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We are given the following conditions:

1. \( f(x, y) \) is a differentiable function with two gradient values:
   - \( \nabla f(0, 1) = 2\mathbf{i} - \mathbf{j} \)
   - \( \nabla f(1, 0) = \mathbf{i} + 3\mathbf{j} \)
   
2. \( \mathbf{r}(t) = (e^t - 1)\mathbf{i} + (e^t - t)\mathbf{j} \)
   
3. \( g(t) = f(\mathbf{r}(t)) \), and we are tasked with finding \( g'(0) \).

### Solution steps:

We begin by applying the chain rule to find \( g'(t) \):
\[
g'(t) = \frac{d}{dt} f(\mathbf{r}(t)) = \nabla f \cdot \frac{d\mathbf{r}(t)}{dt}
\]
This means that \( g'(t) \) is the dot product of the gradient of \( f \) with the derivative of \( \mathbf{r}(t) \) with respect to \( t \).

First, compute the derivative of \( \mathbf{r}(t) \):
\[
\mathbf{r}(t) = (e^t - 1)\mathbf{i} + (e^t - t)\mathbf{j}
\]
Differentiate this with respect to \( t \):
\[
\frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt} \left( (e^t - 1)\mathbf{i} + (e^t - t)\mathbf{j} \right) = e^t \mathbf{i} + (e^t - 1) \mathbf{j}
\]

Now, evaluate this at \( t = 0 \):
\[
\frac{d\mathbf{r}(t)}{dt}\bigg|_{t=0} = e^0 \mathbf{i} + (e^0 - 1) \mathbf{j} = \mathbf{i}
\]

Next, we need to determine \( \nabla f \) at the point \( \mathbf{r}(0) \). Evaluating \( \mathbf{r}(0) \):
\[
\mathbf{r}(0) = (e^0 - 1)\mathbf{i} + (e^0 - 0)\mathbf{j} = 0\mathbf{i} + 1\mathbf{j} = \mathbf{j}
\]
So, \( \mathbf{r}(0) = (0, 1) \), and we are given that:
\[
\nabla f(0, 1) = 2\mathbf{i} - \mathbf{j}
\]

Now, use the chain rule to compute \( g'(0) \):
\[
g'(0) = \nabla f(0, 1) \cdot \frac{d\mathbf{r}(t)}{dt}\bigg|_{t=0} = (2\mathbf{i} - \mathbf{j}) \cdot \mathbf{i}
\]
Since \( \mathbf{i} \cdot \mathbf{i} = 1 \) and \( \mathbf{j} \cdot \mathbf{i} = 0 \), we get:
\[
g'(0) = 2(1) + (-1)(0) = 2
\]

Thus, the answer is \( \boxed{2} \).

Would you like further details or have any questions?

Here are five related questions to think about:
1. How does the chain rule apply to vector-valued functions?
2. What are the key properties of the dot product used here?
3. How would the solution change if \( \mathbf{r}(t) \) were different?
4. How does the gradient vector relate to directional derivatives?
5. How do we interpret the values of \( \nabla f \) geometrically?

**Tip:** When solving problems involving gradients and vector functions, always check the points where you evaluate the function carefully to avoid errors!

Suppose f(x,y) is a differentiable function such that ∇f(0,1)=2i−j and ∇f(1,0)=i+3j. Let r(t)=(e^t−1)i+(e^t−t)j and define g(t)=f(r(t)). Find g′(0).

Learn how to apply the chain rule to compute g′(0) where g(t)=f(r(t)) using vector calculus. This problem involves gradients, differentiable functions, and vector dot products. Suitable for undergraduate-level calculus students.

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