Use the chain rule to find dz dt . z = x2 + y2 + xy, x = sin(t), y = 3et dz dt

To find $\frac{dz}{dt}$ using the chain rule, we need to differentiate $z = x^2 + y^2 + xy$ with respect to $t$, using the given functions $x(t) = \sin(t)$ and $y(t) = 3e^t$.

Step-by-step solution:

1. Differentiate $z$ with respect to $x$ and $y$: $\frac{\partial z}{\partial x} = 2x + y$ $\frac{\partial z}{\partial y} = 2y + x$

2. Use the chain rule: $\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}$

3. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

   • $x(t) = \sin(t)$, so $\frac{dx}{dt} = \cos(t)$
   • $y(t) = 3e^t$, so $\frac{dy}{dt} = 3e^t$

4. Substitute the values: $\frac{dz}{dt} = (2x + y)\cos(t) + (2y + x)3e^t$

5. Now substitute $x = \sin(t)$ and $y = 3e^t$: $\frac{dz}{dt} = (2\sin(t) + 3e^t)\cos(t) + (2(3e^t) + \sin(t))3e^t$

6. Simplify the expression: $\frac{dz}{dt} = (2\sin(t) + 3e^t)\cos(t) + (6e^t + \sin(t))3e^t$ $\frac{dz}{dt} = 2\sin(t)\cos(t) + 3e^t\cos(t) + 18e^{2t} + 3e^t\sin(t)$

Thus, the final expression for $\frac{dz}{dt}$ is: $\boxed{\frac{dz}{dt} = 2\sin(t)\cos(t) + 3e^t\cos(t) + 18e^{2t} + 3e^t\sin(t)}$

Let me know if you need further explanation or have any questions about the process!

Follow-up Questions:

1. How do we apply the product rule in this context for mixed terms like $xy$?
2. Can you provide an example where the chain rule is applied in a multivariable context?
3. How would the chain rule work in cases where $z$ is a function of more than two variables?
4. How can we simplify expressions like $2\sin(t)\cos(t)$ further using trigonometric identities?
5. What happens if $x(t)$ and $y(t)$ are more complex functions, say involving powers or more trigonometric functions?

Tip:

When applying the chain rule, always remember to differentiate each variable in terms of $t$, and then combine the results carefully.