# How do you find a unit vector a)parallel to and b)normal to the graph of f(x) at the indicated point. f(x)= x^3, "at" (1,1)?

Oct 3, 2017

Unit vector patallel to function at $\left(1 , 1\right)$ is $\frac{3}{\sqrt{10}} \hat{i} - \frac{1}{\sqrt{10}} \hat{j}$ and normal vector is $\frac{1}{\sqrt{10}} \hat{i} + \frac{3}{\sqrt{10}} \hat{j}$

#### Explanation:

The slope of tangent at $f \left(x\right) = {x}^{3}$ at $\left(1 , 1\right)$ is $f ' \left(1\right)$ and hence slope of normal is $- \frac{1}{f ' \left(1\right)}$

Now as $f \left(x\right) = {x}^{3}$, $f ' \left(x\right) = 3 {x}^{2}$. Slope of tangent at $\left(1 , 1\right)$ is $3$

and slope of normal is $- \frac{1}{3}$

Therefore equation of tangent is$y - 1 = 3 \left(x - 1\right)$ or $3 x - y = 2$

and unit vector parallel to $f \left(x\right)$ would be $\frac{3 \hat{i} - \hat{j}}{\sqrt{{1}^{+} {3}^{2}}}$

or $\frac{3}{\sqrt{10}} \hat{i} - \frac{1}{\sqrt{10}} \hat{j}$

and equation of normal is $y - 1 = - \frac{1}{3} \left(x - 1\right)$

i.e. $3 y - 3 = - x + 1$ or $x + 3 y = 4$

and unit normal vector would be $\frac{\hat{i} + 3 \hat{j}}{\sqrt{{1}^{+} {3}^{2}}}$

or $\frac{1}{\sqrt{10}} \hat{i} + \frac{3}{\sqrt{10}} \hat{j}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along $x$-axis and $y$-axis respectively.