If 
u(t) = 

sin(5t), cos(6t), t

  and  
v(t) = 

t, cos(6t), sin(5t)

,
 use the theorem given below to find 
d
dt
 

u(t) · v(t)

.
Suppose u and v are differentiable vector functions. Then 
d
dt
 

u(t) · v(t)

 = u ′(t) · v(t) + u(t) · v ′(t).
don't explain, just give me the answer

The derivative of \( u(t) \cdot v(t) \) is:

\[
10t \cos(5t) + 2\sin(5t) - 6\sin(12t)
\]

Do you want more details or have any questions?

Here are 5 related questions:
1. How do you compute the dot product of two vector functions?
2. What is the significance of the product rule in vector calculus?
3. How do you interpret the result geometrically when dealing with vector functions?
4. Can you calculate the second derivative of \( u(t) \cdot v(t) \)?
5. How do vector functions apply to physical systems, like motion or forces?

**Tip:** When differentiating vector functions, apply the product rule to each component individually!

If u(t) = sin(5t), cos(6t), t and v(t) = t, cos(6t), sin(5t), use the theorem given below to find d/dt [u(t) · v(t)]. Suppose u and v are differentiable vector functions. Then d/dt [u(t) · v(t)] = u′(t) · v(t) + u(t) · v′(t).

Learn how to differentiate the dot product of two vector functions u(t) and v(t) using the product rule for vector calculus. This detailed solution covers essential concepts like the dot product, differentiation of vector functions, and the product rule theorem. Suitable for college-level students studying vector calculus.

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