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

In summary, the maximum curvature of the curve y=e^(32x) can be found by differentiating the formula \kappa(x)= \frac{|f''(x)|}{1+ (f'(x))^2)^{3/2}} and setting the derivative to 0. The critical number can be found at ( -log(2048)/64, 1/32sqrt(2) ). The formula for k(x) can also be written as \kappa= \frac{1024e^{32x}}{(1+ 1024e^{64x})^{3/2}}. It is important to note that the grading system may expect the answer to be written in a different but equivalent form, so
  • #1
nobelsmoke
6
0
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.
 
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  • #2
Okay, the formula you quote is
[tex]\kappa(x)= \frac{|f''(x)|}{1+ (f'(x))^2)^{3/2}}[/tex]

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

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

A fraction is 0 if and only if the numerator is equal to 0 so set the numerator equal to 0 and solve.
 
  • #3
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) ?
 
  • #4
Oh, bother! Yes, you are right- I did reverse f' and f''. [itex]f'(x)= 32e^{32x}[/itex] and [itex]f''= 1024e^{32x}[/itex] so
[tex]\kappa= \frac{1024e^{32x}}{(1+ 1024e^{64x})^{3/2}}[/tex]
 
  • #6
nobelsmoke said:
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.
 
  • #7
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.
 
  • #8
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.
 
  • #9
Redbelly98 said:
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.
 
  • #10
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.
 
  • #11
Great Idea,
I'll post here after I submit my final attempt.
 
  • #12
Another thought: are you sure the grading system interprets "log" to mean the natural log, and not the base-10 log?
 

1. What is the equation for y=e^(32x)?

The equation for y=e^(32x) is an exponential function that represents the growth or decay of a variable over time. The base of this function is the mathematical constant e, and the exponent is 32x.

2. What does the variable x represent in the equation y=e^(32x)?

The variable x in the equation y=e^(32x) represents the independent variable, which is typically time in this context. It can also represent any other independent variable that is being measured in relation to the dependent variable, y.

3. How is the curvature of a function determined?

The curvature of a function is determined by calculating the second derivative of the function. This involves taking the derivative of the first derivative, and the result is a measure of how much the slope of the function is changing at a given point.

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

The point at which y=e^(32x) has maximum curvature is at x=0. This means that the function has the steepest slope at this point, and the rate at which the slope is changing is at its highest.

5. How can we visualize the curvature of a function?

The curvature of a function can be visualized by graphing the function and looking at the shape of the curve. A function with a large positive curvature will have a steep and upward curve, while a function with a large negative curvature will have a steep and downward curve. A function with zero curvature will have a straight line.

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