The correct question is:
(a) Find the derivative of r(t) = (2 + t³)i + te^(−t)j + sin(6t)k.
(b) Find the unit tangent vector at the point t = 0.
Answer:
The derivative of r(t) is 3t²i + (1 - t)e^(-t)j + 6cos(6t)k
(b) The unit tangent vector is (j/2 + 3k)
Step-by-step explanation:
Given
r(t) = (2 + t³)i + te^(−t)j + sin(6t)k.
(a) To find the derivative of r(t), we differentiate r(t) with respect to t.
So, the derivative
r'(t) = 3t²i +[e^(-t) - te^(-t)]j + 6cos(6t)k
= 3t²i + (1 - t)e^(-t)j + 6cos(6t)k
(b) The unit tangent vector is obtained using the formula r'(0)/|r(0)|. r(0) is the value of r'(t) at t = 0, and |r(0)| is the modulus of r(0).
Now,
r'(0) = 3t²i + (1 - t)e^(-t)j + 6cos(6t)k; at t = 0
= 3(0)²i + (1 - 0)e^(0)j + 6cos(0)k
= j + 6k (Because cos(0) = 1)
r'(0) = j + 6k
r(0) = (2 + t³)i + te^(−t)j + sin(6t)k; at t = 0
= (2 + 0³)i + (0)e^(0)j + sin(0)k
= 2i (Because sin(0) = 0)
r(0) = 2i
Note: Suppose A = xi +yj +zk
|A| = √(x² + y² + z²).
So |r(0)| = √(2²) = 2
And finally, we can obtain the unit tangent vector
r'(0)/|r(0)| = (j + 6k)/2
= j/2 + 3k
