Question

Find a unit vector that is parallel to the line tangent to the parabola y = x2 at the point (5, 25).

Answer 1

The line tangent to $y=x^2$ at the point (5, 25) as slope equal to $y'$ when $x=5$:

$y=x^2\implies y'=2x\implies\mathrm{slope}=10$

Then the tangent has equation

$y-25=10(x-5)\implies y=10x-25$

But only the slope is important here; any vector $t(1,10)$ is parallel to this line. Then a unit vector would be obtained by dividing this vector by its norm. Let $t=1$ for simplicity; then the unit vector is

$\dfrac{(1,10)}{\sqrt{1^2+10^2}}=\left(\dfrac1{\sqrt{101}},\dfrac{10}{\sqrt{101}}\right)$

Answer 2

A unit vector that is parallel to the line tangent to the parabola at the given points is: $(\frac{1}{\sqrt{101}}, \frac{10}{\sqrt{101}})$

Given the following data;

• Point on the x-axis = 5

• Point on the y-axis = 25

• $y = x^2$

In this exercise, you're required to find a unit vector that is parallel to the line tangent to the parabola at the given points.

First of all, we would differentiate the value of y at point x = 3 in order to find the slope of the line.

$y = x^{2}\\\\y' = 2x$

Substituting the value of x, we have;

$y' = 2(5)\\\\y' = 10$

Therefore, the slope (m) of the tangent line equation is 10.

V = (1, 10) is a vector parallel to the tangent line.

Next, we would find a unit vector by dividing the vector (V) by its norm respectively.

A unit vector parallel to the tangent line is given by the mathematical expression;

$\frac{V}{\sqrt{V^{2} + V^{2}}} = \frac{(1, \; 10)}{\sqrt{1^{2}\; + \;10^{2}}} \\\\= \frac{(1, \; 10)}{\sqrt{1 \;+ \;100}}\\\\= (\frac{1}{\sqrt{101}}, \frac{10}{\sqrt{101}})$

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