@what pt. does y=e^(32x) have max curvature?

[nobelsmoke]

Here is the question:

At what point does the curve y=e^(32x) have maximum curvature?


I have tried this method: http://www.math.washington.edu/~conroy/m126-general/exams/mt2SolMath126Win2006.pdf

Problem 4. Adapting for 32x rather than x

it seems to get a bit lengthier with 32x than with x

i got ( -log(2048)/64, 1/32sqrt(2) )

but this was incorrect .

It seems finding the critical number takes some manipulations.

 

[HallsofIvy]

Okay, the formula you quote is
$$\kappa(x)= \frac{|f''(x)|}{1+ (f'(x))^2)^{3/2}}$$

You have $f(x)= e^{32x}$ so that $f'(x)= 32e^{32x}$ and $f''= 1024e^{32x}$ so
$$\kappa(x)= \frac{32e^{32x}}{(1+ 1048576e^{32x})^{3/2}}$$.

To find a maximum (or minimum) for that function, differentiate and set the derivative to 0:
$$\frac{1024e^{32x}(1+ 1048576e^{32x})^{3/2}- 50331648e^{32x}(1+ 1048576e^{32x}(33554432e^{32x})}{(1+ 1048576e^{32x})^3}= 0$$

A fraction is 0 if and only if the numerator is equal to 0 so set the numerator equal to 0 and solve.

 

[nobelsmoke]

Firstly, HallsofIvy Thank you for responding.

The formula for k(x) you have written is different than the one I quoted. In mine, 1 is inside the parenthesis in the denominator, also raised to the power of (3/2).

Anyway, we know that abs(f''(x)) belongs in the numerator.
And,

abs(f''(x)) = 1024e^(32x)

So, how do we end up with simply 32e^(32x) in the numerator for k(x) ?

 

[HallsofIvy]

Oh, bother! Yes, you are right- I did reverse f' and f''. $f'(x)= 32e^{32x}$ and $f''= 1024e^{32x}$ so
$$\kappa= \frac{1024e^{32x}}{(1+ 1024e^{64x})^{3/2}}$$

 

[nobelsmoke]

What point did you come to?
Here is a link to my two incorrect attempts:
http://i.imgur.com/DDKEO.png

 

[Redbelly98]

What point did you come to?
Here is a link to my two incorrect attempts:
http://i.imgur.com/DDKEO.png

Can you show your actual work? Our forum policy is not to provide students with answers. But if you show your work, we can probably spot where the error is.

 

[Redbelly98]

Wait, never mind. I find that I agree with your 2nd answer given in Post #5. So the grading program may expect you to write it in a different (but equivalent) form.

 

[nobelsmoke]

Thank you for responding RedBelly98
Here's my work:

f'(x) = 32e^(32x)
f''(x) = 1024e^(32x)

k(x) = (1024e^(32x)) / [1+1024e^(64x)]^3/2

k'(x) =
(32768e^(32x) / (1024e^(64x)+1)^(3/2) - (100663296e^(96x)) / (1+1024e^(64x))^(5/2)

when k'(x) = 0 , x = - log(2048)/64

y = e^(32x)
y = e^ (-32log(2048)/64 )
y = e^ (-log(2048)/2)
y = 1 / (32sqrt(2))

The point is at ( -log(2048)/64, 1/32sqrt(2) ).

Somewhere I went wrong.

 

[nobelsmoke]

Wait, never mind. I find that I agree with your 2nd answer given in Post #5. So the grading program may expect you to write it in a different (but equivalent) form.

Ok, thank you. I will keep that in mind.

 

[Redbelly98]

Perhaps if you expressed 2048 as (an integer)^{(another integer)}, then you could simplify the expression -log(2048)/64 and get it in a form that would be accepted.

 

[nobelsmoke]

Great Idea,
I'll post here after I submit my final attempt.

 

[Redbelly98]

Another thought: are you sure the grading system interprets "log" to mean the natural log, and not the base-10 log?