Math Help - Derivative and normal

1. Derivative and normal

find an equation of the normal line to the curve
$y = 2 - (\frac{1}{3})x^2$ paralell to $x - y = 0$

i cant seem to find the slope of the normal line
im confused

answer at the back of book : $4x - 4y = 11$

find an equation of the normal line to the curve
$y = 2 - (\frac{1}{3})x^2$ paralell to $x - y = 0$

i cant seem to find the slope of the normal line
im confused
You have,
$x-y=0$
Thus,
$y=x$
You can write,
$y=1\cdot x+0$
Slope is 1, Intercept is 0.

Thus, you need to find the point(s) where the derivative (slope) of the normal line is -1. The reason why -1 is because if the derivative is tangent to the curve, but you need the normal line. Which is found by taking the reciprocal of the slope and making it change signs. In this case one.

Thus, find all $x$ such that,
$y'=-1$
Where,
$y=2-\frac{1}{3}x^2$

find an equation of the normal line to the curve
$y = 2 - (\frac{1}{3})x^2$ paralell to $x - y = 0$

i cant seem to find the slope of the normal line
im confused

answer at the back of book : $4x - 4y = 11$
Hello,

two lines are perpendicular if the product of their slopes equals -1.

$l_1:\ y=m_1 \cdot x + k_1$. k_1 is the y-intercept of the line l_1
$l_2:\ y=m_2 \cdot x + k_2$. k_2 is the y-intercept of the line l_2

$l_1 \perp l_2\ \Longleftrightarrow\ m_1 \cdot m_2=-1$

With your problem: l_1 : y = x, that means the slope m_1 = 1

thus the perpendicular line l_2 has the slope m_2 = -1

Now find the gradient of your function which has the same value as the slope of the perpendicular line:

$y'=-\frac{2}{3} \cdot x$. y' = -1. Solve for x. You'll get x = 1.5.

Plug in this value into the equation of your function to calculate the y-value: y = 1.25. Now you have a point (1.5, 1.25) and the slope (-1). Use the point-slope-formula to get the equation of the line:
$y=-x+2.75$

I've attached a sketch of your function and the two lines.

EB

I don't agree with their answer . . . Is there a typo?

Find an equation of the normal line to the curve $y \:= \:2 - \frac{1}{3}x^2$ parallel to $x - y \:= \:0$

The line $y = x$ has slope $m = 1.$
. . Our normal will also have slope $1.$

A normal to a curve is perpendicular to the tangent at that point.
. . Hence, the tangent will have slope $-1.$

The slope of a tangent is given by the derivative: . $y' \:=\:-\frac{2}{3}x$
The tangent has slope $-1$ when: . $-\frac{2}{3}x\:=\:-1\;\;\Rightarrow\;\; x = \frac{3}{2}$
. . and: . $y \:=\:2-\frac{1}{2}\left(\frac{3}{2}\right)^2\:=\:\frac{5} {4}$

The normal contains the point $\left(\frac{3}{2},\,\frac{5}{4}\right)$ and has slope $m = 1$.

Its equation is: . $y - \frac{5}{4}\:=\:1\left(x - \frac{3}{2}\right)\quad\Rightarrow\quad 4x - 4y\:=\:1$

5. Originally Posted by Soroban

I don't agree with their answer . . . Is there a typo?

The line $y = x$ has slope $m = 1.$
. . Our normal will also have slope $1.$

A normal to a curve is perpendicular to the tangent at that point.
. . Hence, the tangent will have slope $-1.$

The slope of a tangent is given by the derivative: . $y' \:=\:-\frac{2}{3}x$
The tangent has slope $-1$ when: . $-\frac{2}{3}x\:=\:-1\;\;\Rightarrow\;\; x = \frac{3}{2}$
. . and: . $y \:=\:2-\frac{1}{2}\left(\frac{3}{2}\right)^2\:=\:\frac{5} {4}$

The normal contains the point $\left(\frac{3}{2},\,\frac{5}{4}\right)$ and has slope $m = 1$.

Its equation is: . $y - \frac{5}{4}\:=\:1\left(x - \frac{3}{2}\right)\quad\Rightarrow\quad 4x - 4y\:=\:1$

I agree with Soroban on this one. However, I think there is a conflict between the answer that Soroban got and the answer in the back of the book.

Soroban got: $4x-4y=1$
Back of book: $4x-4y=11$

I started doing the working on paper for this one but I made a mistake on one part of the working after finding the coordinates of intercept! That's why I didn't post....

Following the working of Soroban, I would say it is the back of the book that is incorrect!