Unit vector of a curve

Question

Find a unit normal vector to the curve $y=\cfrac{4}{3}x-\cfrac{2}{3}$ at point $(2,2)$

Would like to know in detail how is this solvable, and how should i present my answer?

Answer 1

You can find the slope of the tangent by taking the derivative. The normal is perpendicular to the tangent, which means the slopes multiply to make $-1,$ so $m_{\text{normal}}=\frac {-1}{m_{\text{tangent}}}$. Given the slope, you need to find a vector in that direction, which is $(1,m_{\text{normal}})$ and normalize it to get the answer.

Added: the tangent has a slope of $\frac 43$. As this is a linear function, that does not depend on the point chosen, but for another function it might. The normal then has a slope of $\frac {-1}{\frac 43}=-\frac 34$. A vector with slope $-\frac 34$ is $(1,-\frac 34)$ but this is not a unit vector so we find the length as $\sqrt {1^2+(\frac 34)^2}=\sqrt{\frac {25}{16}}=\frac 54$ and scale the vector by the reciprocal of this, giving a vector $(\frac 45, -\frac 35)$

Answer 2

The unit normal vector is defined as a unit vector (length $1$) perpendicular (normal) to the tangent vector (the slope).

The slope of the curve is given by derivative of the function. In this case $\left(\frac{4}{3}\right)$, it is also the $m$ in $y = mx + b$.

Then to find the normal vector, it is simply in the perpendicular direction, so the negative inverse of the slope, $-\frac{3}{4}$. This can be represented as $-4\mathbf{i} + 3\mathbf{j}$ or $4\mathbf{i} - 3\mathbf{j}$.

Then in order to convert this to a unit vector, divide by the magnitude, which is $\sqrt{3^2+4^2}=5$.

So your final answer should be something along the lines of $\frac{1}{5}(-4, 3)$ or $-\frac{4}{5}\mathbf{i} + \frac{3}{5}\mathbf{j}$

and the negative is reversible i.e. $\frac{4}{5}\mathbf{i} - \frac{3}{5}\mathbf{j}$.