anonymous
  • anonymous
Find the point of maximum curvature on function y=e^x
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
you would have to know that at the point of maximum curvature, dy/dx=0
anonymous
  • anonymous
what you have to do is to differentiate the equation of the curve you have and equate the result to zero.
anonymous
  • anonymous
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:$$\kappa\left(-\frac12\log2\right)=\frac{e^{-\frac12\log2}}{(1+e^{-\log2})^\frac32}=\frac{\sqrt{1/2}}{(3/2)^\frac32}=\sqrt{\frac12}\cdot\sqrt{\frac8{27}}=\sqrt{\frac4{27}}=\frac2{3\sqrt3}$$

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anonymous
  • anonymous
We can tell its a maximum by recognizing its derivative is positive for \(1-2e^{2t}>0\) and negative for \(1-2e^{2t}>0\).
anonymous
  • anonymous
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)
anonymous
  • anonymous
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:$$\kappa(x)=\frac{y''}{\left(1+(y')^2\right)^\frac32}$$
Loser66
  • Loser66
@oldrin.bataku so, we treat y = e^x as r(t) ? or we have to transfer it back to parameter.?
Loser66
  • Loser66
I don't get how to let y =e^x turn to r(t) in your method
anonymous
  • anonymous
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
Loser66
  • Loser66
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"
Loser66
  • Loser66
or "consider ..."
Loser66
  • Loser66
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 " ?
anonymous
  • anonymous
I'm embedding the curve in the \(xy\)-plane.

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