The velocity is \(\mathbf r'(t)=(2t,\ 2,\ 1/t)\) , and

so for \(t>0\) the speed is \(2t+1/t\) . The distance travelled is \(\int_1^e(2t+1/t)\,dt=\bigl[t^{2}+\ln t\bigr]_1^e=(e^{2}+1)-(1+0)\) , which is \(e^{2}\) .

For the curvature at \(t=1\) : \(\mathbf r'(1)=(2,2,1)\) , of length \(3\) , and \(\mathbf r''(t)=(2,0,-1/t^{2})\) gives \(\mathbf r''(1)=(2,0,-1)\) . The cross product is \((2\cdot(-1)-1\cdot 0,\ 1\cdot 2-2\cdot(-1),\ 2\cdot 0-2\cdot 2)=(-2,4,-4)\) , of length \(6\) . By Theorem 1.10, \(\kappa=6/3^{3}\) , which is \(\tfrac29\) .

Let \(F(x,y,z)=xz^{3}+y^{2}z-2xy\) . At \((1,1,1)\) , \(F=1+1-2=0\) , so the point is on the surface. The partial derivatives are \(F_x=z^{3}-2y=-1\) , \(F_y=2yz-2x=0\) , and \(F_z=3xz^{2}+y^{2}=4\) . Since \(F_z\ne 0\) , the implicit function theorem shows that near the point the surface is the graph of a \(C^1\) function \(z=\varphi(x,y)\) , with

So \(\partial z/\partial x\) is \(\tfrac14\) and \(\partial z/\partial y\) is \(0\) . The tangent plane has normal \(\nabla F=(-1,0,4)\) : \(-(x-1)+4(z-1)=0\) , that is, \(x-4z=-3\) . Solving for \(z\) gives \(z=1+\tfrac14(x-1)\) , consistent with the two partial derivatives.

The partial derivatives are \(f_x=2xy-2y=2y(x-1)\) and \(f_y=x^{2}-2x+2y\) . From \(f_x=0\) , either \(y=0\) or \(x=1\) . If \(y=0\) , then \(f_y=x(x-2)=0\) gives \((0,0)\) and \((2,0)\) . If \(x=1\) , then \(f_y=-1+2y=0\) gives \((1,\tfrac12)\) .

The Hessian is \(\begin{pmatrix}2y&2x-2\\ 2x-2&2\end{pmatrix}\) . At \((0,0)\) and at \((2,0)\) its determinant is \(0\cdot 2-4\) , that is, \(-4\) : both are saddle points. At \((1,\tfrac12)\) the Hessian is diagonal with entries \(1\) and \(2\) , positive definite: a strict local minimum, with value \(\tfrac12-1+\tfrac14\) , that is, \(-\tfrac14\) .

There is no global extremum. Write \(f=y^{2}+y\bigl((x-1)^{2}-1\bigr)\) . With \(y=-1\) this is \(2-(x-1)^{2}\) , which tends to \(-\infty\) as \(x\to\infty\) ; with \(x=1\) it is \(y^{2}-y\) , which tends to \(+\infty\) . The local minimum is not global.

Minimize and maximize \(x^{2}+y^{2}\) subject to \(g=5x^{2}-6xy+5y^{2}=8\) . The constraint set is an ellipse, closed and bounded, so both extremes exist, and \(\nabla g=(10x-6y,\ 10y-6x)\) vanishes only at the origin. The multiplier equations \(2x=\lambda(10x-6y)\) , \(2y=\lambda(10y-6x)\) say that \((x,y)\) is an eigenvector of \(A=\begin{pmatrix}5&-3\\ -3&5\end{pmatrix}\) , the matrix of the quadratic form, with eigenvalue \(1/\lambda\) .

The eigenvalues of \(A\) are \(2\) , with eigenvector \((1,1)\) , and \(8\) , with eigenvector \((1,-1)\) . A point \(s(1,1)\) lies on the ellipse when \(2\cdot 2s^{2}=8\) , so \(s=\pm\sqrt2\) : the points \(\pm(\sqrt2,\sqrt2)\) , at distance \(2\) from the origin. A point \(s(1,-1)\) lies on it when \(8\cdot 2s^{2}=8\) , so \(s=\pm1/\sqrt2\) : the points \(\pm(1/\sqrt2,-1/\sqrt2)\) , at distance \(1\) .

So the farthest points are \(\pm(\sqrt2,\sqrt2)\) and the nearest are \(\pm(1/\sqrt2,-1/\sqrt2)\) . The semi-axes of the ellipse \(\mathbf x^{\top}A\mathbf x=8\) lie along the eigenvectors of \(A\) and have lengths \(\sqrt{8/\mu}\) for the eigenvalues \(\mu\) : the smallest eigenvalue gives the longest axis.

The inner integral cannot be done in elementary terms, so reverse the order. The limits describe \(0\le y\le 1\) , \(\sqrt y\le x\le 1\) : the region below the parabola \(y=x^{2}\) , above the \(x\) -axis, and to the left of \(x=1\) . Sliced vertically it is \(0\le x\le 1\) , \(0\le y\le x^{2}\) . Hence

which is \(\tfrac29\bigl(2\sqrt2-1\bigr)\approx 0.406\) . The inner integration supplied the factor \(x^{2}\) that makes the substitution \(w=1+x^{3}\) possible.

The surfaces meet where \(x^{2}+y^{2}=2y\) , that is, \(x^{2}+(y-1)^{2}=1\) , a circle of radius \(1\) centred at \((0,1)\) . Over the disc \(D\) it bounds, the plane lies above the paraboloid, so the volume is \(\iint_D(2y-x^{2}-y^{2})\,dA\) .

In polar coordinates the circle is \(r^{2}=2r\sin\theta\) , that is, \(r=2\sin\theta\) , traced once for \(0\le\theta\le\pi\) , and the integrand is \(2r\sin\theta-r^{2}\) . With the Jacobian factor,

Since \(\int_0^{\pi}\sin^{4}\theta\,d\theta=\tfrac{3\pi}{8}\) , the volume is \(\pi/2\) .

Put \(u=x+y\) and \(v=x-2y\) . The four lines become \(u=0\) , \(u=2\) , \(v=0\) , \(v=3\) , so \(D\) corresponds to the rectangle \(0\le u\le 2\) , \(0\le v\le 3\) , and the map is linear and one-to-one. Its Jacobian is

which is \(-3\) , so by Proposition 7.3 the factor required is \(\lvert\partial(x,y)/\partial(u,v)\rvert=\tfrac13\) . The integrand is \(x+y=u\) . Therefore

The value is \(2\) . As a check on the factor, the area of \(D\) comes out as \(\tfrac13\cdot 6=2\) , so the average of \(x+y\) over \(D\) is \(1\) , the midpoint of its range \([0,2]\) .

On the disc \(z=1\) , oriented upward, the normal is \(\mathbf k\) and \(\mathbf F\cdot\mathbf k=z+1=2\) , a constant. The disc has area \(\pi\) , so the upward flux through it is \(2\pi\) .

For the closed surface use the divergence theorem. The field is polynomial and \(\nabla\cdot\mathbf F=2x-2x+1=1\) , so the total outward flux is the volume of the cylinder: the outward flux is \(\pi\) .

The pieces can be checked separately. The top contributes \(2\pi\) . On the bottom, \(z=0\) , the outward normal is \(-\mathbf k\) and \(\mathbf F\cdot(-\mathbf k)=-1\) , contributing \(-\pi\) . On the side, the outward normal at \((\cos\theta,\sin\theta,z)\) is \((\cos\theta,\sin\theta,0)\) and \(\mathbf F\cdot\mathbf n=\cos^{3}\theta-2\cos\theta\sin^{2}\theta\) , which integrates to \(0\) over a full turn. The total is \(2\pi-\pi+0\) , as the theorem requires.

Use Stokes’ theorem with \(S\) the part of the plane \(z=2-y\) inside the cylinder, oriented upward; its boundary is \(C\) with the stated orientation. The curl of \(\mathbf F=(-y^{2},x,z^{2})\) is \((0-0,\ 0-0,\ 1+2y)\) , that is, \((0,0,1+2y)\) . The surface is the graph of \(g(x,y)=2-y\) over the unit disc \(D\) , with normal vector \((-g_x,-g_y,1)=(0,1,1)\) , so

since the integral of \(y\) over the disc vanishes by symmetry. The circulation is \(\pi\) . A direct computation with \(\mathbf r(t)=(\cos t,\sin t,2-\sin t)\) leads to the integrand \(\sin^{3}t+\cos^{2}t-(2-\sin t)^{2}\cos t\) , whose integral over a full period is also \(\pi\) , but with more work.

For a unit vector \((a,b)\) and \(t\ne 0\) ,

If \(b\ne 0\) the denominator tends to \(b^{2}\ne 0\) and the quotient tends to \(0\) . If \(b=0\) the numerator is \(0\) for every \(t\) . So every directional derivative at the origin is \(0\) , and in particular \(\nabla f(0,0)=(0,0)\) .

If \(f\) were differentiable at the origin, the quotient \(f(\mathbf h)/\lVert\mathbf h\rVert\) of Definition 2.3 would tend to \(0\) along every path. Along the parabola \(y=x^{2}\) with \(x>0\) , \(f(x,x^{2})=x^{5}/(2x^{4})\) , which is \(x/2\) , while \(\lVert(x,x^{2})\rVert=x\sqrt{1+x^{2}}\) . The quotient is \(1/(2\sqrt{1+x^{2}})\) , which tends to \(\tfrac12\) , not \(0\) . So \(f\) is not differentiable at the origin, although it is continuous there and all its directional derivatives exist and fit the formula \(\nabla f\cdot\mathbf u\) . Even that is not enough: the linear approximation must be good along curved approaches too.

The error is the use of the divergence theorem on the solid ellipsoid, which contains the origin, where \(\mathbf F\) is undefined. The theorem requires the field to be \(C^1\) on an open set containing the whole solid, and the identity \(\nabla\cdot\mathbf F=0\) , valid away from the origin, says nothing about what happens at the singular point.

To find the correct value, choose a sphere of radius \(\varepsilon\) about the origin so small that it lies inside the ellipsoid; any \(\varepsilon<2\) works, since the shortest semi-axis is \(2\) . On the region between the sphere and the ellipsoid the field is \(C^1\) with zero divergence, so by the divergence theorem the outward flux through the ellipsoid minus the outward flux through the sphere is \(0\) . By Example 9.9 the latter is \(4\pi\) . The correct value is \(4\pi\) , the same as for a sphere, and no parametrization of the ellipsoid was needed.