Final answer:
The derivative r'(t), which represents the velocity, is found by taking the time derivative of each component of the position function r(t) = ti + 5tj + t²k, resulting in r'(t) = i + 5j + 2tk.
Step-by-step explanation:
To find the derivative r'(t), the position function r(t) = ti + 5tj + t²k is given. The derivative of the position function with respect to time, t, will give us the velocity function.
The position function r(t) can be written as:
- x(t) = t
- y(t) = 5t
- z(t) = t²
Now, taking the derivative of each component with respect to time, we have:
- x'(t) = dx(t)/dt = 1
- y'(t) = dy(t)/dt = 5
- z'(t) = dz(t)/dt = 2t
Thus, the velocity function r'(t) will be:
r'(t) = x'(t)i + y'(t)j + z'(t)k = i + 5j + 2tk.
This is the functional form of the velocity for the particle at any moment t.