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March 29, 2024, 07:34:27 am

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cotangent

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Methods question
« on: March 27, 2020, 05:56:25 pm »
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Yo does anyone know how to do this:

1) a) What point on the parabola y^2 = 2px is nearest to the point to (a,0), where 0<p<a

I was thinking of using limits like let f(x) = ±(2px)^0.5 then do lim as h approaches 0 (f(x+h), but then I got stuck. Can someone help. thanks  :)
« Last Edit: March 27, 2020, 06:00:38 pm by cotangent »
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cotangent

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Re: Methods question
« Reply #1 on: March 27, 2020, 06:13:06 pm »
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Edit: nvm, i think i got it.
I think u just use distance formula between two points (a,0) and (x,y). then diff that with respect to x and equal that to 0 for min distance.
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fun_jirachi

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Re: Methods question
« Reply #2 on: March 27, 2020, 06:39:47 pm »
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Hey there!

When we want to consider the smallest distance from a point on a curve from a fixed point (a, 0), we consider a moving point on the curve such that the tangent to the curve at the moving point is perpendicular to the line from the moving point to the fixed point (a, 0).

In this case, since you've got that 0<p<a, we can restrict our function to \(y = \pm \sqrt{2px}\). Something also important to note is the parametric form of our moving point, which in this case will be \((pt^2, \sqrt{2}pt)\). Note that when we minimise the distance from the parabola to the fixed point, we can have positive y and negative y for the same x-value - we can thus restrict our function to \(y = \sqrt{2px}\), consider the positive case then have the points \((pt^2, \pm \sqrt{2}pt)\) satisfy the condition.

Now, the tangent to the curve at the moving point has gradient \(\frac{\sqrt{2p}}{2\sqrt{x}} = \frac{\sqrt{2p}}{2\sqrt{pt^2}} = \frac{1}{\sqrt{2}t}\). The gradient of the line between the fixed point and the moving point is \(\frac{y_2-y_1}{x_2-x_1} = \frac{\sqrt{2}pt}{pt^2-a}\).

Now, we want these to be perpendicular to minimise the distance ie. \(\frac{\sqrt{2}pt}{pt^2-a} \times \frac{1}{\sqrt{2}t} = -1\).
Hopefully it's evident that we have \(p = a - pt^2 \implies p(1+t^2) = a \implies t = \sqrt{\frac{a}{p}-1}\). Substituting this value of t into \((pt^2, \pm \sqrt{2}pt)\) will give us the points on the curve that satisfy this condition.

Hope this helps :)
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cotangent

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Re: Methods question
« Reply #3 on: March 27, 2020, 06:43:11 pm »
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Hey there!

When we want to consider the smallest distance from a point on a curve from a fixed point (a, 0), we consider a moving point on the curve such that the tangent to the curve at the moving point is perpendicular to the line from the moving point to the fixed point (a, 0).

In this case, since you've got that 0<p<a, we can restrict our function to \(y = \pm \sqrt{2px}\). Something also important to note is the parametric form of our moving point, which in this case will be \((pt^2, \sqrt{2}pt)\). Note that when we minimise the distance from the parabola to the fixed point, we can have positive y and negative y for the same x-value - we can thus restrict our function to \(y = \sqrt{2px}\), consider the positive case then have the points \((pt^2, \pm \sqrt{2}pt)\) satisfy the condition.

Now, the tangent to the curve at the moving point has gradient \(\frac{\sqrt{2p}}{2\sqrt{x}} = \frac{\sqrt{2p}}{2\sqrt{pt^2}} = \frac{1}{\sqrt{2}t}\). The gradient of the line between the fixed point and the moving point is \(\frac{y_2-y_1}{x_2-x_1} = \frac{\sqrt{2}pt}{pt^2-a}\).

Now, we want these to be perpendicular to minimise the distance ie. \(\frac{\sqrt{2}pt}{pt^2-a} \times \frac{1}{\sqrt{2}t} = -1\).
Hopefully it's evident that we have \(p = a - pt^2 \implies p(1+t^2) = a \implies t = \sqrt{\frac{a}{p}-1}\). Substituting this value of t into \((pt^2, \pm \sqrt{2}pt)\) will give us the points on the curve that satisfy this condition.

Hope this helps :)
Thanks, do u know why using d/dx for the distance formula and equalling it to 0 for min distance doesn't work ?
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fun_jirachi

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Re: Methods question
« Reply #4 on: March 27, 2020, 06:52:54 pm »
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That would logically work - you'd just have to differentiate with respect to the parameter then verify that for the value of the parameter where (in this case d/dt = 0)  is a minimum. You can't use d/dx in the circumstances you describe - you'd have to alter them slightly, because you're determining the distance from a fixed point to a moving point.

Hope this makes some sense :)
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BiggestVCESweat

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Re: Methods question
« Reply #5 on: March 29, 2020, 09:25:14 pm »
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That would logically work - you'd just have to differentiate with respect to the parameter then verify that for the value of the parameter where (in this case d/dt = 0)  is a minimum. You can't use d/dx in the circumstances you describe - you'd have to alter them slightly, because you're determining the distance from a fixed point to a moving point.

Hope this makes some sense :)
You can still use the derivative of the distance, who led you to believe that you can't?

fun_jirachi

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Re: Methods question
« Reply #6 on: March 30, 2020, 02:07:20 am »
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You can still use the derivative of the distance, who led you to believe that you can't?

I'm quite sure you misinterpreted what I said - you have to differentiate the distance wrt the parameter, not using d/dx as the question described (and I quote 'Thanks, do u know why using d/dx for the distance formula and equalling it to 0 for min distance doesn't work ?'). I did also state this 'That would logically work - you'd just have to differentiate with respect to the parameter then verify that for the value of the parameter where (in this case d/dt = 0)  is a minimum. ' The part I actually rejected was using the notation d/dx - you're not differentiating wrt the curve or anything; you're differentiating wrt to the parameter you've set for yourself. I'm sorry that my answer did not come across as clearly the first time around :)
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S_R_K

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Re: Methods question
« Reply #7 on: March 30, 2020, 12:21:19 pm »
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I'm not convinced that the question requires a parameterisation of the parabola. Given a "moving point" with a given set of parametric equations, the locus of that point is given by the parabola \(y^2 = 2px\), and hence the minimum distance from the curve to the point (a, 0) is found by minimising \((x-a)^2 + (\sqrt{2px} - 0)^2\).

Using this method, the point found is at \(x = a-p\), and note that the normal to \(y^2 = 2px\) at that point also passes through (a, 0), thus confirming the distance is minimised.

Edit: also, as we should expect, this gives the same answer as the method given by Fun Jirachi.
« Last Edit: March 30, 2020, 12:23:51 pm by S_R_K »

IThinkIFailed

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Re: Methods question
« Reply #8 on: April 01, 2020, 11:48:33 am »
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Uh... is this in the methods course?
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Re: Methods question
« Reply #9 on: April 01, 2020, 01:45:55 pm »
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Parameterising curves is not in the Methods course. Finding distance between curves / points is in the Methods course.