Solution. \[F(3,1) = 1+\sqrt {4-1} = 1+\sqrt 3 .\] The variable \(x\) does not appear, so it is unrestricted; the square root requires \(4-y^2\geq 0\), that is \(-2\leq y\leq 2\). The domain is therefore the horizontal strip \[\big \{(x,y): -2\leq y\leq 2\big \},\] of infinite extent in \(x\). As \(y\) ranges over \([-2,2]\), \(4-y^2\) ranges over \([0,4]\) and its square root over \([0,2]\), so the range is \([1,3]\).

Solution. Three conditions must hold at once. The square roots require \[x\geq 0,\qquad y\geq 0,\qquad z\geq 0 ,\] and the logarithm requires its argument to be strictly positive: \[4-x^2-y^2-z^2 > 0 \iff x^2+y^2+z^2 < 4 .\] The domain is therefore the part of the open ball of radius \(2\) centred at the origin lying in the first octant — one eighth of the ball, with the flat faces included and the spherical face excluded.

Solution.

(a)

A polynomial is continuous everywhere, so substitute: \(5(1) - (1)(4) = 1\).

(b)

The denominator is \(4+3 = 7\neq 0\) at the point, so the quotient is continuous there and the limit is \(\dfrac {4-2}{7} = \dfrac {2}{7}\).

(c)

Both parts vanish at the origin, so test paths. Along \(y=0\), \[\frac {x^4}{x^2} = x^2\longrightarrow 0 .\] Along \(x=0\), \[\frac {-4y^2}{2y^2} = -2\longrightarrow -2 .\] The two values differ, so the limit does not exist.

(d)

Along \(y=0\) the function is identically \(0\). Along the parabola \(y = x^2\), \[\frac {x^2\cdot x^2e^{x^2}}{x^4+3x^4} = \frac {e^{x^2}}{4}\longrightarrow \frac 14 .\] Different paths give different values, so the limit does not exist. Note that every straight line through the origin gives \(0\) here; the parabola is what exposes it, which is why testing lines alone is never enough.

Solution. Away from the origin the function is a quotient of polynomials with non-zero denominator, hence continuous.

At the origin, note that \(\dfrac {x^2}{2x^2+y^2}\leq \dfrac 12\) for all \((x,y)\neq (0,0)\), so \[\left |f(x,y)\right | = \frac {x^2}{2x^2+y^2}\left |y\right |^3 \leq \frac {\left |y\right |^3}{2}\longrightarrow 0 .\] The limit is therefore \(0\), but \(f(0,0) = 1\). Since the limit exists and differs from the value, \(f\) is discontinuous at the origin and the discontinuity is removable: redefining \(f(0,0) = 0\) makes \(f\) continuous everywhere.

The bound is worth noticing — it settles the limit for every path at once, which testing paths individually can never do.

Solution. Let \(\varepsilon >0\). Since \(y^2\leq x^2+y^2\), \[\left |\frac {5xy^2}{x^2+y^2} - 0\right | = 5\left |x\right |\frac {y^2}{x^2+y^2}\leq 5\left |x\right | \leq 5\sqrt {x^2+y^2} .\] So if \(0<\sqrt {x^2+y^2}<\delta \) with \(\delta = \dfrac {\varepsilon }{5}\), then \[\left |\frac {5xy^2}{x^2+y^2}\right | < 5\delta = \varepsilon ,\] which is exactly the definition. Hence the limit is \(0\).

Solution.

(a)

\[f_x = \frac {1 + \dfrac {x}{\sqrt {x^2+y^2}}}{x+\sqrt {x^2+y^2}} = \frac {1}{\sqrt {x^2+y^2}} ,\] after combining the numerator over \(\sqrt {x^2+y^2}\) and cancelling the common factor \(x+\sqrt {x^2+y^2}\). At \((3,4)\), \(\sqrt {9+16} = 5\), so \(f_x(3,4) = \dfrac 15\).

(b)

\(f_x = 4x^3y^2 - 3x^2y\), so \[f_{xx} = 12x^2y^2 - 6xy = 6xy\big (2xy-1\big ).\] Then \(f_{xy} = 8x^3y - 3x^2\) and \[f_{xyx} = 24x^2y - 6x = 6x\big (4xy-1\big ).\]

Solution. For (a) and (b) use implicit differentiation via \(\dfrac {\partial z}{\partial x} = -\dfrac {F_x}{F_z}\).

(a)

With \(F = e^z-xyz\), \[\frac {\partial z}{\partial x} = \frac {yz}{e^z-xy},\qquad \frac {\partial z}{\partial y} = \frac {xz}{e^z-xy}.\]

(b)

With \(F = yz+x\ln y-z^2\), \[\frac {\partial z}{\partial x} = \frac {-\ln y}{y-2z},\qquad \frac {\partial z}{\partial y} = \frac {-\Big (z+\dfrac {x}{y}\Big )}{y-2z} = \frac {-\big (x+yz\big )}{y\big (y-2z\big )}.\]

(c)

The variables separate completely: \(\dfrac {\partial z}{\partial x} = f'(x)\) and \(\dfrac {\partial z}{\partial y} = g'(y)\).

(d)

By the chain rule with inner function \(\dfrac {x}{y}\), \[\frac {\partial z}{\partial x} = \frac {1}{y}f'\Big (\frac xy\Big ),\qquad \frac {\partial z}{\partial y} = -\frac {x}{y^2}f'\Big (\frac xy\Big ).\] Note that \(x\dfrac {\partial z}{\partial x} + y\dfrac {\partial z}{\partial y} = 0\), which is Euler’s theorem for a function homogeneous of degree \(0\).

Solution. Because mixed partials of a smooth function are independent of order, choose the order that does the least work. Differentiating with respect to \(y\) first kills the inverse sine term outright, since it does not involve \(y\): \[f_y = 2xyz^3 .\] Then \[f_{yx} = 2yz^3,\qquad f_{yxz} = 6yz^2 .\] Hence \(f_{xzy} = 6yz^2\).

Taking the order as literally written would require differentiating \(\sin ^{-1}(x\sqrt z)\) twice before the \(y\) step removes it — considerably more work for the same answer.

Solution. Write \(r = \sqrt {x^2+y^2+z^2}\), so \(u = r^{-1}\) and \(\dfrac {\partial r}{\partial x} = \dfrac {x}{r}\). Then \[u_x = -\frac {1}{r^2}\cdot \frac {x}{r} = -\frac {x}{r^3},\] \[u_{xx} = -\frac {1}{r^3} + \frac {3x}{r^4}\cdot \frac {x}{r} = -\frac {1}{r^3} + \frac {3x^2}{r^5}.\] By symmetry the same holds with \(y\) and \(z\), so \[u_{xx}+u_{yy}+u_{zz} = -\frac {3}{r^3} + \frac {3\big (x^2+y^2+z^2\big )}{r^5} = -\frac {3}{r^3} + \frac {3r^2}{r^5} = 0 ,\] for all \((x,y,z)\neq (0,0,0)\). This \(u\) is the Newtonian potential, and the equation is Laplace’s.

Solution. \[T_x = \frac {-120x}{\big (1+x^2+y^2\big )^2},\qquad T_y = \frac {-120y}{\big (1+x^2+y^2\big )^2}.\] At \((2,1)\) the denominator is \((1+4+1)^2 = 36\), so \[T_x(2,1) = \frac {-240}{36} = -\frac {20}{3}\approx -6.67\ ^{\circ }\text {C/m},\] \[T_y(2,1) = \frac {-120}{36} = -\frac {10}{3}\approx -3.33\ ^{\circ }\text {C/m}.\] Both are negative: the plate cools in every outward direction, and twice as fast in \(x\) as in \(y\) because the point is twice as far out along \(x\).

Solution.

(a)

\(z_x = \dfrac {y}{2\sqrt {xy}}\) and \(z_y = \dfrac {x}{2\sqrt {xy}}\), both equal to \(\dfrac 12\) at \((1,1)\). Hence \[z = 1 + \frac 12(x-1) + \frac 12(y-1) = \frac {x+y}{2}.\]

(b)

\(z_x = \dfrac {1}{x-2y}\) and \(z_y = \dfrac {-2}{x-2y}\), which at \((3,1)\) are \(1\) and \(-2\). Hence \[z = 0 + 1(x-3) - 2(y-1) = x-2y-1 .\]

Solution. At \((1,1)\): \(\cos \pi = -1\), so \(f(1,1) = 1+1 = 2\). Now \[f_x = -y\cos \pi y \longrightarrow 1,\] \[f_y = -x\cos \pi y + \pi xy\sin \pi y \longrightarrow 1 + 0 = 1 ,\] since \(\sin \pi = 0\). Hence \[L(x,y) = 2 + 1(x-1) + 1(y-1) = x+y .\] Therefore \[f(1.02, 0.97)\approx 1.02+0.97 = 1.99 ,\] against the true value \(1.98501\), an error of about \(0.005\).

Solution. \[z_x = 2x-y \longrightarrow 7,\qquad z_y = -x+6y \longrightarrow -9 ,\] at \((3,-1)\). With \(dx = -0.04\) and \(dy = 0.05\), \[dz = 7(-0.04) + (-9)(0.05) = -0.28 - 0.45 = -0.73 .\] The exact change is \[\triangle z = z(2.96,-0.95) - z(3,-1) = 12.1391 - 12.8571\ldots = -0.7189 .\] The differential is within \(0.011\) of the true change, and the discrepancy is second order in the increments, as expected.

Solution. First locate the parameter values: \(\textbf {r}_1(0) = \langle 2,1,3\rangle \) and \(\textbf {r}_2(1) = \langle 2,1,3\rangle \), so \(t=0\) and \(u=1\).

Both tangent vectors lie in the tangent plane: \[\textbf {r}_1{}'(t) = \langle 3,\ -2t,\ -4+2t\rangle \longrightarrow \langle 3,0,-4\rangle ,\] \[\textbf {r}_2{}'(u) = \langle 2u,\ 6u^2,\ 2\rangle \longrightarrow \langle 2,6,2\rangle .\] A normal is therefore \[\underline {n} = \langle 3,0,-4\rangle \times \langle 2,6,2\rangle = \langle 24,\ -14,\ 18\rangle ,\] which may be divided by \(2\) to give \(\langle 12,-7,9\rangle \). The plane through \(P(2,1,3)\) is \[12(x-2) - 7(y-1) + 9(z-3) = 0 \implies 12x - 7y + 9z = 44 .\] The surface itself is never needed: two curves through the point supply the plane.

Solution. At \((0,2)\) the intermediate variables take the values \[u = 0,\qquad v = 2,\qquad w = 1 ,\] so \(P = \sqrt {0+4+1} = \sqrt 5\). The chain rule gives \[\frac {\partial P}{\partial x} = \frac {u u_x + v v_x + w w_x}{\sqrt {u^2+v^2+w^2}},\] and similarly for \(y\). The partials of the intermediate variables at \((0,2)\) are \[u_x = e^y = e^2,\quad v_x = ye^x = 2,\quad w_x = ye^{xy} = 2,\] \[u_y = xe^y = 0,\quad v_y = e^x = 1,\quad w_y = xe^{xy} = 0 .\] Hence \[\frac {\partial P}{\partial x} = \frac {0\cdot e^2 + 2\cdot 2 + 1\cdot 2}{\sqrt 5} = \frac {6}{\sqrt 5} = \frac {6\sqrt 5}{5}\approx 2.683 ,\] \[\frac {\partial P}{\partial y} = \frac {0\cdot 0 + 2\cdot 1 + 1\cdot 0}{\sqrt 5} = \frac {2}{\sqrt 5} = \frac {2\sqrt 5}{5}\approx 0.894 .\] The term with \(u = 0\) drops out, which is why \(e^2\) never appears in the answer.

Solution. Implicit differentiation of \(F = 0\) gives each partial as a ratio of the partials of \(F\), with a minus sign, the variable held fixed being the third one in each case: \[\frac {\partial z}{\partial x} = -\frac {F_x}{F_z},\qquad \frac {\partial x}{\partial y} = -\frac {F_y}{F_x},\qquad \frac {\partial y}{\partial z} = -\frac {F_z}{F_y}.\] Multiplying, \[\frac {\partial z}{\partial x}\frac {\partial x}{\partial y} \frac {\partial y}{\partial z} = \Big (-\frac {F_x}{F_z}\Big )\Big (-\frac {F_y}{F_x}\Big )\Big (-\frac {F_z}{F_y}\Big ) = -\frac {F_xF_yF_z}{F_xF_yF_z} = -1 ,\] the three minus signs producing one. The hypothesis that none of the partials vanishes is what allows the cancellation.

The result is a standard warning: partial derivatives are not fractions, and treating them as such would give \(+1\) here.

Solution. Put \(F(x,y) = x^2-3xy+y^3-7\). First, the point lies on the curve: \[F(4,3) = 16 - 36 + 27 - 7 = 0 .\] Next, \(F\) is a polynomial, hence continuously differentiable, and \[F_y = -3x+3y^2 ,\qquad F_y(4,3) = -12+27 = 15 \neq 0 .\] By the implicit function theorem these two facts guarantee that near \((4,3)\) the equation determines \(y\) uniquely as a differentiable function of \(x\).

Differentiating, \[\frac {dy}{dx} = -\frac {F_x}{F_y} = -\frac {2x-3y}{-3x+3y^2} ,\] and at \((4,3)\), \(F_x = 8-9 = -1\), so \[\frac {dy}{dx} = -\frac {-1}{15} = \frac {1}{15}.\]

Solution.

(a)

\[f_x = -2-2x = 0 \implies x = -1,\qquad f_y = 4-8y = 0 \implies y = \frac 12 .\] The second partials are constant: \(f_{xx} = -2\), \(f_{yy} = -8\), \(f_{xy} = 0\), so \[D = (-2)(-8) - 0 = 16 > 0 ,\] and \(f_{xx}<0\). Hence \(\Big (-1,\dfrac 12\Big )\) is a local maximum, with value \[f\Big (-1,\frac 12\Big ) = 9+2+2-1-1 = 11 .\]

(b)

\[f_x = \cos x\sin y = 0,\qquad f_y = \sin x\cos y = 0 .\] Inside the square the solutions are the four points \(\Big (\pm \dfrac {\pi }{2},\pm \dfrac {\pi }{2}\Big )\), together with \((0,0)\) and points on the boundary.

The second partials are \(f_{xx} = -\sin x\sin y\), \(f_{yy} = -\sin x\sin y\) and \(f_{xy} = \cos x\cos y\), so \(D = \sin ^2x\sin ^2y - \cos ^2x\cos ^2y\).

At \(\Big (\dfrac {\pi }{2},\dfrac {\pi }{2}\Big )\) and \(\Big (-\dfrac {\pi }{2},-\dfrac {\pi }{2}\Big )\): \(D = 1>0\) and \(f_{xx} = -1<0\), so these are local maxima with \(f = 1\).

At \(\Big (\dfrac {\pi }{2},-\dfrac {\pi }{2}\Big )\) and \(\Big (-\dfrac {\pi }{2},\dfrac {\pi }{2}\Big )\): \(D = 1>0\) and \(f_{xx} = 1>0\), so these are local minima with \(f = -1\).

At \((0,0)\): \(D = -1<0\), so it is a saddle point, with \(f = 0\).

Solution. On a closed bounded region the extremes occur either at an interior critical point or on the boundary, so both must be examined.

Interior

\(f_x = 2x-2 = 0\) and \(f_y = 2y = 0\) give the single point \((1,0)\), which lies inside the triangle, with \(f(1,0) = -1\).

The edge \(x=0\), \(-2\leq y\leq 2\)

Here \(f = y^2\), which runs from \(0\) at \((0,0)\) up to \(4\) at \((0,\pm 2)\).

The edge from \((2,0)\) to \((0,2)\), where \(y = 2-x\)

\[f = x^2+(2-x)^2-2x = 2x^2-6x+4 ,\] minimised where \(4x-6 = 0\), that is \(x = \dfrac 32\), giving \(f = -\dfrac 12\). The endpoints give \(f(2,0) = 0\) and \(f(0,2) = 4\).

The edge from \((2,0)\) to \((0,-2)\), where \(y = x-2\)

By the symmetry of \(f\) in \(y\), this gives the same values.

Conclusion

Comparing all candidates, the absolute minimum is \(-1\) at \((1,0)\), and the absolute maximum is \(4\), attained at both \((0,2)\) and \((0,-2)\).