Question

Find the derivative, r'(t), of the vector function. r(t) = t sin(8t), t², t cos(9t).

A) r'(t) = (sin(8t) + 8t cos(8t))i + 2tj - (9t² sin(9t))k
B) r'(t) = (sin(8t) + 8t cos(8t))i + 2tj + (9t² cos(9t))k
C) r'(t) = (cos(8t) + 8t sin(8t))i + 2tj - (9t² sin(9t))k
D) r'(t) = (cos(8t) + 8t sin(8t))i + 2tj + (9t² cos(9t))k

Answer 1

The derivative of the vector function r(t) = t sin(8t), t², t cos(9t) is (sin(8t) + t(8cos(8t)))i + 2tj + (cos(9t) + t(−9sin(9t)))k.

Step-by-step explanation:

To find the derivative of a vector function, we need to take the derivative of each component separately. Let's find the derivatives of each component of the given vector function:

r(t) = t sin(8t), t², t cos(9t)

First component:

r'(t) = (d/dt)(t sin(8t)) = sin(8t) + t(8cos(8t))

Second component:

r'(t) = (d/dt)(t²) = 2t

Third component:

r'(t) = (d/dt)(t cos(9t)) = cos(9t) + t(−9sin(9t))

Therefore, the derivative of the vector function r(t) = t sin(8t), t², t cos(9t) is:
r'(t) = (sin(8t) + t(8cos(8t)))i + 2tj + (cos(9t) + t(−9sin(9t)))k