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The image contains a vector calculus problem, which describes a curve by the vector equation \(\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j}\), where \(t > 0\). There are three parts to the question:

1. **Calculate \(\mathbf{r}'(t)\)**, which is the derivative of the vector equation.
2. **Calculate the gradient of the curve at the point where \(t = 1\)**.
3. **Determine a unit vector \(\hat{s}\) parallel to the tangent to the curve where \(t = 1\)**.

Let’s solve each part step by step:

### 1. Calculate \(\mathbf{r}'(t)\)
The vector equation is:
\[
\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j}
\]

To find \(\mathbf{r}'(t)\), differentiate each component with respect to \(t\):
\[
\mathbf{r}'(t) = \frac{d}{dt}(t\mathbf{i}) + \frac{d}{dt}(t^2\mathbf{j})
\]
\[
\mathbf{r}'(t) = \mathbf{i} + 2t\mathbf{j}
\]

So, \(\mathbf{r}'(t) = \mathbf{i} + 2t\mathbf{j}\).

### 2. Calculate the gradient at \(t = 1\)
The gradient of the curve is given by \(\mathbf{r}'(t)\), so we substitute \(t = 1\) into \(\mathbf{r}'(t)\):
\[
\mathbf{r}'(1) = \mathbf{i} + 2(1)\mathbf{j} = \mathbf{i} + 2\mathbf{j}
\]

Thus, the gradient at \(t = 1\) is \(\mathbf{i} + 2\mathbf{j}\).

### 3. Determine a unit vector \(\hat{s}\) parallel to the tangent at \(t = 1\)
To find a unit vector, we need to normalize \(\mathbf{r}'(1)\). The formula for the magnitude of a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\) is:
\[
|\mathbf{v}| = \sqrt{a^2 + b^2}
\]

For \(\mathbf{r}'(1) = \mathbf{i} + 2\mathbf{j}\), we have:
\[
|\mathbf{r}'(1)| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}
\]

Now, the unit vector \(\hat{s}\) is:
\[
\hat{s} = \frac{\mathbf{r}'(1)}{|\mathbf{r}'(1)|} = \frac{\mathbf{i} + 2\mathbf{j}}{\sqrt{5}} = \frac{1}{\sqrt{5}}\mathbf{i} + \frac{2}{\sqrt{5}}\mathbf{j}
\]

Thus, the unit vector \(\hat{s}\) parallel to the tangent to the curve at \(t = 1\) is:
\[
\hat{s} = \frac{1}{\sqrt{5}}\mathbf{i} + \frac{2}{\sqrt{5}}\mathbf{j}
\]

Would you like further details or clarification on any step? Here are five follow-up questions to consider:

1. What is the geometric interpretation of \(\mathbf{r}'(t)\)?
2. How would you calculate the tangent vector at any arbitrary point \(t = t_0\)?
3. Can you explain the significance of the unit vector \(\hat{s}\)?
4. How does the magnitude of \(\mathbf{r}'(t)\) relate to the speed of motion along the curve?
5. What would change if the curve were described in three dimensions?

**Tip**: The unit vector always has a magnitude of 1, which makes it useful in defining direction without scaling effects.

A curve is described by the vector equation r(t) = t*i + t^2*j, t > 0. Calculate r'(t). Calculate the gradient of the curve at the point where t = 1. Determine a unit vector s parallel to the tangent to the curve where t = 1.

This problem involves finding the derivative of the vector equation r(t) = t*i + t^2*j, calculating the gradient at t = 1, and determining the unit vector parallel to the tangent. Ideal for students studying vector calculus at the university level.

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