Minimum distance

**Naples** · January 10th 2010, 08:33 PM

Find the minimum distance from the origin to the curve y=e^x
a) 0.72
b) 0.74
c) 0.76
d) 0.78
e) 0.80

I used the distance formula using the pts. (0,0) and (x, e^x) to get:

d = sqrt(x^2 + e^2x)

However, I'm not sure where to go from there...

Any help is greatly appreciated.

**bandedkrait** · January 10th 2010, 09:31 PM

It can be said that the minimum distance of this curve from the origin is the distance of the point from the curve along the normal drawn to the curve passing through the given point,

$y=e^x$

$\frac{\mathrm{d} y}{\mathrm{d} x} = e^x$

Now consider a point on this curve, $(t,e^t)$

The equation of the tangent to the curve passing through this point is thus,

$(y-e^t)=e^t(x-t) \therefore y=e^t + e^t(x-t)$

Now since the normal to the curve at the given point has to be made to pass through the origin,

Its equation would be of the form , $y=mx$

Now since it's normal to the given tangent,
$m=-e^{-t}$

Now the intersection of the tangent and normal is at the point t, so that,

$-t = e^{2t}$

Now it's required to solve this above equation.

(Is use of a calculator allowed?)

If yes, then by a little inspection, we can estimate that the root for the above relation lies roughly between -.4265 and -.426.

Since the answer is required upto 2 significant digits only, -.426 can be assumed to be the correct value of t in the above case.

So use x=-.426 in the equation for distance, the answer comes out be approximately 0.78

**Amer** · January 10th 2010, 09:32 PM

Originally Posted by Naples

Find the minimum distance from the origin to the curve y=e^x
a) 0.72
b) 0.74
c) 0.76
d) 0.78
e) 0.80

I used the distance formula using the pts. (0,0) and (x, e^x) to get:

d = sqrt(x^2 + e^2x)

However, I'm not sure where to go from there...

Any help is greatly appreciated.

$D = \sqrt{x^2 + e^{2x}}$

find the critical point then find when the function have the minimum value

$D' = \frac{2x + 2e^{2x}}{2\sqrt{x^2+e^{2x}}}$

critical point

$2x+e^{2x} = 0$

$x=-.4263027510069$

and this critical point is minimum value
and the image of is is

$.0779767136088$

and the nearest solution is .78

**bandedkrait** · January 10th 2010, 09:36 PM

Originally Posted by Amer

and this critical point is minimum value
and the image of is is

$.0779767136088$

and the nearest solution is .78

How did you solve that, did you use a calculator? As in how did you solve the equation $e^{2x}+x = 0$

**Drexel28** · January 10th 2010, 09:37 PM

I mean, I am not encouraging this whatsoever, but if you have your function that you want to minimize and you are given the only five valid values...why not just plug them in? Once again, that is a dishonest way of doing it, but it might give you an idea of where to go.

**Amer** · January 10th 2010, 09:41 PM

Originally Posted by bandedkrait

How did you solve that, did you use a calculator?

sure I used calculator it is not easy to find it without calculator I think,

**Drexel28** · January 10th 2010, 09:43 PM

Originally Posted by bandedkrait

How did you solve that, did you use a calculator? As in how did you solve the equation $e^{2x}+x = 0$

Doubt? The equation $e^{2x}+x=0$ cannot be solved in terms of usual numerical representation. It can only be approximated as Amer showed.

**pomp** · January 10th 2010, 09:44 PM

Originally Posted by Amer

sure I used calculator it is not easy to find it without calculator I think,

True, the exact value of the x coordinate of the critical point is

$x = - \frac{\Omega}{2}$ ,

where $\Omega$ is the omega constant:

Omega Constant -- from Wolfram MathWorld

Not really anything to do with the problem, but you may be interested.

**bandedkrait** · January 10th 2010, 09:54 PM

Originally Posted by Drexel28

Doubt? The equation $e^2x+x=0$ cannot be solved in terms of usual numerical representation. It can only be approximated as Amer showed.

Okay, I tried to solve it graphically but couldn't get anything meaningful, so I approximated it using a calculator, considering $f(x)=x+e^{2x}$ , and then using Newton-Raphson method...

**HallsofIvy** · January 11th 2010, 04:33 AM

By the way, for distance functions generally, that is $\sqrt{(x-a)^2+ (y-b)^2}$ , perhaps with other conditions, "minimizing" the distance is exactly the same as minimzing the square of the distance. So, instead of taking the derivative of $\sqrt{x^2 + e^{2x}}$ , you could have just minimized $x^2+ e^{2x}$ . Then, instead of setting $\frac{2x + 2e^{2x}}{2\sqrt{x^2+e^{2x}}}= 0$ you would have just $2x+ 2e^{2x}= 0$ . Of course, that what you get anyway after multiplying through by the denominator but the derivative is a little easier.

**HallsofIvy** · January 11th 2010, 04:35 AM

Originally Posted by Drexel28

I mean, I am not encouraging this whatsoever, but if you have your function that you want to minimize and you are given the only five valid values...why not just plug them in? Once again, that is a dishonest way of doing it, but it might give you an idea of where to go.

If the problem is multiple choice, that is not at all "cheating". It may not be the way the person setting the problem intended, but that is their fault!

**bandedkrait** · January 11th 2010, 05:12 AM

Originally Posted by HallsofIvy

By the way, for distance functions generally, that is $\sqrt{(x-a)^2+ (y-b)^2}$ , perhaps with other conditions, "minimizing" the distance is exactly the same as minimzing the square of the distance. So, instead of taking the derivative of $\sqrt{x^2 + e^{2x}}$ , you could have just minimized $x^2+ e^{2x}$ . Then, instead of setting $\frac{2x + 2e^{2x}}{2\sqrt{x^2+e^{2x}}}= 0$ you would have just $2x+ 2e^{2x}= 0$ . Of course, that what you get anyway after multiplying through by the denominator but the derivative is a little easier.

Ha, I used the other approach just to avoid dealing with the square root!

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