# Proving that a distance is always equal to 1

• Oct 13th 2009, 12:51 PM
Lord Darkin
Proving that a distance is always equal to 1
Hi guys, this is the second problem I have here in the math forums (and should be my last for a while, hopefully). (Speechless)

I am given a question that says I need to prove that the distance between a and c is always equal to one. The question also says that line "l" is the tangent line to the graph of the function y = e^x (of course, at the point where (a,b). C is when that line touches the x-axis.

The image below is what I made to try and make things clear for you.

http://i960.photobucket.com/albums/a...-9Calculus.jpg

So, I think I have an idea of how to prove it. Should I be trying to do something with the pythagorean theorem where the vertical line is the dotted line from point (a,b) to point (a) on the x axis and the other leg of the triangle is from a to c. The a - to - c leg is what I'm looking for to equal one.

x^2 + y^2 = c^2

Where x is a-to-c. Thus ...

x^2 = 1, so how do I prove this?
• Oct 13th 2009, 01:01 PM
redsoxfan325
Quote:

Originally Posted by Lord Darkin
Hi guys, this is the second problem I have here in the math forums (and should be my last for a while, hopefully). (Speechless)

I am given a question that says I need to prove that the distance between a and c is always equal to one. The question also says that line "l" is the tangent line to the graph of the function y = e^x (of course, at the point where (a,b). C is when that line touches the x-axis.

The image below is what I made to try and make things clear for you.

http://i960.photobucket.com/albums/a...-9Calculus.jpg

So, I think I have an idea of how to prove it. Should I be trying to do something with the pythagorean theorem where the vertical line is the dotted line from point (a,b) to point (a) on the x axis and the other leg of the triangle is from a to c. The a - to - c leg is what I'm looking for to equal one.

x^2 + y^2 = c^2

Where x is a-to-c. Thus ...

x^2 = 1, so how do I prove this?

I would approach it like this. The derivative of $\displaystyle e^x$ at the point $\displaystyle x=a$ is $\displaystyle e^a$. So the line tangent to $\displaystyle e^x$ at $\displaystyle (a,e^a)$ is given by $\displaystyle e^a=e^a\cdot a+b$, where $\displaystyle b$ is the y-intercept. Solving the equation for $\displaystyle b$ gives $\displaystyle b=e^a(1-a)$.

So the equation of your tangent is $\displaystyle y=e^a x+e^a(1-a)$; in other words, $\displaystyle y=e^a(x+1-a)$.

From this equation, it's easy to see that it is equal to $\displaystyle 0$ only when $\displaystyle x=a-1$, which is exactly what you were trying to prove. $\displaystyle \square$
• Oct 13th 2009, 01:06 PM
Lord Darkin
Quote:

Originally Posted by redsoxfan325
I would approach it like this. The derivative of $\displaystyle e^x$ at the point $\displaystyle x=a$ is $\displaystyle e^a$. So the line tangent to $\displaystyle e^x$ at $\displaystyle (a,e^a)$ is given by $\displaystyle e^a=e^a\cdot a+b$, where $\displaystyle b$ is the y-intercept. Solving the equation for $\displaystyle b$ gives $\displaystyle b=e^a(1-a)$.

So the equation of your tangent is $\displaystyle y=e^a x+e^a(1-a)$; in other words, $\displaystyle y=e^a(x+1-a)$.

From this equation, it's easy to see that it is equal to $\displaystyle 0$ only when $\displaystyle x=a-1$, which is exactly what you were trying to prove. $\displaystyle \square$

Thanks. :) I understand now. This site is awesome, hope I can donate once I get a credit card.
• Oct 13th 2009, 01:36 PM
Lord Darkin
By the way, did you use the point slope formula to get

$\displaystyle e^a=e^a\cdot a+b$

If not, how did you get that?
• Oct 13th 2009, 01:39 PM
redsoxfan325
It was worded sort of poorly by me. I was using the slope intercept equation for a line: $\displaystyle y=mx+b$. I plugged in $\displaystyle e^a$ for $\displaystyle y$, $\displaystyle e^a$ for $\displaystyle m$, and $\displaystyle a$ for $\displaystyle x$, and solved for $\displaystyle b$.