Page 1 of 1
0568 Question 23
Posted: Thu Nov 06, 2008 11:42 pm
by newb
For what value of b is the line y = 10x tangent to the curve y = e^bx at some point in the xy-plane?
1. 10/e
2. 10
3. 10e
4. e^10
5. e
The answer was 1.
I tried finding where the two eqns intersect.
I also tried taking the derivative of y = e^bx and setting that = 10
How should I approach this problem? I am finding it difficult to simplify 10/e when working backwords from the answer.
Thank you very much in advance.
Posted: Fri Nov 07, 2008 12:25 am
by Nameless
use the form of the tangent line at x0
y=f'(x0)(x-x0)+f(x0)
We have f(x)= e^(bx) so f'(x)=be^(bx) then f'(x0)=be^(bx0)=10 (1)
Moreover since y=10x so we have f'(x0)x0=f(x0) (2)
from (1) &(2) you can solve for b=e/10
Re: 0568 Question 23
Posted: Sat Mar 27, 2010 8:20 am
by thmsrhn
I didn t understand, Please try explaining one more time no?
Re: 0568 Question 23
Posted: Sun Mar 28, 2010 1:30 am
by mathQ
At the point where line is tangent to curve, 10x = e^bx ------ (1)
at that point, dy/dx = b e^bx = 10 -------- (2)
from (1) put value of e^bx in (2)
so, b 10x = 10 ==> bx =1
put this in (1), so b = 10/e
Re: 0568 Question 23
Posted: Sat Apr 13, 2013 6:29 pm
by elyd
Hi. I have a way that hopefully clears things up.
First, we set (i) 10x=e^bx because this is the point at which the two lines, y=10x and y=e^bx, will intersect in the xy-plane.
I solved for 10, and got (ii) 10=(e^bx)/x.
The derivative of y=e^bx is y'=be^bx. The derivative will equal 10 at a certain point, thus (iii)10=be^bx.
I set the two 10's equal to each other, thus (iv) (e^bx)/x=b(e^bx). If you solve for b on the right hand side, you get (v) b=1/x.
Plug (1/x) in for b in equation (i) and we find that x=e/10.
By subbing in e/10 for x in equation (v) it's clear that b=10/e.