Linda's Stationary points question from Y!A

Question

2)(a) The deflection \(y\) of a structural member is given by the formula:

\[y=x^3-2x^2 –x\]

Where x is the distance from one end.

(i) Find the value of \(x\) for which a turning point occur 

(ii) Determine the value of \(y\) at the turning point

(b)(i) Find the stationary points of the surface

\[z=x^3 +xy+y^2\]

(ii)Determine their nature

Answers:
2.a The stationary points are given by solutions to \(y'(x)=0\), but from the nature of the question we may assume we are only interested in positive solutions.
\[y'(x)=3x^2-4x-1\]Descartes' rule of signs tells us that this quadratic has exactly one positive root (and in consequence also one negative root, but we are not interested in that). We can find this using the quadratic formula:
\[x=\frac{4\pm\sqrt{16+12}}{6}=\frac{2\pm\sqrt{7}}{3}\] and we are interested only in the positive root:\[x=\frac{2+\sqrt{7}}{3}\]

So for part (i) we have the stationary point of interest occurs when \(\displaystyle x=(2+\sqrt{7})/3\) which gives for part (ii) the deflection at the stationary point is:
\[ \begin{aligned} y&=\frac{{\left( \sqrt{7}+2\right) }^{3}}{27}-\frac{2\,{\left( \sqrt{7}+2\right) }^{2}}{9}-\frac{\sqrt{7}+2}{3}\\ &=-\frac{2\times {7}^{\frac{3}{2}}+34}{27}\end{aligned} \]

2.b.i The stationary points are given by solutions to: \[ \frac{\partial z}{\partial x}=\frac{\partial z}{\partial y}=0\], which give the pair of equations:\[\begin{aligned}  \frac{\partial z}{\partial x}&=3x^2+y=0 \\  \frac{\partial z}{\partial y}&=x+2y=0\end{aligned} \]
From the second of these we find that \(y=-x/2\) which we substitute into the first to get \(3x^2-x/2=0\) which has solutions \(x=0, 1/6\) with corresponding \(y=0,-1/12\), so the stationary points are \( (0,0) \) and \( (\frac{1}{6},-\, \frac{1}{12}) \)

2.b.ii The classification of the stationary points can start with \( (0,0)\), where since along the line \(y=0\) function behaves like \(z=x^3\) which has a point of inflection at \(x=0\), we conclude that  \( (0,0)\) is a saddle point.

For the other stationary point we use the classification rule:

\[D=\left[  \frac{\partial^2z}{\partial x^2}\frac{\partial^2z}{\partial y^2}-\left(\frac{\partial^2z}{\partial x \partial y} \right) \right]_{ (\frac{1}{6},-\, \frac{1}{12})}\].

If \(D \lt 0\) then we have a saddle point.

If \(D \gt 0\) and \(\partial^2z/\partial x^2 \gt 0 \) (or  \(\partial^2z/\partial y^2 \gt 0 \) ) then we have a local minimum

If \(D \gt 0\) and \(\partial^2z/\partial x^2 \lt 0 \)  (or  \(\partial^2z/\partial y^2 \lt 0 \) ) then we have a local maximum

Otherwise the test is inconclusive.

I will leave the application of this test to the point \( (\frac{1}{6},-\, \frac{1}{12}) \) as an exercises for the reader.

(note I treated the stationary point \((0,0)\) differently because (a) it was easier, (b) the above test would have been inconclusive.)

Louis's Question from YahooAnswers:Fp1 Polynomial and roots question! Help!?

Question:
1.Find the range of values of \(a\) for which \[(2-3a)x^2+(4-a)x+2=0\]has real roots.

2. If the roots of the equation \(4x^3+7x^2-5x-1=0\) are \(\alpha\) , \(\beta\)  and \( \gamma\),find the equation whose roots are:
(a) \( \alpha+1,\beta+1\) and \(\gamma+1\)
(b) \(\alpha^2 \beta^2\) and \( \gamma^2 \)

Answer:
Part 1.
For a quadratic \(ax^2+bx+c\) the discriminant is \(b^2-4ac\) this is the term that appears under the square root sign in the quadratic formula: \[x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}\]for the solution of \(ax^2+bx+c=0\). The quadratic equation has real roots precisely when the discriminant is non-negative.

The discriminant of \((2-3a)x^2+(4-a)x+2\) is \((4-a)^2-4(2-3a)2=a^2+16a\). Because the discriminant is positive for large \(|a|\)  the discriminant is negative between the roots of \(a^2+16a\) which is on the interval \( (-16,0)\). Hence the discriminant is non-negative when \(a \lt -16\) or \(a \gt 0\) and so  \((2-3a)x^2+(4-a)x+2=0\) has real roots when \(a \lt -16\) or \(a \gt0\).

Part 2.
You may know Vieta's relations for the roots and coefficients of polynomials or cubics in particular or we can derive them:

For a cubic \(ax^3+bx^2+cx+d\) with roots \(\alpha,\beta\) and \(\beta\) we have:\[\begin{aligned}a(x-\alpha)(x-\beta)(x-\gamma)&=ax^3-a(\alpha+\beta+\gamma)x^2+a(\alpha\beta+\alpha\gamma+\beta\gamma)x-a(\alpha\beta\gamma)\\ &=ax^3+bx^2+cx+d \end{aligned}\] means that \(b=-a(\alpha+\beta+\gamma)\), \(c=a(\alpha\beta+\alpha\gamma+\beta\gamma)\) and \(d=-a(\alpha\beta\gamma)\).

So for the cubic \(4x^3+7x^2-5x-1\) we have:
\(\alpha+\beta+\gamma=-7/4\)
\(\alpha\beta+\alpha\gamma+\beta\gamma=-5/4\)
\(\alpha\beta\gamma=1/4\)

(a) Let the equation with roots  \( \alpha+1,\beta+1\) and \(\gamma+1\) be:\[x^3+ux^2+vx+w=0\] (we can of course multiply this by any constant we want if we want the left hand side to not be monic). Now from the relations between coefficients and roots we know that \[\begin{aligned}w&=-( \alpha+1)(\beta+1)(\gamma+1)\\ &=-(\alpha \beta \gamma + \alpha\beta + \alpha \gamma + \beta \gamma +\alpha+\beta+\gamma +1) \\ &= - \frac{1}{4} + \frac{5}{4} + \frac{7}{4} -1=\frac{7}{4} \end{aligned}\]
\[\begin{aligned}v&=( \alpha+1)(\beta+1)+(\alpha+1)(\gamma+1)+(\beta+1)(\gamma+1) \\
&=\alpha\beta+\alpha+\beta+1+\alpha\gamma+\alpha+\gamma+1+\beta\gamma+\beta+\gamma+1\\
&=(\alpha\beta+\alpha\gamma+\beta\gamma)+2(\alpha+\beta+\gamma)+3\\
&=\dots
\end{aligned}\]

(There is at least one other method of doing part (a), but I use Viete's relations here because they will be needed for part (b) and I don't want to provide two different approaches for the two parts.

...to be continued? Probably not, there is enough here to indicate how to complete the solution.

Ken's Question from Yahoo Answers: Probability Question For Verification?

Question:
"A point \({\rm{P}}(x,y)\) is chosen at random in a unit disc, centred at \((0,0)\).

The probability required is that the point chosen is such that both \(| x -y| \lt 1\) and \(|x+y| \lt 1\) .

Is the answer \(2/\pi\) or \(1-2/\pi\)?

Thank you."

Answer:
I take disc to be a disc of radius 1 centred at the origin

The region defined by the inequalities \(|x-y| \lt 1\) and \(|x+y| \lt 1\) is an inscribed square to the circle, which has side \(\sqrt{2}\) and hence area \(2\). The area of the circle is \(\pi\), so the probability that a point sampled uniformly on the unit disc satisfies the inequalities is the ratio of these two area: \(2/\pi\).

To convince yourself that the required region is the interior of the square rather than the exterior consider the point \((0,0)\), does it satisfy the inequalities. It it does then you want the interior of the square rather than the exterior.

Below is a scatter plot showing random points uniformly sampled on the unit disc and in black those satisfying the inequalities:

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CrazyCat's MATHS QUESTION HELP 3? from Yahoo Answers

CrazyCat's Question:
Find the discrimnant of \(kx^2 - 4x + k\) in terms of \(k\), hence find possible values of \(k\) given that \(kx^2 -4x + k = 0\) has equal roots.


Answer:
For a quadratic \(ax^2+bx+c\) the discriminant is \(b^2-4ac\) this is the term that appears under the square root sign in the quadratic formula:
\[x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}\]for the solution of  \(ax^2+bx+c=0\).  The quadratic equation has equal roots precisely when the discriminant is zero.

Now for the problem at hand \(a=k\), \(b=-4\) and \(c=k\) so the discriminant is \(D=b^2-4ac=16-4k^2\), and when \(D=0\) we have \(16-4k^2=0\) which we may solve for \(k\) to find: \(k=\pm2\).

http://www.localtutor.co.uk/Tuition/Havant_Maths_Tuition.html

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Q3 from STEP 1998 Paper1

Which of the following are true and which false? Justify your answers

(i) \(a^{\ln(b)}=b^{\ln(a)}\),  for all \(a,b>0\).

(ii) \(\cos(\sin(\theta))=\sin(\cos(\theta))\), for all real \(\theta\).

(iii) There exists a polynomial \(P\) such that \(|P(\theta)-\cos(\theta)| \lt 10^{ -6 } \) for all real \(\theta\)

(iv) \(x^4+3+x^{-4} \ge 5\) for all \(x\gt 0\).

                                                                             

Solution:
(i) True, since we know for \(a,b \gt 0\) that \( \ln(b)\ln(a)=\ln(a)\ln(b) \) and by the laws of logarithms this may be rewritten: \(\ln(a^{\ln(b)})=\ln(b^{\ln(a)})\) and as the equality of logarithms implies the equality of the arguments we can conclude that \(a^{\ln(b)}=b^{\ln(a)}\) for all \(a,b \gt 0\).

(ii) False, putting \(\theta=0\) the left hand side is \(1\) and the right hand side is \(\sin(1)\ne 1\).

(iii) False, polynomials are unbounded on the reals while \(|\cos(\theta)| \le 1\).

(iv) True, The function \(f(x)=x^4+3+x^{-4}\) goes to infinity as \(x\) goes to \(0\) from above, and as \(x\) goes to infinity. It also in continuous and differentiable and has one stationary point on \((0,\infty)\). This stationary point therefore must be a global minimum and it occurs at \(x=1\), and \(f(1)=5\).