Question

Find the derivative, r'(t), of the vector function r(t) = t sin(3t), t², t cos(5t)?

Answer 1

The derivative
`\(r'(t)\)` of the vector function
`\(r(t) = t \sin(3t), t^2, t \cos(5t)\)` is given by
`\[r'(t) = (\sin(3t) + 3t\cos(3t))\mathbf{i} + 2t\mathbf{j} + (\cos(5t) - 5t\sin(5t))\mathbf{k}.\]`

Step-by-step explanation:

To find the derivative
`\(r'(t)\)` of the vector function
`\(r(t) = t \sin(3t), t^2, t \cos(5t)\)`, we differentiate each component of the vector separately with respect to (t). Let
`\(\mathbf{i}, \mathbf{j},\)`and
`\(\mathbf{k}\)` represent the unit vectors in the (x), (y), and (z) directions, respectively.

The first component,
`\(t \sin(3t)\)`, is differentiated using the product rule and the chain rule, resulting in
`\(\sin(3t) + 3t\cos(3t)\)` for the (x)-component. The second component,
`\(t^2\)`, is simply differentiated to (2t) for the (y)-component. The third component,
`\(t \cos(5t)\)`, is differentiated using the product and chain rules, giving
`\(\cos(5t) - 5t\sin(5t)\)` for the (z)-component.

Therefore, the derivative
`\(r'(t)\)` of the vector function
`\(r(t)\)` is expressed as
`\[(\sin(3t) + 3t\cos(3t))\mathbf{i} + 2t\mathbf{j} + (\cos(5t) - 5t\sin(5t))\mathbf{k}.\]` This result represents the rate of change of each component with respect to (t) and provides the velocity vector for the given position vector
`\(r(t)\)`.

Answer 2

To find the derivative of the vector function r(t), we need to find the derivative of each component separately using the product rule and power rule. The first component is t sin(3t), the second component is t², and the third component is t cos(5t).

Step-by-step explanation:

To find the derivative of the vector function r(t) = t sin(3t), t², t cos(5t), we need to find the derivative of each component separately. Let's start with the first component, t sin(3t):

1. Apply the product rule: (f.g)' = f'g + fg'
2. Derivative of t is 1.
3. Derivative of sin(3t) is 3cos(3t) using the chain rule.
4. Combine the derivatives: (t sin(3t))' = 1*sin(3t) + t*3cos(3t) = sin(3t) + 3tcos(3t).

Now let's find the derivative of the second component, t², which is a simple power rule:

1. Derivative of t² is 2t.

Finally, let's find the derivative of the third component, t cos(5t):

1. Again, apply the product rule: (f.g)' = f'g + fg'.
2. Derivative of t is 1.
3. Derivative of cos(5t) is -5sin(5t) using the chain rule.
4. Combine the derivatives: (t cos(5t))' = 1*cos(5t) + t*(-5sin(5t)) = cos(5t) - 5tsin(5t).

Let's summarize the derivatives of each component:

• The derivative of t sin(3t) is sin(3t) + 3tcos(3t).
• The derivative of t² is 2t.
• The derivative of t cos(5t) is cos(5t) - 5tsin(5t).