August 19, 2017, 05:13:52 pm

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#### beau77bro

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##### Re: 3U Maths Question Thread
« Reply #2025 on: May 19, 2017, 10:12:38 pm »
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Hi! Very rusty with locus and param, need someone to help with these

1) Find the gradients of the common tangents of the parabola y= x(x-4) and the circle (x-2)^2 + y^2 = 4 (Im not quite sure, do i diff both then simultaneous or?)

ive been trying to help my friend do this question all day. would these be the tangents we're finding? i think the question makes sense-ish but is this what common tangents means or is it when it's a tangent when they intesect? either way geogebra (even tho roughly done seems to think that the gradients are the answers) assuming the other side is identical.

but how do u actually derive the answer? i've tried equatting the gradients, trying to do sub in y=mx +b and then making the discriminant = 0 and trying to do simultaneous but it goes no where? i dont know if im adding any thing relevant but maybe the diagram will help legorgo, or someone else work it out?

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #2026 on: May 20, 2017, 12:27:22 am »
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ive been trying to help my friend do this question all day. would these be the tangents we're finding? i think the question makes sense-ish but is this what common tangents means or is it when it's a tangent when they intesect? either way geogebra (even tho roughly done seems to think that the gradients are the answers) assuming the other side is identical.

but how do u actually derive the answer? i've tried equatting the gradients, trying to do sub in y=mx +b and then making the discriminant = 0 and trying to do simultaneous but it goes no where? i dont know if im adding any thing relevant but maybe the diagram will help legorgo, or someone else work it out?
Those are just the tangents to each curve, at the points of intersection.

A common tangent is when the tangents to both curves (at their points of intersection) are the same line. The curves are essentially tangential to each other when this appears.
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#### jamonwindeyer

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##### Re: 3U Maths Question Thread
« Reply #2027 on: May 20, 2017, 12:43:58 am »
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Those are just the tangents to each curve, at the points of intersection.

A common tangent is when the tangents to both curves (at their points of intersection) are the same line. The curves are essentially tangential to each other when this appears.

Nah, I think that diagram is showing lines that happen to be tangents of both the circle and parabola, but not at the same point. It's not what I'd define as a common tangent either but it makes the question answerable at least!

As to finding it - I'm honestly not sure! On first thought, you need a point on the parabola $x_1,y_1$, and a point on the parabola $x_2,y_2$, such that the gradient of the interval joining the two points is equal to the gradient of the curve at both points. To me that's two equations with four variables. You could then use the equations of the curves (for example, $y_1=x_1(x_1-4)$) to get rid of two of the variables, leaving two equations in two variables - Simulteneously solve

(I anticipate that algebra to be horrendous)

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #2028 on: May 20, 2017, 12:46:01 am »
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Nah, I think that diagram is showing lines that happen to be tangents of both the circle and parabola, but not at the same point. It's not what I'd define as a common tangent either but it makes the question answerable at least!

As to finding it - I'm honestly not sure! On first thought, you need a point on the parabola $x_1,y_1$, and a point on the parabola $x_2,y_2$, such that the gradient of the interval joining the two points is equal to the gradient of the curve at both points. To me that's two equations with four variables. You could then use the equations of the curves (for example, $y_1=x_1(x_1-4)$) to get rid of two of the variables, leaving two equations in two variables - Simulteneously solve

(I anticipate that algebra to be horrendous)
I tried to reverse engineer it to see if that's what they meant, but I couldn't quite get it to work.

Maybe I had to do it from first principles.
$\text{I gave a hint that two of the solutions of}\\ (x-2)^2 + [x(x-4)]^2=4\\\text{ must}\textit{ by inspection be }x=0, x=4\\ \text{which makes the equation much easier to solve (just do polynomial division or something)}$
$\text{But it's still a lot more work to do}\\ \text{Plus I really prefer just using implicit differentiation rather than dealing with square roots}$
« Last Edit: May 20, 2017, 12:49:24 am by RuiAce »
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#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #2029 on: May 20, 2017, 12:52:09 am »
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\text{Actually, here's a clever way of solving the resultant quartic}\\ \begin{align*}(x-2)^2 + [x(x-4)]^2 &=4\\ (x^2-4x+4) + [x(x-4)]^2&=4\\ x(x-4) + x(x-4)x(x-4)&=0\\ x(x-4)(1+x(x-4))&=0\\ x(x-4)(1-4x+x^2)&=0\end{align*}
« Last Edit: May 20, 2017, 09:58:39 am by RuiAce »
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#### beau77bro

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##### Re: 3U Maths Question Thread
« Reply #2030 on: May 20, 2017, 09:54:38 am »
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\text{Actually, here's a clever way of solving the resultant quartic}\\ \begin{align*}(x-2)^2 + [x(x-4)]^2 &=4\\ (x^2-4x+4) + [x(x-4)]^2&=4\\ x(x-4) - x(x-4)x(x-4)&=0\\ x(x-4)(1-x(x-4))&=0\\ x(x-4)(1+4x-x^2)&=0\end{align*}
sorry to ask this but where does the negative come from in the third line?
sorry to ask this but where does the negative come from in the third lin

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #2031 on: May 20, 2017, 09:58:47 am »
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Fixed
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#### beau77bro

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##### Re: 3U Maths Question Thread
« Reply #2032 on: May 20, 2017, 02:23:59 pm »
+1

I've gotten to here but Idk where to go from there

#### beau77bro

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##### Re: 3U Maths Question Thread
« Reply #2033 on: May 20, 2017, 02:39:17 pm »
+1
I forgot a negative

#### beau77bro

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##### Re: 3U Maths Question Thread
« Reply #2034 on: May 20, 2017, 02:49:37 pm »
+1

Happy to say I finally got it

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #2035 on: May 20, 2017, 03:01:33 pm »
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Oh wow. I can't believe I forgot the possibility of something like this. I owe some people a big apology here. Here's a diagram to explain what was actually going on.

I had to zoom in to make it obvious that there was actually a tangent going on

Since the quartic is reducible to a quadratic, the quadratic formula can be used to avoid factorisation
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#### beau77bro

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##### Re: 3U Maths Question Thread
« Reply #2036 on: May 20, 2017, 03:06:14 pm »
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wait rui how did u work out f(x) was there a shortcut?

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #2037 on: May 20, 2017, 03:12:58 pm »
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wait rui how did u work out f(x) was there a shortcut?
Nah not really. I'd stick to the discriminant method.
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#### bsdfjnlkasn

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##### Re: 3U Maths Question Thread
« Reply #2038 on: May 20, 2017, 03:41:18 pm »
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Hey there,

I'm pretty sure that this is an easy probability question and I somehow managed to stumble to the correct answer but I would probably prefer a method by which to get the answer as I don't think my guessing is a good way to start and continue with the topic haha

50 tagged fish were released into a dam known to contain fish. Later a sample of thirty fish was netted from this dam, of which eight were found to be tagged. Estimate the total number of fish in the dam just prior to the sample of thirty being removed.

The answer is 187 or 188

Thank you!

#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #2039 on: May 20, 2017, 03:57:23 pm »
+1
Hey there,

I'm pretty sure that this is an easy probability question and I somehow managed to stumble to the correct answer but I would probably prefer a method by which to get the answer as I don't think my guessing is a good way to start and continue with the topic haha

50 tagged fish were released into a dam known to contain fish. Later a sample of thirty fish was netted from this dam, of which eight were found to be tagged. Estimate the total number of fish in the dam just prior to the sample of thirty being removed.

The answer is 187 or 188

Thank you!
Ironically this question appears from just Google searching the first sentence.
$\text{Let }x\text{ be the total number of fish in the dam}\\ \textit{just prior to the sample of 30 being removed}\\ \text{i.e. after 50 tagged, before 30 removed}\\ \text{This question is then just a ratio analysis question}$
$\text{At any time, the ratio }\frac{\text{No. of tagged fish}}{\text{No. of untagged fish}}\text{ is equal (for this estimate)}\\ \text{Hence, }\frac{8}{22} = \frac{50}{x-50}\\ \text{The solution is }x=187.5$
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