Cumulative assessment: Calculus III

Substitute the whole parameterization into the plane

The line has $x=1+t$, $y=2t$, $z=3t$. Substitution gives $2(1+t)-2t+3t=4$, hence $t=2/3$. The intersection is $\boxed{(5/3,4/3,2)}$. The plane normal is $(2,-1,1)$, of length $\sqrt6$. Its distance from the origin is $\boxed{4/\sqrt6}$. The normal’s dot product with the line direction is $3\ne0$, so the line is not parallel to the plane and the intersection is unique.

Keep position, tangent direction, and bending separate

Velocity is $(-2\sin t,2\cos t,1)$ with norm $\boxed{\sqrt5}$. Acceleration is $(-2\cos t,-2\sin t,0)$. Their cross product is $(2\sin t,-2\cos t,4)$, with norm $2\sqrt5$, so curvature is $\boxed{\kappa=2/5}$. At zero, position is $(2,0,0)$ and velocity is $(0,2,1)$. A tangent line is $\boxed{\vv\ell(s)=(2,0,0)+s(0,2,1)}$. The new parameter $s$ labels points on the line and need not equal the curve parameter away from the touching point.

A stronger remainder bound proves differentiability

Let $r=\sqrt{x^2+y^2}$. Since $(x^2-y^2)^2\ge0$, $4x^2y^2\le(x^2+y^2)^2=r^4$. Therefore, for $r>0$, $|f(x,y)|\le r^2/4\to0$, proving continuity at the defined value zero. Along either coordinate axis, $f$ is zero, so both partial derivatives at the origin are zero. To prove full differentiability with the zero linear map, examine

Thus the derivative is $\boxed{Df(0,0)=0}$, equivalently gradient $(0,0)$. Partial derivatives alone would not have established this; the uniform first-order remainder bound does.

The same gradient gives a plane and directional rates

The partial derivatives are $f_x=ye^{xy}$ and $f_y=xe^{xy}$. At $(1,0)$, $f=1$ and $\grad f=(0,1)$. The tangent plane is $\boxed{z=1+y}$. Normalize $(3,4)$ to $(3/5,4/5)$; the directional derivative is $(0,1)\cdot(3/5,4/5)=\boxed{4/5}$. Using the unnormalized direction would multiply the rate by the vector’s length and answer a different parameter-speed question.

Account for both changing inputs

The partial derivatives are $z_x=2xy+\sin y$ and $z_y=x^2+x\cos y$. At $t=1$, $(x,y)=(1,0)$, so $z_x=0$ and $z_y=2$. Also $x'=2t=2$ and $y'=1/t=1$ at that instant. Thus

A direct check is available from $z=t^4\ln t+t^2\sin(\ln t)$: its derivative at $1$ is $1+1=2$.

A closed region requires interior and boundary reasoning

The continuous function attains extrema on the compact ball. Its gradient $(1,1,1)$ is never zero, so there is no interior stationary candidate. On the boundary sphere, the multiplier equations are $(1,1,1)=2\lambda(x,y,z)$. They force equal coordinates; using the sphere gives points $\pm(1,1,1)/\sqrt3$, with values $\pm\sqrt3$. An independent global check is Cauchy–Schwarz:

Equality is attained at those two points. The greatest value is $\boxed{\sqrt3}$ and least $\boxed{-\sqrt3}$.

A changed slice width supplies the missing differential factor

The region is $0\le x\le2$ and $x/2\le y\le1$, equivalently $0\le y\le1$ and $0\le x\le2y$. Thus

With $u=y^3$, $\dd u=3y^2\dd y$, this becomes $\tfrac23\int_0^1\cos u\dd u=\boxed{(2/3)\sin1}$. The new region is the same triangle, not the rectangle obtained by leaving the old bounds unchanged.

A cylindrical Jacobian multiplies the integrand

Use $0\le\theta\le2\pi$, $0\le r\le1$, and $0\le z\le4-r^2$, with $\dd V=r\dd z\dd r\dd\theta$. Therefore

The integrand is nonnegative on the solid. Treating the height bound as a constant would replace the paraboloid by a cylinder and change the object being integrated.

Change the function and the area element

The Jacobian matrix is $\begin{pmatrix}1&-1\\2&1\end{pmatrix}$ with determinant $3$. It is nonzero, so this linear map is one-to-one. Also $x+y=3u$. Hence

The four corner images are $(0,0)$, $(1,2)$, $(-2,2)$, and $(-1,4)$. The parallelogram’s area is $6$ and the affine integrand’s value at its center is $3/2$, giving the independent check $6\cdot3/2=9$.

An explicit potential is stronger evidence than curl alone

The function $\phi(x,y,z)=xyz$ has gradient $(yz,xz,xy)=\vv F$ throughout $\R^3$. For any piecewise smooth path between the endpoints, the fundamental theorem for line integrals gives

The explicit globally defined potential proves path independence. A zero curl on a punctured or otherwise unsuitable domain would not by itself justify the same conclusion without further domain reasoning.

Use a compatible orientation and then check the boundary directly

The upward normal induces counterclockwise orientation when viewed from above. The field is smooth everywhere and its curl is $(0,0,2)$. Stokes’ theorem lets us replace the cap by the unit disk in $z=0$ with upward normal, obtaining $\oint\vv F\cdot\dd\vv r=\iint_D2\dd A=\boxed{2\pi}$. For a direct check, parameterize the boundary by $(\cos t,\sin t,0)$, $0\le t\le2\pi$. The field and tangent vector there are both $(-\sin t,\cos t,0)$, so their dot product is $1$. Its integral is again $2\pi$. Reversing the boundary orientation would reverse the sign.

Close the surface, compute the total, then remove the cap

The surfaces meet at $r=\sqrt2$. Cylindrical bounds for the solid are $0\le\theta\le2\pi$, $0\le r\le\sqrt2$, and $r^2\le z\le2$. Its volume is

The field is continuously differentiable everywhere and has divergence $3$, so total outward flux through the closed boundary is $3V=6\pi$. The top disk has outward normal $+\vv k$, so its flux is $2$ times disk area $2\pi$, or $4\pi$. The requested curved flux is therefore $\boxed{6\pi-4\pi=2\pi}$. The solid lies above the paraboloid, so the curved boundary’s outward normal points downward. For a direct check its oriented area vector is $(2x,2y,-1)\dd A$. On $z=x^2+y^2=r^2$, the dot product with $\vv F$ is $2r^2-r^2=r^2$, whose disk integral is $2\pi$. Upward orientation would have produced the negative of the requested curved flux.