'(1) = 2e²i - 1/ej + 2k
∥r'∥ = √(4e^4 + 1 + 4)
So, the unit vector at (1,1,4) = 1/√(4e^4 + 1 + 4) 2e²i - 1/ej + 2k
Rats. t=0
r'(0) = (2e²,-1,4)
you take it from there.
RATS!
r'(0) = (2,-1,0)
unit vector = 1/√5 (2,-1,0)
how does 1/e equal to 1? ∥r'∥ = √(4e^4 + 1 + 4)
oh ok, now i get it.
To find a unit tangent vector to the given vector function at a specific point, you can follow these steps:
Step 1: Evaluate the derivative of the vector function.
- Differentiate each component of the vector function with respect to t.
- For the given vector function r(t) = e^(2t)i + e^(-1)j + (t^2 + 4)k, its derivative is r'(t) = 2e^(2t)i - e^(-1)j + 2t*k.
Step 2: Substitute the value of t for the given point.
- To find the unit tangent vector at the point (1, 1, 4), substitute t = 1 into the derivative.
- r'(1) = 2e^(2*1)i - e^(-1)j + 2(1)k = 2e^2i - e^(-1)j + 2k.
Step 3: Calculate the magnitude of the tangent vector.
- The magnitude (length) of a vector is found using the formula: |r'(t)| =sqrt( (component1)^2 + (component2)^2 + (component3)^2 )
- So, |r'(1)| = sqrt((2e^2)^2 + (-e^(-1))^2 + 2^2) = sqrt(4e^4 + e^(-2) + 4).
Step 4: Divide the tangent vector by its magnitude to get the unit tangent vector.
- The unit tangent vector, T(t), is given by: T(t) = r'(t) / |r'(t)|.
- In this case, the unit tangent vector at (1, 1, 4) is: T(1) = (2e^2i - e^(-1)j + 2k) / sqrt(4e^4 + e^(-2) + 4).
Therefore, the unit tangent vector to r(t) = e^(2t)i + e^(-1)j + (t^2 + 4)k at the point (1, 1, 4) is (2e^2i - e^(-1)j + 2k) / sqrt(4e^4 + e^(-2) + 4).