Thread: Proving that a distance is always equal to 1

1. 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).

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.

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?

2. 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).

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.

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 $e^x$ at the point $x=a$ is $e^a$. So the line tangent to $e^x$ at $(a,e^a)$ is given by $e^a=e^a\cdot a+b$, where $b$ is the y-intercept. Solving the equation for $b$ gives $b=e^a(1-a)$.

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

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

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

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

From this equation, it's easy to see that it is equal to $0$ only when $x=a-1$, which is exactly what you were trying to prove. $\square$
Thanks. I understand now. This site is awesome, hope I can donate once I get a credit card.

4. By the way, did you use the point slope formula to get

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

If not, how did you get that?

5. It was worded sort of poorly by me. I was using the slope intercept equation for a line: $y=mx+b$. I plugged in $e^a$ for $y$, $e^a$ for $m$, and $a$ for $x$, and solved for $b$.