To find a tangent vector at a given point specified by the parameter \( t \), we start by differentiating the vector function \( \mathbf{r}(t) \) with respect to \( t \).
Given the vector function:
\( \mathbf{r}(t) = 7t^2 \mathbf{i} + t^2 \mathbf{j} + t \mathbf{k} \)
We differentiate it:
\( \mathbf{r}'(t) = \frac{d}{dt}(7t^2) \mathbf{i} + \frac{d}{dt}(t^2) \mathbf{j} + \frac{d}{dt}(t) \mathbf{k} \)
This results in:
\( \mathbf{r}'(t) = 14t \mathbf{i} + 2t \mathbf{j} + 1 \mathbf{k} \)
Now, to evaluate the tangent vector at a specific value of \( t \), we substitute that value into \( \mathbf{r}'(t) \). Let’s say we want to find the tangent vector at \( t = a \) (you can replace \( a \) with your actual value).
For \( t = a \):
\( \mathbf{r}'(a) = 14a \mathbf{i} + 2a \mathbf{j} + 1 \mathbf{k} \)
Next, we need to convert this tangent vector into a unit vector. The length of the vector \( \\mathbf{r}'(a)\) can be calculated using the formula:
\( || \mathbf{r}'(a) || = \sqrt{(14a)^2 + (2a)^2 + (1)^2} = \sqrt{196a^2 + 4a^2 + 1} = \sqrt{200a^2 + 1} \)
Now we divide each component of \( \mathbf{r}'(a) \) by its length to get the unit tangent vector:
\( \mathbf{T}(a) = \frac{1}{|| \mathbf{r}'(a) ||} \mathbf{r}'(a) = \frac{14a \mathbf{i} + 2a \mathbf{j} + 1 \mathbf{k}}{\sqrt{200a^2 + 1}} \)
This unit tangent vector will represent the direction of the curve at the point specified by the parameter \( t = a \).