# eqn of normal to the curve

• Oct 7th 2008, 04:48 AM
maybeline9216
eqn of normal to the curve
Find the eqn of the normal to the curve y=e^(3x+2) at the point where x=-1, leaving your answer in terms of e.

For this qn, im not sure of the simplified version of the answer since you have to leave the answer in terms of e.
• Oct 7th 2008, 05:09 AM
earboth
Quote:

Originally Posted by maybeline9216
Find the eqn of the normal to the curve y=e^3x+2 at the point where x=-1, leaving your answer in terms of e.

For this qn, im not sure of the simplified version of the answer since you have to leave the answer in terms of e.

1. The point $\displaystyle P(-1,e^{-3}+2)$ belongs to the graph of the function and the normal.

2. Slope of the tangent m = y':

$\displaystyle y'=3 \cdot e^{3x}$ and therefore $\displaystyle m = \dfrac{3}{e^{3}}$

3. The direction of the normal is $\displaystyle r = -\dfrac{e^3}3$

4. Use point-slope-formula of a straight line:

$\displaystyle y-(e^{-3}+2)=-\dfrac{e^3}3 \cdot (x-(-1))$ . Solve for y:

5. Equation of the normal:

$\displaystyle y = -\dfrac{e^3}3 \cdot x -\dfrac{e^3}3+\dfrac1{e^3} + 2$
• Oct 7th 2008, 05:18 AM
maybeline9216
umm sorry shld be y= e^(3x+2)
• Oct 7th 2008, 05:26 AM
earboth
Quote:

Originally Posted by maybeline9216
umm sorry shld be y= e^(3x+2)

Ähemm!

The way to solve your question is the same, so I'll give you the results only:

$\displaystyle P(-1, e^{-1})$

$\displaystyle y'=3 \cdot e^{3x+2}$

$\displaystyle m = \dfrac3{e}$

$\displaystyle r = -\dfrac e3$

Equation of the normal:

$\displaystyle y-e^{-1}=-\dfrac e3 \cdot (x+1)$

$\displaystyle \boxed{y=-\dfrac e3 \cdot x -\dfrac e3 +e^{-1}}$
• Oct 7th 2008, 05:33 AM
ticbol
Quote:

Originally Posted by maybeline9216
Find the eqn of the normal to the curve y=e^3x+2 at the point where x=-1, leaving your answer in terms of e.

For this qn, im not sure of the simplified version of the answer since you have to leave the answer in terms of e.

Let me guess, "qn" is question.

So what if you have to leave the answer in terms of e? That's okay.
That's what the qn is asking for.

I bet you know that the y' is the slope of the tangent line to anywhere on the y-curve.
And you know too that -1/(y') is the slope of the normal line to anywhere on the y-curve.

So, you find the y'.
It should be
y' = 3e^(3x +2)

The slope of the normal line?
Why, m = -1 / 3e^(3x +2), of course.

Nww you find the point on the y-curve where the normal line is supposed to inject the curve.
It is given as the point where x = -1.
So, the y?
y = e^(3(-1) +2) = e^(-1) = 1/e

Thus, the point is (-1,1/e)

To find the equation of the normal line, you can use the point-slope form of the line
(y -y1) = m(x -x1)

You aleady have the (x1,y1) ....the (-1,1/e)

What about the m where x = -1?
m = -1 / 3e^(3(-1) +2)
m = -1 / 3e^(-1)
m = -e/3

Hence, the normal line is
(y -1/e) = (-e/3)(x -(-1))
y -1/e = (-e/3)x -e/3
y = (-e/3)x -e/3 +1/e ------answer.
• Oct 7th 2008, 05:34 AM
maybeline9216
Can this be simplified any further???
• Oct 7th 2008, 05:40 AM
earboth
Quote:

Originally Posted by maybeline9216
Can this be simplified any further???

You can collect the constant summands:

$\displaystyle \boxed{y=-\dfrac e3 \cdot x -\dfrac e3 +e^{-1} = -\dfrac e3 \cdot x +\dfrac {3-e^2}{3e}}$