Question

Find the point of maximum curvature on function y=e^x

Answer 1

you would have to know that at the point of maximum curvature,
dy/dx=0

Answer 2

what you have to do is to differentiate the equation of the curve you have and equate the result to zero.

Answer 3

A curvature question -- neat! Let's consider a vector-valued function yielding the same curve embedded in three-dimensional space with a straightforward parameterization \(x=t,y=e^t,z=0\):$$\mathbf{r}(t)=\left(t,e^t,0\right)$$Now, we know that we can determine our curvature as follows: $$\kappa(t)=\frac{\|\mathbf{r'}(t)\times\mathbf{r''}(t)\|}{\|\mathbf{r}'(t)\|^3}$$We determine straightforwardly (differentiating component-wise) that \(\mathbf{r}'(t)=(1,e^t,0),\mathbf{r}''(t)=(0,e^t,0)\). We can compute the cross product rather simply:$$\mathbf{r}'(t)\times\mathbf{r}''(t)=(0-0,0-0,e^t-0)=(0,0,e^t)$$Now we compute magnitudes:$$\|\mathbf{r}'(t)\|=\sqrt{1^2+(e^t)^2}=\sqrt{1+e^{2t}}\\\|\mathbf{r}'(t)\times\mathbf{r}''(t)\|=\sqrt{(e^t)^2}=e^t$$... so our curvature is given by $$\kappa(t)=\frac{e^t}{(1+e^{2t})^\frac32}$$We wish to maximize our curvature, so differentiate:$$\kappa'(t)=\frac{e^t(1+e^{2t})^\frac32-e^t(3e^{2t}\sqrt{1+e^{2t}})}{(1+e^{2t})^3}$$We're interesting in where this vanishes, so only the numerator matters:$$e^t(1+e^{2t})^\frac32-3e^{3t}\sqrt{1+e^{2t}}=0\\\sqrt{1+e^{2t}}\left(e^t(1+e^{2t})-3e^{3t}\right)=0$$To maximize, we expect either to be \(0\),$$1+e^{2t}=0\\e^{2t}=-1$$... but \(e^{2t}\) is positive.$$e^t(1+e^{2t})-3e^{3t}=0\\e^t\left(1+e^{2t}-3e^{2t}\right)=0\\1+e^{2t}-3e^{2t}=0\\1=2e^{2t}\\\frac12=e^{2t}\\2t=\log\frac12=-\log2\\t=-\frac12\log 2$$Now that we don't really know whether that's a maximum, but recognize that our derivative is positive to the left and negative to the right of this value.

Now then, we evaluate our curvature at this point:$$\kappa\left(-\frac12\log2\right)=\frac{e^{-\frac12\log2}}{(1+e^{-\log2})^\frac32}=\frac{\sqrt{1/2}}{(3/2)^\frac32}=\sqrt{\frac12}\cdot\sqrt{\frac8{27}}=\sqrt{\frac4{27}}=\frac2{3\sqrt3}$$

Answer 4

We can tell its a maximum by recognizing its derivative is positive for \(1-2e^{2t}>0\) and negative for \(1-2e^{2t}>0\).

Answer 5

Whoa.

What do you think of this method?
y=e^(x)
y'=e^(x)
y''=e^(x)
k=(y'')/(1+(y')^2)^(3/2)=e^(x)/(1+e^(2…
Now you have the curvature as a function of x.
To find the max., differentiate the curvature and set it equal to zero.
k'=-(2e^(3x)-e^(x))*(e^(2x)+1)/(1+e^(2…
The only term that can equal zero is the first term.
2e^(3x)=e(x)
2e^(2x)=1
e^(2x)=1/2
2x=ln(1/2)
x=(1/2)ln(1/2)

Answer 6

That method is really just the same as mine -- the difference being that I computed \(\kappa(t)\) for an equivalent space curve and "rederived" that formula you used:$$\kappa(x)=\frac{y''}{\left(1+(y')^2\right)^\frac32}$$

Answer 7

@oldrin.bataku so, we treat y = e^x as r(t) ? or we have to transfer it back to parameter.?

Answer 8

I don't get how to let y =e^x turn to r(t) in your method

Answer 9

He just added the time parameter to it.
As we see from our equation y = e^x. When our x is 1, our y is e^1, and Z is not a part of the problem,
so for time 1: <1,e^1,0>
Time 2: <2,e^2,0>
Following our initial equation

Answer 10

I can see the link, just want to know whether there is any logic to put it in neat or just "see it" or "guess it"

Answer 11

or "consider ..."

Answer 12

for example: if I say obviously 3/6 =1/2 , to a lower level student , he will ask exactly my question "how" " I know but how " ?

Answer 13

I'm embedding the curve in the \(xy\)-plane.