Question

Find a unit vector that is parallel to the line tangent to the parabola y = x2 at the point (4, 16).

Answer 1

The slope of the line tangent to the graph of $y=x^2$, at the point $(4,16)$ is given by:

As we know that the slope of the tangent is obtained by differentiating the function at the desired point:

So the slope is given by:

$\frac{dy}{dx} |_{x=4}$

$\frac{d(x^2)}{dx} |_{x=4}$

$2x|_{x=4}=8$

So, a parallel vector is $\hat i +8\hat j$ since this vector has the same slope as the obtained slope.

Length of the vector is $\sqrt{1^2+8^2}=\sqrt{1+64}=\sqrt{65}$

Therefore, a unit vector parallel to the tangent line is:

$\frac{1}{\sqrt{65}}(\hat i+8 \hat j)$