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 Post subject: 0568 Question 23
PostPosted: Thu Nov 06, 2008 11:42 pm 
Joined: Wed Oct 29, 2008 4:02 am

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Joined: Wed Oct 29, 2008 4:02 am
Posts: 6
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.


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 Post subject:
PostPosted: Fri Nov 07, 2008 12:25 am 
Joined: Sun Aug 31, 2008 4:42 pm

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Joined: Sun Aug 31, 2008 4:42 pm
Posts: 129
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


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 Post subject: Re: 0568 Question 23
PostPosted: Sat Mar 27, 2010 8:20 am 
Joined: Fri Mar 26, 2010 7:18 am

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Joined: Fri Mar 26, 2010 7:18 am
Posts: 17
I didn t understand, Please try explaining one more time no?


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 Post subject: Re: 0568 Question 23
PostPosted: Sun Mar 28, 2010 1:30 am 
Joined: Thu Mar 25, 2010 12:14 am

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Joined: Thu Mar 25, 2010 12:14 am
Posts: 41
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


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