Solution.

(a)

The square root needs \(x^2+y^2\geq 9\) and the denominator must not vanish, so \[\text {domain} = \big \{(x,y): x^2+y^2\geq 9,\ y\neq x\big \},\] the closed exterior of the circle of radius \(3\) with the line \(y=x\) removed. On that set the numerator takes every value in \([0,\infty )\) and the denominator every non-zero value of either sign, so the range is all of \(\mathbb {R}\).

(b)

The square root must be defined and non-zero, so \(x^2-y>0\), that is \[\text {domain} = \big \{(x,y): y < x^2\big \},\] the open region below the parabola \(y = x^2\). Since \(x^2-y\) takes every value in \((0,\infty )\), the range is \((0,\infty )\).

(c)

Only \(x=0\) is excluded, so the domain is the whole plane less the \(y-\)axis. The inverse tangent takes values in \(\Big (-\dfrac {\pi }{2}, \dfrac {\pi }{2}\Big )\), and every such value is attained, so that is the range.

Solution. A level curve is the set where \(z\) takes a fixed value \(k\); the surface is recovered by stacking them at their heights.

(a)

\(k = \sqrt {13-x^2-y^2}\) gives \(x^2+y^2 = 13-k^2\), circles of radius \(\sqrt {13-k^2}\) for \(0\leq k\leq \sqrt {13}\). They shrink to a point at the top: the surface is the upper half of the sphere of radius \(\sqrt {13}\).

(b)

\(x^2+y^2 = k+1\), circles of radius \(\sqrt {k+1}\) existing only for \(k\geq -1\). Equally spaced values of \(k\) give circles that crowd together as \(k\) grows: a circular paraboloid opening upward with vertex \((0,0,-1)\).

(c)

\(x^2+y^2 = 7-k\), circles for \(k\leq 7\): a paraboloid opening downward with vertex \((0,0,7)\).

(d)

\(2x^2 = k+1\) gives \(x = \pm \sqrt {\dfrac {k+1}{2}}\), pairs of vertical lines. \(y\) is absent, so the surface is a parabolic cylinder: the parabola \(z = 2x^2-1\) translated along the \(y-\)axis.

(e)

\(x^2+y^2 = \tan k\), circles for \(0\leq k<\dfrac {\pi }{2}\). The radii grow without bound as \(k\) approaches \(\dfrac {\pi }{2}\), so the surface rises steeply near the axis and then flattens towards the horizontal plane \(z = \dfrac {\pi }{2}\), which it never reaches.

Solution. Along \(y=0\) the function is identically \(0\), so that path gives \(0\).

Now approach along the curve \(x = y^3\), which passes through the origin. There \[\frac {4\big (y^3\big )^2y^3}{\big (y^3\big )^3+y^9} = \frac {4y^9}{y^9+y^9} = \frac {4y^9}{2y^9} = 2 ,\] constant, so that path gives \(2\).

Two paths give different values, so no limit exists.

Note that every straight line \(y = mx\) gives \(0\) here, so testing lines alone would have suggested the limit was \(0\). The cubic path is what exposes it, and choosing it is guided by making the two terms of the denominator the same order.

Solution. \[\nabla h = \langle -2x, -2y\rangle \ \longrightarrow \ \nabla h(10,20) = \langle -20, -40\rangle .\]

(a)

Steepest increase is along the gradient, \(\langle -20,-40\rangle \), or as a unit vector \(-\dfrac {1}{\sqrt 5}\langle 1,2\rangle \) — towards the summit, as expected since the peak is at the origin.

(b)

Steepest decrease is the opposite direction, \(\dfrac {1}{\sqrt 5}\langle 1,2\rangle \).

(c)

No change occurs perpendicular to the gradient, along \(\pm \dfrac {1}{\sqrt 5}\langle 2,-1\rangle \). On a contour map these are the directions along the contour through the point, while (a) and (b) cross the contours at right angles.

Solution. \[V_x = 10x-3y+yz,\qquad V_y = -3x+xz,\qquad V_z = xy ,\] so at \((3,4,5)\), \[\nabla V = \langle 30-12+20,\ -9+15,\ 12\rangle = \langle 38, 6, 12\rangle .\]

(a)

The unit vector in the given direction is \(\dfrac {1}{\sqrt 3}\langle 1,1,-1\rangle \), so \[D_{\textbf {u}}V = \nabla V\cdot \textbf {u} = \frac {38+6-12}{\sqrt 3} = \frac {32}{\sqrt 3} = \frac {32\sqrt 3}{3}\approx 18.5 .\]

(b)

Most rapidly along \(\nabla V = \langle 38,6,12\rangle \) itself.

(c)

The maximum rate is \(\left |\nabla V\right | = \sqrt {1444+36+144} = \sqrt {1624} = 2\sqrt {406} \approx 40.3\).

Solution. \[f_x = 6x^2-2xy = 2x(3x-y) = 0,\qquad f_y = -x^2+2y = 0 .\] The second gives \(y = \dfrac {x^2}{2}\). Substituting into the first, \(2x\Big (3x-\dfrac {x^2}{2}\Big ) = 0\), so \(x = 0\) or \(x = 6\). The stationary points are \((0,0)\) and \((6,18)\).

The second partials are \(f_{xx} = 12x-2y\), \(f_{yy} = 2\), \(f_{xy} = -2x\), so \[D = 2\big (12x-2y\big ) - 4x^2 .\] At \((6,18)\): \(D = 2(72-36)-144 = -72 < 0\), a saddle point.

At \((0,0)\): \(D = 0\), so the test is inconclusive. Examine the function directly. Along \(y=0\), \(f = 2x^3\), which changes sign at the origin, taking both positive and negative values arbitrarily close to it. Hence \((0,0)\) is neither a maximum nor a minimum: it too is a saddle.

Solution. Let the base be \(a\) by \(b\) and the height \(c\). With no lid the surface area is \[ab + 2ac + 2bc = 12 ,\] and the volume \(V = abc\) is to be maximised. By Lagrange multipliers, \(\nabla V = \lambda \nabla g\) gives \[bc = \lambda (b+2c),\qquad ac = \lambda (a+2c),\qquad ab = \lambda (2a+2b).\] Multiplying the first by \(a\) and the second by \(b\) and subtracting, \(\lambda \big (ab+2ac-ab-2bc\big ) = 0\), so \(a = b\) (taking \(\lambda \neq 0\)). Substituting \(a=b\) into the constraint and the third equation gives \[a = b = 2,\qquad c = 1 ,\] so the maximum volume is \(V = 2\cdot 2\cdot 1 = 4\ \text {m}^3\). Note the base is square and the height is half the base, which is the general shape for a lidless box of fixed area.

Solution. Write \(f = e^{g}\) with \(g = -\dfrac 13x^3+x-y^2\). Since the exponential is never zero and always positive, \(\nabla f = f\nabla g\) vanishes exactly where \(\nabla g\) does, and \(f\) has a maximum or minimum precisely where \(g\) does. \[g_x = -x^2+1 = 0 \implies x = \pm 1,\qquad g_y = -2y = 0 \implies y = 0 .\] For \(g\): \(g_{xx} = -2x\), \(g_{yy} = -2\), \(g_{xy} = 0\), so \(D = 4x\).

At \((1,0)\): \(D = 4>0\) and \(g_{xx} = -2<0\), a local maximum, with \(f = e^{2/3}\).

At \((-1,0)\): \(D = -4<0\), a saddle point, with \(f = e^{-2/3}\).

Working through \(g\) rather than \(f\) avoids differentiating the exponential twice, and is legitimate because \(t\mapsto e^t\) is strictly increasing.

Solution. The function depends only on \(r^2 = x^2+y^2\), so write \(G(r) = 2r^2e^{-r^2}\) with \(r\geq 0\).

(a)

\[G'(r) = 4re^{-r^2} - 4r^3e^{-r^2} = 4re^{-r^2}\big (1-r^2\big ) = 0\] gives \(r=0\) or \(r=1\). So the stationary points are the origin and every point of the circle \(x^2+y^2 = 1\) — a whole curve of them, which is why the second derivative test is not the tool here.

(b)

Never. Both \(2r^2\) and \(e^{-r^2}\) are non-negative, so \(g\geq 0\) everywhere, with equality only at the origin.

(c)

\(G(r) = \dfrac {2r^2}{e^{r^2}}\rightarrow 0\) as \(r\rightarrow \infty \), since the exponential outgrows any power.

(d)

By (b), \(g\geq 0\) with equality only at the origin, so the origin is the absolute minimum. On the circle \(r=1\), \(g = 2e^{-1}\approx 0.736\); since \(g\) increases from \(0\) to that value as \(r\) goes from \(0\) to \(1\) and decreases back towards \(0\) afterwards, every point of the circle is an absolute maximum.

(e)

A surface of revolution: a dip to zero at the origin, rising to a circular crest of height \(2e^{-1}\) above the unit circle, then falling away asymptotically to the plane \(z=0\) — the shape of a volcano rim.

Solution. By symmetry the box is centred at the origin, so if one corner is \((x,y,z)\) in the first octant the box has edges \(2x\), \(2y\), \(2z\) and volume \(V = 8xyz\). Maximise subject to the corner lying on the ellipsoid. Lagrange gives \[8yz = \frac {2\lambda x}{a^2},\qquad 8xz = \frac {2\lambda y}{b^2}, \qquad 8xy = \frac {2\lambda z}{c^2}.\] Multiplying the first by \(x\), the second by \(y\) and the third by \(z\) shows all three of \(\dfrac {x^2}{a^2}\), \(\dfrac {y^2}{b^2}\), \(\dfrac {z^2}{c^2}\) equal, so each is \(\dfrac 13\) by the constraint. Hence \[x = \frac {a}{\sqrt 3},\qquad y = \frac {b}{\sqrt 3},\qquad z = \frac {c}{\sqrt 3},\] the box has dimensions \(\dfrac {2a}{\sqrt 3}\times \dfrac {2b}{\sqrt 3}\times \dfrac {2c}{\sqrt 3}\), and \[V = \frac {8abc}{3\sqrt 3} = \frac {8\sqrt 3}{9}abc\approx 1.54\,abc .\] For a sphere (\(a=b=c\)) this is the cube, as it must be.

Solution. The conditions \(\nabla f = \lambda \nabla g\) read \[yz = 2\lambda x,\qquad xz = 4\lambda y,\qquad xy = 6\lambda z .\] Multiplying each by the corresponding variable gives \[xyz = 2\lambda x^2 = 4\lambda y^2 = 6\lambda z^2 .\] If \(\lambda = 0\) then \(xyz = 0\), giving the value \(0\). Otherwise \[x^2 = 2y^2 = 3z^2 ,\] and substituting into the constraint, \(x^2+x^2+x^2 = 6\), so \(x^2 = 2\) and \[x = \pm \sqrt 2,\qquad y = \pm 1,\qquad z = \pm \sqrt {\tfrac 23}.\] At such a point \[\left |xyz\right | = \sqrt 2\cdot 1\cdot \sqrt {\tfrac 23} = \frac {2}{\sqrt 3} = \frac {2\sqrt 3}{3}\approx 1.155 .\] Choosing the signs so the product is positive gives the maximum \(\dfrac {2\sqrt 3}{3}\), and so the product is negative gives the minimum \(-\dfrac {2\sqrt 3}{3}\).

Solution.

The derivatives at \((1,1)\)

\[f_x = y^2 + \frac {y}{1+xy},\qquad f_y = 2xy + \frac {x}{1+xy},\] so \(f(1,1) = 1+\ln 2\), \(f_x(1,1) = 1+\dfrac 12 = \dfrac 32\) and \(f_y(1,1) = 2+\dfrac 12 = \dfrac 52\). Differentiating again, \[f_{xx} = -\frac {y^2}{(1+xy)^2} \rightarrow -\frac 14,\qquad f_{yy} = 2x - \frac {x^2}{(1+xy)^2} \rightarrow \frac 74,\] \[f_{xy} = 2y + \frac {1}{(1+xy)^2} \rightarrow \frac 94 .\]

(a)

With \(h = x-1\) and \(k = y-1\), \[f \approx 1+\ln 2 + \frac 32h + \frac 52k + \frac 12\Big [-\frac 14h^2 + 2\cdot \frac 94hk + \frac 74k^2\Big ]\] \[= 1+\ln 2 + \frac 32h + \frac 52k - \frac {h^2}{8} + \frac {9hk}{4} + \frac {7k^2}{8}.\]

(b)

The tangent plane is the linear part of that expansion: \[z = 1+\ln 2 + \frac 32(x-1) + \frac 52(y-1).\]

(c)

Stationary points satisfy \[y^2 + \frac {y}{1+xy} = 0,\qquad 2xy + \frac {x}{1+xy} = 0 .\] Both hold when \(x=y=0\), so \((0,0)\) is stationary. There \[f_{xx}(0,0) = 0,\qquad f_{yy}(0,0) = 0,\qquad f_{xy}(0,0) = 1 ,\] so \(D = 0\cdot 0 - 1 = -1 < 0\) and \((0,0)\) is a saddle point.

Solution.

(a)

\[f_x = 3x^2-3y = 0 \implies y = x^2,\qquad f_y = 2y-3x+1 = 0 .\] Substituting, \(2x^2-3x+1 = 0\), so \((2x-1)(x-1) = 0\) and \(x = \dfrac 12\) or \(x = 1\). The stationary points are \(\Big (\dfrac 12,\dfrac 14\Big )\) and \((1,1)\).

With \(f_{xx} = 6x\), \(f_{yy} = 2\), \(f_{xy} = -3\), the discriminant is \(D = 12x-9\).

At \(\Big (\dfrac 12,\dfrac 14\Big )\): \(D = -3<0\), a saddle, value \(\dfrac {1}{16}\).

At \((1,1)\): \(D = 3>0\) and \(f_{xx} = 6>0\), a local minimum, value \(0\).

(b)

Both stationary points lie in \(R\), contributing \(\dfrac {1}{16}\) and \(0\). Now the four edges.

On \(y=0\): \(f = x^3\), running from \(0\) to \(8\).

On \(y=2\): \(f = x^3-6x+6\), with \(f' = 3x^2-6 = 0\) at \(x = \sqrt 2\), giving \(6-4\sqrt 2\approx 0.34\); the endpoints give \(6\) and \(2\).

On \(x=0\): \(f = y^2+y\), from \(0\) to \(6\).

On \(x=2\): \(f = 8+y^2-6y+y = y^2-5y+8\), minimised at \(y = \dfrac 52\) which is outside \([0,2]\), so the endpoints give \(8\) and \(2\).

Comparing every candidate, the global minimum is \(0\), attained at \((1,1)\) and at the origin, and the global maximum is \(8\), at \((2,0)\).

Solution.

Interior

\(z_x = 2x-1 = 0\) and \(z_y = 4y = 0\) give \(\Big (\dfrac 12, 0\Big )\), inside the disc, with \(z = \dfrac 14-\dfrac 12 = -\dfrac 14\).

Boundary

On \(x^2+y^2 = 1\) substitute \(y^2 = 1-x^2\): \[z = x^2+2\big (1-x^2\big )-x = 2-x^2-x ,\qquad -1\leq x\leq 1 .\] Then \(\dfrac {dz}{dx} = -2x-1 = 0\) at \(x = -\dfrac 12\), giving \(z = 2-\dfrac 14+\dfrac 12 = \dfrac 94\). The endpoints give \(z = 0\) at \(x=1\) and \(z = 2\) at \(x=-1\).

Conclusion

The absolute minimum is \(-\dfrac 14\) at \(\Big (\dfrac 12,0\Big )\), and the absolute maximum is \(\dfrac 94\) at \(\Big (-\dfrac 12, \pm \dfrac {\sqrt 3}{2}\Big )\).

Solution. With two constraints, \(\nabla f = \lambda \nabla g + \mu \nabla h\) where \(g = x+y+z\) and \(h = y^2+z^2\): \[1 = \lambda ,\qquad 2 = \lambda + 2\mu y,\qquad 0 = \lambda + 2\mu z .\] From the first, \(\lambda = 1\). The second gives \(2\mu y = 1\) and the third \(2\mu z = -1\), so \(\mu \neq 0\) and \[y = \frac {1}{2\mu },\qquad z = -\frac {1}{2\mu },\qquad \text {hence } z = -y .\] Substituting into \(y^2+z^2 = 4\) gives \(2y^2 = 4\), so \(y = \pm \sqrt 2\) and \(z = \mp \sqrt 2\). The first constraint then gives \(x = 1-y-z = 1\).

The two candidate points are \(\big (1, \sqrt 2, -\sqrt 2\big )\) and \(\big (1, -\sqrt 2, \sqrt 2\big )\), where \[f = 1+2\sqrt 2 \approx 3.83 \qquad \text {and}\qquad f = 1-2\sqrt 2\approx -1.83 .\] These are the maximum and minimum respectively.

Solution. The extreme points lie on the line through the centre and the given point, which saves the Lagrange calculation. The point is at distance \[\left |\langle 3,1,-1\rangle \right | = \sqrt {9+1+1} = \sqrt {11}\approx 3.317\] from the origin, which exceeds the radius \(2\), so it lies outside the sphere.

The unit vector towards it is \(\dfrac {1}{\sqrt {11}}\langle 3,1,-1\rangle \), so the nearest point is \[\frac {2}{\sqrt {11}}\langle 3,1,-1\rangle = \Big (\frac {6}{\sqrt {11}}, \frac {2}{\sqrt {11}}, -\frac {2}{\sqrt {11}}\Big ),\] at distance \(\sqrt {11}-2\approx 1.32\), and the farthest is the antipode \[-\frac {2}{\sqrt {11}}\langle 3,1,-1\rangle ,\] at distance \(\sqrt {11}+2\approx 5.32\).

Solution.

(a)

\(f(0,0) = -8\); \(f_x = 3x^2+6x\) and \(f_y = 3y^2-6y\) both vanish at the origin; \(f_{xx} = 6x+6\rightarrow 6\), \(f_{yy} = 6y-6\rightarrow -6\), \(f_{xy} = 0\). Hence \[f \approx -8 + \frac 12\big (6x^2 - 6y^2\big ) = -8 + 3x^2 - 3y^2 .\] The observation is that this is simply the part of \(f\) of degree at most two, read off from the original expression. A polynomial is its own Taylor expansion about any point, so no work was necessary.

(b)

By the same observation, the third-degree expansion is the whole polynomial: \[f = -8 + 3x^2 - 3y^2 + x^3 + y^3 ,\] exact rather than approximate.

(c)

Here the expansion must be recomputed, since the powers of \(x\) and \(y\) are not powers of \(x-1\) and \(y-2\). At \((1,2)\), \[f = 1+8+3-12-8 = -8,\qquad f_x = 3+6 = 9,\qquad f_y = 12-12 = 0,\] \[f_{xx} = 12,\qquad f_{yy} = 6,\qquad f_{xy} = 0 .\] With \(h = x-1\) and \(k = y-2\), \[f \approx -8 + 9h + \frac 12\big (12h^2 + 6k^2\big ) = -8 + 9h + 6h^2 + 3k^2 .\]

Solution. Write \(s = 1+x^2+y^2\), so \(g = s^{-1}\).

(a)

\(g_x = -\dfrac {2x}{s^2}\) and \(g_y = -\dfrac {2y}{s^2}\), so \[g_{xx} = \frac {6x^2-2y^2-2}{s^3},\qquad g_{yy} = \frac {6y^2-2x^2-2}{s^3},\qquad g_{xy} = \frac {8xy}{s^3}.\]

(b)

At the origin \(s=1\), so \(g = 1\), both first derivatives vanish, and \(g_{xx} = g_{yy} = -2\) with \(g_{xy} = 0\). Hence \[g \approx 1 + \frac 12\big (-2x^2 - 2y^2\big ) = 1 - x^2 - y^2 .\] The same result follows faster from the geometric series: \(\dfrac {1}{1+t} = 1-t+t^2-\cdots \) with \(t = x^2+y^2\).

(c)

The quadratic part \(-x^2-y^2\) is negative for every \((x,y)\neq (0,0)\), so \(g\) decreases in every direction away from the origin: a local maximum, value \(1\). It is in fact the absolute maximum, since \(s\geq 1\) everywhere.

(d)

The function depends only on \(r^2 = x^2+y^2\), so the surface is a surface of revolution: a single peak of height \(1\) above the origin, falling away in all directions and tending to \(0\) as \(r\) grows, without ever reaching it.

Solution. With \(F = x^3+y^3+z^3+6xyz-1\), \[F_x = 3x^2+6yz,\qquad F_y = 3y^2+6xz,\qquad F_z = 3z^2+6xy ,\] so \[\frac {\partial z}{\partial x} = -\frac {F_x}{F_z} = -\frac {x^2+2yz}{z^2+2xy},\qquad \frac {\partial z}{\partial y} = -\frac {F_y}{F_z} = -\frac {y^2+2xz}{z^2+2xy},\] valid wherever \(z^2+2xy\neq 0\), which is the condition of the implicit function theorem here.

Solution. Rather than differentiate, check homogeneity. Replacing \(x\) by \(tx\) and \(y\) by \(ty\), \[F(tx,ty) = (tx)^4(ty)^2\sin ^{-1}\frac {ty}{tx} = t^6x^4y^2\sin ^{-1}\frac {y}{x} = t^6F(x,y),\] the inverse sine being unchanged because its argument is a ratio. So \(F\) is homogeneous of degree \(6\), and Euler’s theorem gives \[xF_x + yF_y = 6F\] at once. Differentiating directly gives the same result with considerably more labour.

Solution.

(a)

\[\frac {\partial (x,y)}{\partial (u,v)} = \begin {vmatrix} 1 & 4\\ 3 & -2\end {vmatrix} = -2-12 = -14 ,\] so the inverse transformation has Jacobian \(-\dfrac {1}{14}\), the two being reciprocal.

(b)

\[\frac {\partial x}{\partial u} = \frac {v}{(u+v)^2},\quad \frac {\partial x}{\partial v} = \frac {-u}{(u+v)^2},\quad \frac {\partial y}{\partial u} = \frac {-v}{(u-v)^2},\quad \frac {\partial y}{\partial v} = \frac {u}{(u-v)^2},\] so \[\frac {\partial (x,y)}{\partial (u,v)} = \frac {uv}{(u+v)^2(u-v)^2} - \frac {uv}{(u+v)^2(u-v)^2} = 0 .\] The Jacobian vanishes identically, so this transformation has no inverse and the second part of the question cannot be answered as asked. A zero Jacobian means the two functions are dependent, and indeed \[x + 2xy - y = 1\] for all admissible \(u,v\): the image is a curve in the plane, not a region.

(c)

\[\frac {\partial (x,y,z)}{\partial (u,v,w)} = \begin {vmatrix} v & u & 0\\ 0 & w & v\\ w & 0 & u \end {vmatrix} = v(wu-0) - u(0-vw) + 0 = 2uvw ,\] so the inverse has Jacobian \(\dfrac {1}{2uvw}\), defined wherever none of \(u,v,w\) is zero.

Solution. \[\frac {\partial (x,y,z)}{\partial (\rho ,\phi ,\theta )} = \begin {vmatrix} \sin \phi \cos \theta & \rho \cos \phi \cos \theta & -\rho \sin \phi \sin \theta \\ \sin \phi \sin \theta & \rho \cos \phi \sin \theta & \rho \sin \phi \cos \theta \\ \cos \phi & -\rho \sin \phi & 0 \end {vmatrix}.\] Expanding along the last row and simplifying with \(\cos ^2+\sin ^2 = 1\) throughout gives \[\frac {\partial (x,y,z)}{\partial (\rho ,\phi ,\theta )} = \rho ^2\sin \phi .\] This is the factor that appears in the volume element for spherical coordinates, \(dV = \rho ^2\sin \phi \,d\rho \,d\phi \,d\theta \), and it is why that factor is what it is rather than something to be memorised.

Solution. Take the edges in turn.

On \(v=0\), \(0\leq u\leq 1\): \(x = u^2\), \(y = 0\), giving the segment \(0\leq x\leq 1\) of the \(x-\)axis.

On \(u=1\), \(0\leq v\leq 1\): \(x = 1-v^2\) and \(y = 2v\), so \(v = \dfrac y2\) and \[x = 1-\frac {y^2}{4},\qquad 0\leq y\leq 2 ,\] an arc of a leftward parabola from \((1,0)\) to \((0,2)\).

On \(v=1\), \(0\leq u\leq 1\): \(x = u^2-1\) and \(y = 2u\), so \[x = \frac {y^2}{4}-1,\qquad 0\leq y\leq 2 ,\] an arc of a rightward parabola from \((-1,0)\) to \((0,2)\).

On \(u=0\), \(0\leq v\leq 1\): \(x = -v^2\), \(y = 0\), the segment \(-1\leq x\leq 0\).

The image is therefore the region bounded by the two parabolic arcs and the segment \([-1,1]\) of the \(x-\)axis. The transformation is \(z\mapsto z^2\) in complex notation, which is why the straight edges become parabolas.

Solution.

(a)

The map is linear, so it takes the rectangle to a parallelogram and it is enough to map the four corners: \[(0,0)\mapsto (0,0),\quad (3,0)\mapsto (6,3),\quad (3,2)\mapsto (12,1),\quad (0,2)\mapsto (6,-2).\] The image is the parallelogram with those vertices. Its area is \(\left |\det \right | = \left |2(-1)-3(1)\right | = 5\) times the area \(6\) of \(S\), that is \(30\), which the Jacobian predicts without any drawing.

(b)

The triangle has edges \(v=1\), \(u=0\) and \(v=u\). Under \(x = u^2\), \(y = v\):

\(v=1\) with \(0\leq u\leq 1\) becomes \(y=1\), \(0\leq x\leq 1\).

\(u=0\) becomes \(x=0\), \(0\leq y\leq 1\).

\(v=u\) becomes \(y = u\) and \(x = u^2\), that is \(x = y^2\) with \(0\leq y\leq 1\).

The image is the region bounded above by \(y=1\), on the left by the \(y-\)axis and on the right by the parabola \(x = y^2\). The straight edge \(v=u\) has been bent, because this transformation is not linear.

Solution. A map is linear when every component is a homogeneous polynomial of degree one in the variables — no powers, no products, no constants.

First map: not linear.

The components contain \(x^3\) and \(x^2\). Failure is confirmed concretely: \(F(2,0) = (-8,4)\) but \(2F(1,0) = (-2,2)\), so \(F(2\textbf {v})\neq 2F(\textbf {v})\).

Second map: linear.

Every component is a linear combination, and the map is multiplication by \[\begin {pmatrix}1 & 1\\ 2 & 1\\ 0 & 1\end {pmatrix}.\]

Third map: not linear.

The second component \(xz\) is a product of two variables. Concretely, doubling \((x,y,z)\) multiplies \(xz\) by four, not two.

Solution. The inverse function theorem says a continuously differentiable map is locally invertible wherever its Jacobian determinant is non-zero, and that the Jacobian of the inverse is then the reciprocal.

(a)

\[J = \begin {vmatrix}2 & -3\\ 7 & 2\end {vmatrix} = 4+21 = 25 \neq 0 ,\] constant, so the map is invertible everywhere and \(J_{f^{-1}} = \dfrac {1}{25}\).

(b)

\[J = \begin {vmatrix}1 & -2\\ -3 & 1\end {vmatrix} = 1-6 = -5\neq 0 ,\] again constant, so \(J_{f^{-1}} = -\dfrac 15\) at every point.

(c)

\[J = \begin {vmatrix} 2xy & x^2-z & -y\\ \dfrac {2}{z} & \dfrac {1}{z} & -\dfrac {2x+y}{z^2}\\ 0 & 0 & -6z \end {vmatrix}.\] At \((3,0,1)\) the entries are \[\begin {vmatrix} 0 & 8 & 0\\ 2 & 1 & -6\\ 0 & 0 & -6 \end {vmatrix} = -6\begin {vmatrix}0 & 8\\ 2 & 1\end {vmatrix} = -6(0-16) = 96 \neq 0 ,\] expanding along the last row. So the map is locally invertible at that point and \(J_{f^{-1}}(3,0,1) = \dfrac {1}{96}\).

Solution. Functions are dependent exactly when their Jacobian vanishes identically.

(a)

The Jacobian determinant is identically zero, so they are dependent. The relation is the familiar identity \[u^2 = v + 2w ,\] since \((x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx)\).

(b)

The Jacobian vanishes identically, so they are dependent, and the relation is \[u = \tan v ,\] which is the addition formula \(\tan (A+B) = \dfrac {\tan A+\tan B}{1-\tan A\tan B}\) with \(A = \tan ^{-1}x\) and \(B = \tan ^{-1}y\).

(c)

Both functions depend on \(x\) and \(y\) only through the combination \(s = x+y\): \[u = s^2+2s,\qquad v = e^{s}.\] So the Jacobian vanishes and they are dependent, with \[u = \big (\ln v\big )^2 + 2\ln v .\]