e^x and tangent through the origin?

[apple2357]

Hi, I am exploring ways to think about this problem and would welcome any graphical insights. I can prove that this happens algebraically but i am trying to see whether there is some kind of visual explanation? My sense is there ought to be something along the lines of transformations and stretches but can't piece it together so any thoughts would be much appreciated!

Take y= e^x and consider a point on the curve so that the tangent at that point passes through the origin:


The x coordinate of Point P is 1

Moving further and now consider y= e^(2x). Where would you expect the point P to be?




It turns out that the x coordinate of point P is now 1/2

And for y= e^(3x) and x coordinate of point P is 1/3

etc..



etc... Is there a geometric explanation?

 

[Steven G]

1st of all, the y value you have for the three points P are wrong. In the last graph you have the wrong x value.
Look carefully at the derivative of each of your three functions. That should tell you your answer.

 

[HallsofIvy]

The derivative of e^x is e^x. At x= a that is e^a so the tangent line to y= e^x at x= a is y= e^a(x- a)+ e^a= e^a(x- a+ 1). When x= 0, y= e^a(1- a). In order that the tangent line pass through the origin we must have y= e^a(1- a)= 0 so a= 1, not e.

 

[apple2357]

1st of all, the y value you have for the three points P are wrong. In the last graph you have the wrong x value.
Look carefully at the derivative of each of your three functions. That should tell you your answer.

I dont understand this? In the first one the x coordinates of P is 1 so the y coordinate is e^1 os P is (1,e) ?
Do you mean the software has rounded it off?

 

[Dr.Peterson]

Hi, I am exploring ways to think about this problem and would welcome any graphical insights. I can prove that this happens algebraically but i am trying to see whether there is some kind of visual explanation? My sense is there ought to be something along the lines of transformations and stretches but can't piece it together so any thoughts would be much appreciated!

Take y= e^x and consider a point on the curve so that the tangent at that point passes through the origin:

View attachment 25434
The x coordinate of Point P is 1

Moving further and now consider y= e^(2x). Where would you expect the point P to be?
View attachment 25435



It turns out that the x coordinate of point P is now 1/2

And for y= e^(3x) and x coordinate of point P is 1/3

etc..

View attachment 25436

etc... Is there a geometric explanation?

The geometric explanation is simple, isn't it? The function y = e^{kx} is obtained from y = e^x by a horizontal compression by a factor of k. If you do the same transformation to the tangent line, it remains tangent at the same value of y, at x = 1/k, while it still passes through the origin.

I'm not sure what Halls is saying is wrong; his work gives the same point you show, (1, e), for the point of tangency, though of course the software doesn't show an exact value. But you knew that, right?

 

[apple2357]

The geometric explanation is simple, isn't it? The function y = e^{kx} is obtained from y = e^x by a horizontal compression by a factor of k. If you do the same transformation to the tangent line, it remains tangent at the same value of y, at x = 1/k, while it still passes through the origin.

I'm not sure what Halls is saying is wrong; his work gives the same point you show, (1, e), for the point of tangency, though of course the software doesn't show an exact value. But you knew that, right?

Yes i did know that the software was rounding off the value, For example i did write x=1/3 ( not 0.33) for the third graph
Thanks for the transformations explanation. I was along those lines but for some reason was getting into a muddle!

 

[Steven G]

I dont understand this? In the first one the x coordinates of P is 1 so the y coordinate is e^1 os P is (1,e) ?
Do you mean the software has rounded it off?

I mean that e and 2.72 are not equal. I also did not know that the software gave the numbers which I saw.