Question

How am I supposed to know if the slope of a normal line is perpendicular to the tangent line? Attached example.

Answer 1

I almost got this right, but missed the negative that should have been present due to it being a perpendicular line. Trying to understand how I can spot this.

Answer 2

when you have perpendicular lines the product of the slope is negative 1

Answer 3

line 1 slope = m
line 2 is perpendicular if slope is -1/m

Answer 4

did you take the 1st derivative of the curve to get the slope of the tangent line?
slope of normal would be -1/slope of tangent
to get the equation use point slope format with the point given

Answer 5

I did get the derivative of the tangent line and ended up with \[\frac{ e^x }{ 2\sqrt(x) } + \sqrt(x)e^x\]

Answer 6

So next I plugged in 1, and ended up with the slope of \[\frac{ 3e }{ 2 }\]

Answer 7

So conceptually I understand it's perpendicular if the slope of one is m and the other is -1/m. However, I don't see where I am supposed to recognize that in this particular problem. Thank you so much for taking the time to assist.

Answer 8

\(\color{purple}{\displaystyle f(x)=\sqrt{x}e^x}\)

\(\color{purple}{\displaystyle f'(x)=\left(\frac{1}{2\displaystyle\sqrt{x}}+\sqrt{x}\right)e^x}\)

\(\color{purple}{\displaystyle f'(1)=\left(\frac{1}{2\sqrt{1}}+\sqrt{1}\right)e^1=\frac{3e}{2}}\)

The equation of the TANGENT would then be,
\(\color{blue }{\displaystyle y-e=\frac{3e}{2}(x-1)}\)

are we correct so far?

Answer 9

Well, then we apply our algebra.

We know that if a line has a slope \(m\) then a perpendicular line (a normal line) will have a slope \(-1/m\). \(\color{red}{{\rm (Or{\tiny~~~}if{\tiny~~~}the{\tiny~~~}line{\tiny~~~}has{\tiny~~~}a{\tiny~~~}slope{\tiny~~~}}a/b,{\tiny~~~~}{\rm then{\tiny~~~}a{\tiny~~~}perpendicular{\tiny~~~}line}}\)
\(\color{red}{{\rm will{\tiny~~~}have{\tiny~~~}a{\tiny~~~}slope}{\tiny~~~}-b/a.)}\)

Answer 10

That's the answer I got, but the actual answer is \[y-e = -\frac{ 3e }{ 2 }(x-1)\]

Because it's perpendicular to the regular line according to the answer I looked at.

Answer 11

The equation you wrote above is NOT the TANGENT, NOR THE NORMAL line!

Answer 12

Oh shoot maybe I misread the answer then. UGH I'm on a roll right now...

Answer 13

\(\color{blue }{\displaystyle y-e=\frac{3e}{2}(x-1)}\)

is the tangent, and then (using the algebra as I mentioned above, and as you will find on almost any other site, and as I am pretty sure you know yourself),

The NORMAL line is,
\(\color{blue }{\displaystyle y-e=-\frac{2}{3e}(x-1)}\)

Answer 14

https://www.desmos.com/calculator/vlv3pygm5q

Answer 15

Omg I see it now, they gave me the normal line. I found the equation for THAT line (I thought that was the tangent line) but since the tangent line is PERPENDICULAR to that line we invert the slope and make it negative?

Answer 16

Thank you so much!

Answer 17

No problem, Dan!