9.3 Workshop Solutions 3

True or False?

i) True, since $x=r\cos\theta$ and $y=r\sin\theta$.

ii) False. The set of points given by $\theta=\frac{\pi}{4}$ is only half of the line $y=x$: the points $(x,x)$ with $x<0$ are given by $\theta=\frac{5\pi}{4}$.

iii) False. The gradient is $-\frac{f_{x}}{f_{y}}$. To see this, we differentiate $f(x,y)=0$ with respect to $x$ to get $f_{x}+f_{y}\frac{dy}{dx}=0$, so that $f_{y}\frac{dy}{dx}=-f_{x}$ and hence $\frac{dy}{dx}=-\frac{f_{x}}{f_{y}}$.

iv) False: we also require $f_{xx}f_{yy}-f_{xy}^{2}>0$ (see slide 4.11 in the notes).

W3.1. We have $x=\cosh t\geq 0$ and

$$x^{2}-y^{2}=\cosh^{2}t-\sinh^{2}t=1,$$

so $(x,y)$ lies on the right branch of the hyperbola.

(ii) ${{dx}\over{dt}}=\sinh t$ and ${{dy}\over{dt}}=\cosh t$ so

$${{dy}\over{dx}}={{dy/dt}\over{dx/dt}}={{\cosh t}\over{\sinh t}}=\coth t.$$

W3.2 (i) With $x=at^{2}$ and $y=2at$ we have

$$y^{2}=4a^{2}t^{2}=4ax,$$

so $(x,y)$ lies on $C$.

(ii) We have

$${{dx}\over{dt}}=2at,\qquad{{dy}\over{dt}}=2a,$$

so the tangent has gradient

$${{dy}\over{dx}}={{dy/dt}\over{dx/dt}}={{1}\over{t}},$$

and the tangent has equation

$$y-2at={{1}\over{t}}(x-at^{2}),\quad{\hbox{or (rearranging)}}\quad y={{x}\over{t}}+at;$$

whereas the normal has gradient

$$-{{dx}\over{dy}}=-t,$$

so the normal has equation

$$y-2at=-t(x-at^{2}).$$

(iii) To determine the arc length we integrate $\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}$ over $t$, from $t=0$ to $t=1$; another possibility is to integrate $\sqrt{1+\left(\frac{dx}{dy}\right)^{2}}$ over $y$, from $y=0$ to $y=2a$. If we try to integrate $\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}$ over $x$ then we obtain a slightly awkward integral of $\sqrt{1+\frac{a}{x}}$, which needs a substitution.

Since $\frac{dx}{dt}=2at$ and $\frac{dy}{dt}=2a$, we see that the arc length is

$$\int_{0}^{1}\sqrt{4a^{2}+4a^{2}t^{2}}\,dt=2a\int_{0}^{1}\sqrt{1+t^{2}}\,dt.$$

Now we make the substitution $t=\sinh u$, so that $\sqrt{1+t^{2}}=\cosh u$ and $dt=\cosh u\,du$. Thus the integral becomes

$$2a\int_{0}^{\sinh^{-1}1}\cosh^{2}u\,du=a\int_{0}^{\sinh^{-1}1}(\cosh 2u+1)\,du=a\left[\frac{1}{2}\sinh 2u+u\right]_{0}^{\sinh^{-1}1}.$$

We therefore get the awkward result that the arc length is $a(\sinh^{-1}1+\frac{1}{2}\sinh(2\sinh^{-1}1))$, but this can be simplified somewhat. Firstly, $\sinh(2y)=2\sinh y\cosh y=2\sinh y\sqrt{1+\sinh^{2}y}$, so $\frac{1}{2}\sinh(2\sinh^{-1}1)=\sqrt{2}$. Secondly, we can express $\sinh^{-1}x$ uniquely in terms of $x$ by solving the equation $\frac{e^{y}-e^{-y}}{2}=x$, so $e^{2y}-2xe^{y}-1=0$, whence $e^{y}=\frac{2x\pm\sqrt{4x^{2}+4}}{2}$, so $y=\log(x+\sqrt{x^{2}+1})$. In particular, $\sinh^{-1}=\log(1+\sqrt{2})$. So the arc length from $t=0$ to $t=1$ is $a(\sqrt{2}+\log(1+\sqrt{2}))$.

W3.3 (i) We let $x=-\cos^{2}\theta$ and $y=-\sin\theta\cos^{2}\theta$, then

$$x^{2}(x+1)=\cos^{4}\theta(1-\cos^{2}\theta)=\cos^{4}\theta\sin^{2}\theta=y^{2};$$

hence $(x,y)$ lies on $T$; moreover

$${{dx}\over{d\theta}}=2\cos\theta\sin\theta,\quad{{dy}\over{d\theta}}=-\cos^{3}\theta+2\sin^{2}\theta\cos\theta;$$

so

$${{dy}\over{dx}}={{dy/d\theta}\over{dx/d\theta}}={{-\cos^{3}\theta+2\sin^{2}\theta\cos\theta}\over{2\cos\theta\sin\theta}}={{2\sin^{2}\theta-\cos^{2}\theta}\over{2\sin\theta}}.$$

(ii) Now let $x=\tan^{2}\theta$ and $y={\hbox{sec}}\,\theta\tan^{2}\theta$; then

$$x^{2}(x+1)=\tan^{4}\theta(\tan^{2}\theta+1)=\tan^{4}\theta{\hbox{sec}}^{2}\theta=y^{2};$$

hence $(x,y)$ lies on $T$; moreover,

$${{dx}\over{d\theta}}=2\tan\theta{\hbox{sec}}^{2}\theta,\quad{{dy}\over{d\theta}}={\hbox{sec}}\theta\tan^{3}\theta+2\tan\theta{\hbox{sec}}^{3}\theta;$$

so

$${{dy}\over{dx}}={{dy/d\theta}\over{dx/d\theta}}={{{\hbox{sec}}\theta\tan^{3}\theta+2\tan\theta{\hbox{sec}}^{3}\theta}\over{2\tan\theta{\hbox{sec}}^{2}\theta}}={{\tan^{2}\theta+2\sec^{2}\theta}\over{2{\hbox{sec}}\,\theta}}.$$

W3.4. (i) We observe that $x^{3}=t^{6}=y^{2}$, so $(x,y)$ lies on $C$, and $x,y\geq 0$, so the point is in the first quadrant.

(ii) We calculate

$${{dx}\over{dt}}=2t,\qquad{{dy}\over{dt}}=3t^{2},$$

so arclength is given by

$$\Bigl({{ds}\over{dt}}\Bigr)^{2}=\Bigl({{dx}\over{dt}}\Bigr)^{2}+\Bigl({{dy}\over{dt}}\Bigr)^{2}=4t^{2}+9t^{4}$$

so

$$L_{a}=\int_{0}^{a}{{ds}\over{dt}}dt=\int_{0}^{a}t\sqrt{4+9t^{2}}\,dt$$

so we substitute $t^{2}=u$ and $du/dt=2t$, so get

$$L_{a}={{1}\over{2}}\int_{0}^{a^{2}}\sqrt{4+9u}\,du={{1}\over{2}}\Bigl[{{2}\over{3\cdot 9}}\bigl(4+9u\bigr)^{3/2}\Bigr]_{0}^{a^{2}}={{1}\over{27}}\Bigl(\bigl(4+9a^{2})^{3/2}-4^{3/2}\Bigr).$$

This calculation was first carried out by Neil, and the simple form of the answer caused much surprise at the time.

(iii) When $a=1$, the precise value of $L_{a}$ is $\frac{13^{3/2}-8}{27}$, which is $1.440$ to three decimal places. To apply Simpson’s rule to the integral $\int_{0}^{1}t\sqrt{4+9t^{2}}\,dt$, we first have to evaluate the function $t\sqrt{4+9t^{2}}$ at the three points $t=0$, $t=\frac{1}{2}$ and $t=1$. The respective values are $0$, $\frac{1}{2}\sqrt{4+\frac{9}{4}}=\frac{5}{4}$, and $\sqrt{13}$. Thus Simpson’s rule produces an estimate of

$$\frac{1-0}{6}\left(1.0+4.\frac{5}{4}+1.\sqrt{13}\right)=\frac{5+\sqrt{13}}{6}.$$

This gives us a value of $1.434$, less than $1\%$ from the precise value.

W3.5. Let $f(x,y)=4x^{2}+xy+6y^{2}-4$; so that $f(x,y)=0$ gives the equation for the ellipse and defines $y$ implicitly as a function of $x$. To find $\frac{dy}{dx}$ we can either use the formula $\frac{dy}{dx}=-\frac{f_{x}}{f_{y}}$, or we can differentiate the equation $f(x,y)=0$ with respect to $x$. The second approach is more instructive, so this is how we’ll do it here. (However, both methods amount to the same thing.)

Since $4x^{2}+xy+6y^{2}-4=0$, when we differentiate with respect to $x$ we get:

$$8x+y+x\frac{dy}{dx}+12y\frac{dy}{dx}=0.$$

Now we collect terms together, so that

$$(x+12y)\frac{dy}{dx}=-8x-y,\;\;\mbox{hence}\;\;\frac{dy}{dx}=-\frac{8x+y}{x+12y}.$$

(Note that $f_{x}=8x+y$ and $f_{y}=x+12y$, which tallies with $\frac{dy}{dx}=-\frac{f_{x}}{f_{y}}$.)

W3.6. (i) We suppose that the formula

$$x+y=\tan^{-1}y$$

defines $y$ implicitly as a function of $x$, and we differentiate to obtain

$$1+{{dy}\over{dx}}={{1}\over{1+y^{2}}}{{dy}\over{dx}},$$

so that

$$(1+y^{2})+(1+y^{2}){{dy}\over{dx}}={{dy}\over{dx}},$$

and hence

$$1+y^{2}+y^{2}{{dy}\over{dx}}=0.$$

Alternatively, let $f(x,y)=x+y+\tan^{-1}y$ with first-order partial derivatives

$${{\partial f}\over{\partial x}}=1,\quad{{\partial f}\over{\partial y}}=1-{{1}\over{1+y^{2}}}={{y^{2}}\over{1+y^{2}}}.$$

Then the required derivative is

$${{dy}\over{dx}}=-{{\partial f/\partial x}\over{\partial f/\partial y}}=-{{1+y^{2}}\over{y^{2}}}.$$

(ii) We can differentiate the formula

$$\tan^{-1}{{y}\over{x}}={{1}\over{2}}\log\bigl(x^{2}+y^{2}\bigr)$$

with respect to $x$, and obtain by the chain rule

$${{1}\over{1+y^{2}/x^{2}}}{{d}\over{dx}}{{y}\over{x}}={{1}\over{2}}{{1}\over{x^{2}+y^{2}}}{{d}\over{dx}}\bigl(x^{2}+y^{2}\bigr)$$

$${{1}\over{1+y^{2}/x^{2}}}{{x{{dy}\over{dx}}-y}\over{x^{2}}}={{1}\over{x^{2}+y^{2}}}\Bigl(x+y{{dy}\over{dx}}\Bigr)$$

which reduces, when we multiply by $x^{2}+y^{2}$, to

$$x{{dy}\over{dx}}-y=x+y{{dy}\over{dx}},$$

$${{dy}\over{dx}}={{x+y}\over{x-y}}.$$

Alternatively, we can introduce

$$f(x,y)=\tan^{-1}(y/x)-{{1}\over{2}}\log(x^{2}+y^{2})-1$$

and calculate the first-order partial derivatives

$$f_{x}={{1}\over{1+y^{2}/x^{2}}}\bigl(-y/x\bigr)-{{x}\over{x^{2}+y^{2}}}=-{{y+x}\over{x^{2}+y^{2}}},$$

$$f_{y}={{1}\over{1+y^{2}/x^{2}}}\bigl(1/x\bigr)-{{y}\over{x^{2}+y^{2}}}={{x-y}\over{x^{2}+y^{2}}};$$

hence

$${{dy}\over{dx}}=-{{f_{x}}\over{f_{y}}}={{x+y}\over{x-y}}.$$

W3.7.(i) When $x=1$, we have $8y^{2}=2$; so that $y=\pm 1/2$.

(ii) We assume that $y$ is defined implicitly as a function of $x$ and we differentiate the relation

$$8y^{2}=x^{2}(x+1)$$

to obtain

$$16y{{dy}\over{dx}}=2x(x+1)+x^{2}=3x^{2}+2x;$$

hence

$${{dy}\over{dx}}={{3x^{2}+2x}\over{16y}}.$$

Alternatively, let $f(x,y)=8y^{2}-x^{2}(x+1)$ so that $f_{x}=-2x(x+1)-x^{2}$ and $f_{y}=16y$; then

$${{dy}\over{dx}}=-{{f_{x}}\over{f_{y}}}={{3x^{2}+2x}\over{16y}}.$$

From the formula above, we see that the tangents to the curve have gradients

$${{dy}\over{dx}}=\begin{cases}5/8,&at$(1,1/2)$;\cr-5/8,&at$(1,-1/2)$.\cr\end{cases}$$

For the equations of the tangents, we have:

$$y-\frac{1}{2}=\frac{5}{8}\left(x-1\right)\;\;\mbox{and}\;\;y-(-\frac{1}{2})=-\frac{5}{8}\left(x-1\right)$$

so we get the two lines $y=\pm\left(\frac{5}{8}x-\frac{1}{8}\right)$ at the respective points $(1,\pm\frac{1}{2})$.

W3.8. (i) We calculate the first-order partial derivatives of $g(x,y)=x^{2}-2xy+y^{3}$ to be

$${{\partial g}\over{\partial x}}=2x-2y,\quad{{\partial g}\over{\partial y}}=-2x+3y^{2};$$

the second-order partial derivatives are

$$g_{xx}=2,\quad g_{yy}=6y,\quad g_{xy}=-2.$$

(ii) The stationary points occur where

$$2x-2y=0\quad{\hbox{and}}\quad-2x+3y^{2}=0.$$

From the first equation, we deduce that $x=y$, so the second equation becomes $-2x+3x^{2}=0$, or $x(-2+3x)=0$, with roots $x=0$ and $x=2/3$; hence the stationary points are

$$(0,0)\quad{\hbox{and}}\quad(2/3,2/3).$$

(iii) The Hessian discriminant is

$$\Delta=g_{xx}g_{yy}-g_{xy}^{2}=12y-4.$$

The stationary points are classified in the following table.

$$\begin{matrix}{\hbox{stationary point}}&\Delta&g_{xx}&{\hbox{nature}}\cr(0,0)&-4<0&*&{\hbox{saddle}}\cr(2/3,2/3)&4&2>0&{\hbox{local minimum}}\cr\end{matrix}$$

W3.9. (i) We have $f_{x}=1-ye^{x}$ and $f_{y}=3y^{2}-e^{x}$, so both partial derivatives are zero exactly when $y=e^{-x}$ and $3y^{2}=e^{x}$. Multiplying, we obtain: $y.3y^{2}=e^{-x}.e^{x}=1$, hence $y=\frac{1}{\sqrt[3]{3}}$. It follows that $e^{x}=\frac{3}{\sqrt[3]{3}^{2}}=\sqrt[3]{3}$, so $x=\frac{1}{3}\log 3$. Thus there is one stationary point $(\frac{1}{3}\log 3,\frac{1}{\sqrt[3]{3}})$.

(ii) Here $g_{x}=y^{3}-\frac{1}{(x+y)^{2}}$ and $g_{y}=3xy^{2}-\frac{1}{(x+y)^{2}}$, so if both partial derivatives are zero we must have $y^{3}=\frac{1}{(x+y)^{2}}=3xy^{2}$, so $y=3x$. Substituting back into the equation $y^{3}=\frac{1}{(x+y)^{2}}$, we get $27x^{3}=\frac{1}{(4x)^{2}}$, so $x^{5}=\frac{1}{27.16}=\frac{1}{432}$. Over ${\mathbb{R}}$, there is only one solution here: $x=\frac{1}{\sqrt[5]{432}}$, and so $y=\frac{3}{\sqrt[5]{432}}=\sqrt[5]{\frac{9}{16}}$. Thus there is one turning point $\frac{1}{\sqrt[5]{432}}(1,3)$.

(iii) Here $h_{x}=3x^{2}-3y^{4}$ and $h_{y}=12y^{3}-12xy^{3}=12y^{3}(1-x)$. Thus if $h_{y}=0$ then either $y=0$ or $x=1$; if $h_{x}=0$ then $x^{2}=y^{4}$, so $x=\pm y^{2}$. Combining these, if $y=0$ and $x^{2}=y^{4}$ then $x=y=0$; if $x=1$ and $x=\pm y^{2}$ then $y=\pm 1$. Thus there are three stationary points: $(0,0)$, $(1,1)$ and $(1,-1)$.

Showing existence of saddle points

i) In this case we have $f_{xx}=-ye^{x}$, $f_{xy}=-e^{x}$ and $f_{yy}=6y$ so that $\Delta=f_{xx}f_{yy}-f_{xy}^{2}=-6y^{2}e^{x}-e^{2x}$. This is clearly negative for all values of $x,y$ so that any stationary point is a saddle point. (It only remains to recall that we found one stationary point for $f(x,y)$.)

ii) We recall that $g(x,y)$ has one stationary point $\frac{1}{\sqrt[5]{432}}(1,3)$. In particular, $y=3x$ at the stationary point. Now $g_{xx}=\frac{2}{(x+y)^{3}}$, $g_{xy}=3y^{2}+\frac{2}{(x+y)^{3}}$, $g_{yy}=6xy+\frac{2}{(x+y)^{3}}$. If $y=3x$ then we have $g_{xx}=\frac{2}{64x^{3}}=\frac{1}{32x^{3}}$, $g_{xy}=27x^{2}+\frac{1}{32x^{3}}$ and $g_{yy}=18x^{2}+\frac{1}{32x^{3}}$. Then

$$\Delta=g_{xx}g_{yy}-g_{xy}^{2}=\frac{1}{32x^{3}}(18x^{2}+\frac{1}{32x^{3}})-(27x^{2}+\frac{1}{32x^{3}})^{2}=-729x^{4}-\frac{9}{8x}.$$

Thus $\Delta<0$ for all $x>0$, and in particular for the unique stationary point. Hence $\frac{1}{\sqrt[5]{432}}(1,3)$ is a saddle point for $g(x,y)$.

iii) Recall that $h(x,y)$ has three stationary points: $(0,0)$, $(1,1)$ and $(1,-1)$. We have $h_{xx}=6x$, $h_{xy}=-12y^{3}$ and $h_{yy}=36y^{2}(1-x)$. Therefore $\Delta=h_{xx}h_{yy}-h_{xy}^{2}=216xy^{2}(1-x)-144y^{6}$. We have $\Delta=-144$ at the points $(1,\pm 1)$ so that both of these are saddle points. (We have $\Delta=0$ at the point $(0,0)$ (so that our existing methods do not allow us to determine the nature of this stationary point.)

W3.10. To find stationary points, we first look for simultaneous solutions of the equations $f_{x}=0$, $f_{y}=0$. Since

$$\frac{\partial f}{\partial x}=4x^{3}-4y,\;\;\;\mbox{and}\;\;\;\frac{\partial f}{\partial y}=4y^{3}-4x$$

the stationary points are the points $(x,y)$ satisfying $x^{3}=y$ and $y^{3}=x$. Then $y=x^{3}=(y^{3})^{3}=y^{9}$, so $y=0,1$ or $-1$. Then in these three cases we have $x=y^{3}=0,1$ or $-1$. So there are three stationary points: $(0,0)$, $(1,1)$ and $(-1,-1)$. To determine what types these points are, we calculate the second order partial derivatives: we have

$$\frac{\partial^{2}f}{\partial x^{2}}=12x^{2},\;\;\;\frac{\partial^{2}f}{\partial x\partial y}=-4\;\;\;\mbox{and}\;\;\;\frac{\partial^{2}f}{\partial y^{2}}=12y^{2}.$$

Thus $\Delta=144x^{2}y^{2}-16$. In particular, at the points $(1,1)$ and $(-1,-1)$, $\Delta$ and $f_{xx}$ are positive; at $(0,0)$, $\Delta$ is negative. So $(1,1)$ and $(-1,-1)$ are local minima, while $(0,0)$ is a saddle point.