To differentiate the given function with respect to $t$, let's start by breaking down the equation:

$y = a \cos(t) + t^2 \sin(t)$

Step-by-step differentiation:

1. Differentiate $a \cos(t)$ with respect to $t$: Since $a$ is a constant, we use the derivative of $\cos(t)$, which is $-\sin(t)$: $\frac{d}{dt} \left( a \cos(t) \right) = -a \sin(t)$

2. Differentiate $t^2 \sin(t)$ with respect to $t$: This is a product of two functions, $t^2$ and $\sin(t)$. We will apply the product rule: $\frac{d}{dt} \left( t^2 \sin(t) \right) = \frac{d}{dt}(t^2) \cdot \sin(t) + t^2 \cdot \frac{d}{dt}(\sin(t))$

   • The derivative of $t^2$ is $2t$.
   • The derivative of $\sin(t)$ is $\cos(t)$. So: $\frac{d}{dt} \left( t^2 \sin(t) \right) = 2t \sin(t) + t^2 \cos(t)$

Putting it all together:

$\frac{dy}{dt} = -a \sin(t) + \left( 2t \sin(t) + t^2 \cos(t) \right)$

Thus, the derivative of $y$ with respect to $t$ is: $\boxed{\frac{dy}{dt} = -a \sin(t) + 2t \sin(t) + t^2 \cos(t)}$

Would you like a more detailed explanation of any part of the process?

Relative Questions:

1. What is the product rule in differentiation and when is it used?
2. Can you explain why the derivative of $\cos(t)$ is $-\sin(t)$?
3. How would the answer change if $a$ were a function of $t$ instead of a constant?
4. What happens if we differentiate $t^2 \sin(t)$ multiple times?
5. How do you apply the chain rule if there is a composite function like $\cos(2t)$ in place of $\cos(t)$?

Tip:

Always remember to apply the product rule when differentiating products of functions like $t^2 \sin(t)$, and ensure constants are handled properly in differentiation.