Let g(x, y) be a function such that partial g partial x (-1,3)=7; partial g partial y (-1,3)=-3. Let overline r (t)= langle t^ 2 -5,3+ln(t-1) rangle Compute d dt (g vec r ) at t = 2 (You need to use one of the special case of the chain rule.)

We are asked to compute \(\frac{d}{dt}\left(g(\mathbf{r}(t))\right)\) at \(t = 2\), where \(\mathbf{r}(t) = \langle t^2 - 5, 3 + \ln(t - 1) \rangle\) and the function \(g(x, y)\) has partial derivatives given at the point \((-1, 3)\).

To solve this, we will use the multivariable chain rule, which for a function \(g(x, y)\) composed with a vector-valued function \(\mathbf{r}(t)\), gives the following formula:

\[
\frac{d}{dt}(g(\mathbf{r}(t))) = \frac{\partial g}{\partial x} \frac{dx}{dt} + \frac{\partial g}{\partial y} \frac{dy}{dt}
\]

Here, \(\mathbf{r}(t)\) is a vector-valued function where:
- \(x(t) = t^2 - 5\)
- \(y(t) = 3 + \ln(t - 1)\)

### Step 1: Compute the derivatives of \(x(t)\) and \(y(t)\)

1. \( \frac{dx}{dt} = \frac{d}{dt}(t^2 - 5) = 2t \)
2. \( \frac{dy}{dt} = \frac{d}{dt}(3 + \ln(t - 1)) = \frac{1}{t - 1} \)

### Step 2: Evaluate at \(t = 2\)

At \(t = 2\):
- \( x(2) = 2^2 - 5 = 4 - 5 = -1 \)
- \( y(2) = 3 + \ln(2 - 1) = 3 + \ln(1) = 3 \)

Thus, \(\mathbf{r}(2) = \langle -1, 3 \rangle\), which matches the point where the partial derivatives of \(g(x, y)\) are given:
- \(\frac{\partial g}{\partial x}(-1, 3) = 7\)
- \(\frac{\partial g}{\partial y}(-1, 3) = -3\)

Also, evaluate the derivatives of \(x(t)\) and \(y(t)\) at \(t = 2\):
- \(\frac{dx}{dt}\bigg|_{t=2} = 2(2) = 4\)
- \(\frac{dy}{dt}\bigg|_{t=2} = \frac{1}{2 - 1} = 1\)

### Step 3: Apply the chain rule

Now, using the chain rule:
\[
\frac{d}{dt}(g(\mathbf{r}(t))) = \frac{\partial g}{\partial x} \frac{dx}{dt} + \frac{\partial g}{\partial y} \frac{dy}{dt}
\]
Substitute the values at \(t = 2\):
\[
\frac{d}{dt}(g(\mathbf{r}(2))) = 7(4) + (-3)(1) = 28 - 3 = 25
\]

### Final Answer:
\[
\frac{d}{dt}(g(\mathbf{r}(t))) \bigg|_{t=2} = 25
\]

Would you like further details or clarification on any step?

Here are 5 related questions:
1. How does the chain rule apply in higher dimensions?
2. What are the conditions for applying the chain rule in multivariable calculus?
3. How do partial derivatives represent changes in multivariable functions?
4. Can you extend this to vector-valued functions for vector calculus?
5. How would the result change if we altered the function \(g(x, y)\)?

**Tip:** When using the chain rule, ensure that all variables and their derivatives are evaluated at the correct points before substitution.

Learn how to compute the derivative of g(r(t)) at t = 2 using the multivariable chain rule. This solution involves calculating partial derivatives and applying the chain rule to vector-valued functions, a key topic in multivariable calculus. Suitable for university-level calculus students.

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