Multivariable calculus – university Year 1 maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Treat \(y\) as a constant for \(f_x\): \(f_x=\) \(2xy\). Treat \(x\) as a constant for \(f_y\): \(f_y=\) \(x^2+9y^2\).
   2. \(g_x=2xy^3\), \(g_y=3x^2y^2\). Then \(g_{xx}=2y^3\), \(g_{yy}=6x^2y\), \(g_{xy}=\dfrac\partial{\partial y}(2xy^3)=6xy^2\) and \(g_{yx}=\dfrac\partial{\partial x}(3x^2y^2)=6xy^2\).
   3. \(h_x=y\cos(xy)\) and \(h_y=x\cos(xy)\). Then \(h_{xy}=\cos(xy)-xy\sin(xy)\) and \(h_{yx}=\cos(xy)-xy\sin(xy)\). They are equal, as Clairaut's theorem predicts for continuous second derivatives.

2. Answer to Question 2

   1. \(\nabla f=(2xy,\,x^2)\), which is \((4,1)\) at \((1,2)\). The unit vector is \(\mathbf u=\left(\frac35,\frac45\right)\). So \(D_{\mathbf u}f=4\cdot\frac35+1\cdot\frac45=\dfrac{16}5=\) 3.2.
   2. \(\nabla f=(yz,\,xz,\,xy)=(4,2,2)\) at \((1,2,2)\). The greatest rate of increase is \(|\nabla f|=\sqrt{16+4+4}=\sqrt{24}\approx\) 4.90, in the direction of the gradient.

3. Answer to Question 3

   1. \(z_x=2x=2\) and \(z_y=2y=2\) at \((1,1)\). The tangent plane is \(z-2=2(x-1)+2(y-1)\), that is \(z=2x+2y-2\).
   2. Chain rule: \(\dfrac{dz}{dt}=z_x\dfrac{dx}{dt}+z_y\dfrac{dy}{dt}=2xy\cdot2t+x^2\cdot3t^2=2t^5\cdot2t+t^4\cdot3t^2=4t^6+3t^6=7t^6\). Directly: \(z=t^4\cdot t^3=t^7\), so \(\dfrac{dz}{dt}=\) \(7t^6\). The two methods agree.
   3. \(\dfrac{dw}{dt}=2x(-\sin t)+2y\cos t=-2\cos t\sin t+2\sin t\cos t=\) 0. The point moves on the unit circle where \(w=x^2+y^2=1\) is constant.

4. Answer to Question 4

   1. \(f_x=2x-2=0\) and \(f_y=2y+4=0\) give \((1,-2)\). \(f_{xx}=2\), \(f_{yy}=2\), \(f_{xy}=0\), so \(D=f_{xx}f_{yy}-f_{xy}^2=4>0\) with \(f_{xx}>0\): a local minimum with value \(1+4-2-8=-5\).
   2. \(g_x=2x\), \(g_y=-2y\) give \((0,0)\). \(g_{xx}=2\), \(g_{yy}=-2\), so \(D=-4<0\): a saddle point.
   3. \(k_x=3x^2-3=0\) gives \(x=\pm1\) and \(k_y=2y=0\) gives \(y=0\). Critical points \((1,0)\), \((-1,0)\). \(k_{xx}=6x\), \(k_{yy}=2\), \(k_{xy}=0\). At \((1,0)\): \(D=12>0\), \(k_{xx}=6>0\): minimum (\(k=-2\)). At \((-1,0)\): \(D=-12<0\): saddle (\(k=2\)).

5. Answer to Question 5

   1. Set \(\nabla f=\lambda\nabla g\) with \(g=x+y\): \((y,x)=\lambda(1,1)\), so \(y=x=\lambda\). The constraint gives \(x=y=5\) and the maximum is 25.
   2. \((1,1)=\lambda(2x,2y)\) gives \(x=y=\dfrac1{2\lambda}\). Then \(2x^2=1\), so \(x=y=\pm\dfrac1{\sqrt2}\). The maximum is \(\dfrac2{\sqrt2}=\) \(\sqrt2\) at \(\left(\frac1{\sqrt2},\frac1{\sqrt2}\right)\), and the minimum is \(-\sqrt2\) at the opposite point.

6. Answer to Question 6

   1. Inner: \(\int_0^2xy\,dy=2x\). Outer: \(\int_0^12x\,dx=\) 1.
   2. Inner: \(\left[xy+\dfrac{y^2}2\right]_0^x=x^2+\dfrac{x^2}2=\dfrac{3x^2}2\). Outer: \(\int_0^1\dfrac{3x^2}2dx=\) \(\dfrac12\).
   3. The region is \(0\le y\le x\le1\). Reversing: \(\int_0^1\int_0^xe^{x^2}dy\,dx=\int_0^1xe^{x^2}dx=\left[\tfrac12e^{x^2}\right]_0^1=\) \(\dfrac{e-1}2\).
   4. Polar: \(\int_0^{2\pi}\int_0^2r^2\cdot r\,dr\,d\theta=2\pi\cdot\dfrac{2^4}4=\) \(8\pi\).

7. Answer to Question 7

   1. \(h(3,2)=100-9-8=\) 83 m.
   2. \(\nabla h=(-2x,-4y)=(-6,-8)\) at \((3,2)\). The steepest ascent is in the direction \((-6,-8)\), that is towards the south-west, towards the summit at the origin.
   3. \(|\nabla h|=\sqrt{36+64}=\) 10 metres of height per hundred metres horizontally (gradient \(0.1\)).
   4. Walking east is the direction \((1,0)\): \(D=\nabla h\cdot(1,0)=\) \(-6\), so the height drops by 6 m per 100 m.

8. Answer to Question 8

   1. \(P=10\times100^{0.6}\times50^{0.4}\approx\) 757.9 units.
   2. \(P_L=6L^{-0.4}K^{0.4}\approx\) 4.55 and \(P_K=4L^{0.6}K^{-0.6}\approx\) 6.06. One extra labour hour adds about 4.55 units of output; one extra machine hour adds about 6.06.
   3. \(\Delta P\approx P_L\,\Delta L+P_K\,\Delta K=4.55\times2+6.06\times1\approx\) 15.2 units.
   4. \(LP_L+KP_K=6L^{0.6}K^{0.4}+4L^{0.6}K^{0.4}=10L^{0.6}K^{0.4}=P\). Numerically \(100(4.547)+50(6.063)=454.7+303.1=757.8\).

9. Answer to Question 9

   1. The base has area \(x^2\) and the four sides have total area \(4xz\). The constraint is the volume.
   2. \(z=\dfrac{32}{x^2}\), so \(S=x^2+\dfrac{128}x\). \(S'=2x-\dfrac{128}{x^2}=0\) gives \(x^3=64\), so \(x=4\). \(S''=2+\dfrac{256}{x^3}>0\): a minimum.
   3. With \(S=x^2+4xz\) and \(g=x^2z\): \((2x+4z,\,4x)=\lambda(2xz,\,x^2)\) in \((x,z)\). The second component gives \(\lambda=4/x\); the first then gives \(2x+4z=8z\), so \(x=2z\). With \(x^2z=32\): \(4z^3=32\), \(z=2\), \(x=4\). The same solution.
   4. The base is \(4\times4\) m and the height is \(2\) m. \(S=16+4\times4\times2=\) 48 m\(^2\), costing \(48\times12=\) £576.

10. Answer to Question 10

    1. \(V=\pi(25)(12)=300\pi\approx\) 942.5 cm\(^3\).
    2. \(V_r=2\pi rh=120\pi\) and \(V_h=\pi r^2=25\pi\).
    3. \(|dV|\le120\pi(0.1)+25\pi(0.2)=12\pi+5\pi=17\pi\approx\) 53.4 cm\(^3\).
    4. \(\dfrac{17\pi}{300\pi}\approx\) 5.7%. The radius contributes \(12\pi\) of the \(17\pi\), so it is the more important measurement to improve.

11. Answer to Question 11

    1. \(P=\dfrac{8.31\times300}{0.025}=\) 99 720 Pa.
    2. \(\dfrac{\partial P}{\partial T}=\dfrac RV=332.4\) Pa/K (pressure rises 332 Pa for each kelvin at fixed volume). \(\dfrac{\partial P}{\partial V}=-\dfrac{RT}{V^2}=-3.99\times10^6\) Pa/m\(^3\) (pressure falls when volume increases).
    3. \(dP\approx332.4(2)+(-3.989\times10^6)(-0.001)=664.8+3989=\) 4 654 Pa.
    4. From \(PV=RT\): \(\dfrac{\partial P}{\partial V}=-\dfrac PV\), \(\dfrac{\partial V}{\partial T}=\dfrac RP\), \(\dfrac{\partial T}{\partial P}=\dfrac VR\). The product is \(-\dfrac PV\cdot\dfrac RP\cdot\dfrac VR=-1\). It is *not* \(+1\) as the cancellation of "fractions" might suggest.

12. Answer to Question 12

    1. Total spending is \(3x+2y\), which must equal 60.
    2. \(\nabla U=\lambda\nabla g\) gives \((y,x)=\lambda(3,2)\), so \(y=3\lambda\) and \(x=2\lambda\). Substituting into the budget: \(3(2\lambda)+2(3\lambda)=12\lambda=60\), so \(\lambda=5\), \(x=10\), \(y=15\). Satisfaction \(U=\) 150.
    3. \(\lambda=5\) means that one extra pound of budget increases the maximum satisfaction by about 5 units (the marginal value of money).
    4. With £61: \(12\lambda=61\), \(\lambda=\dfrac{61}{12}\), \(x=\dfrac{61}6\), \(y=\dfrac{61}4\), \(U=\dfrac{61^2}{24}\approx155.04\). The increase is \(5.04\), close to the multiplier 5.

13. Answer to Question 13

    1. \(r(0,0)=\) 2 mm and \(r(4,3)=2+4+3=\) 9 mm.
    2. Inner: \(\int_0^3(2+x+y)\,dy=6+3x+\dfrac92=10.5+3x\). Outer: \(\int_0^4(10.5+3x)\,dx=42+24=\) 66.
    3. \(66\times1000=\) 66 000 m\(^3\) of water.
    4. The district has area \(12\) km\(^2\), so the mean depth is \(\dfrac{66}{12}=\) 5.5 mm.

14. Answer to Question 14

    1. The region is a disc and the height depends only on \(x^2+y^2=r^2\). So \(V=\displaystyle\int_0^{2\pi}\!\!\int_0^2(4-r^2)\,r\,dr\,d\theta\).
    2. Inner: \(\int_0^2(4r-r^3)\,dr=\left[2r^2-\dfrac{r^4}4\right]_0^2=8-4=4\). Outer: \(2\pi\times4=\) \(8\pi\approx25.1\) m\(^3\).
    3. The maximum height is \(z=4\) m at the centre. The base has radius 2 m.
    4. The base area is \(\pi\cdot2^2=4\pi\), so the mean height is \(\dfrac{8\pi}{4\pi}=\) 2 m.

15. Answer to Question 15

    1. \(f_x=3x^2-3y=0\) and \(f_y=3y^2-3x=0\). So \(y=x^2\) and \(x=y^2\), giving \(x=x^4\), that is \(x(x^3-1)=0\). Critical points: \((0,0)\) and \((1,1)\).
    2. \(f_{xx}=6x\), \(f_{yy}=6y\), \(f_{xy}=-3\), so \(D=36xy-9\).
    3. At \((0,0)\): \(D=-9<0\), a saddle (the pass). At \((1,1)\): \(D=27>0\) and \(f_{xx}=6>0\), a local minimum (the valley) with height \(-1\) m.
    4. As \(x\to-\infty\) with \(y=0\), \(f=x^3\to-\infty\), so the surface has no global minimum. At \((-1,-1)\) the value is \(-1-1-3=-5\), as in the table, but \((-1,-1)\) is not a critical point: the surface keeps falling beyond it.

16. Answer to Question 16

    1. Minimising \(f=x^2+y^2\) is the same as minimising the distance, since distance is the square root.
    2. \((2x,2y)=\lambda(3,4)\) gives \(x=\dfrac{3\lambda}2\), \(y=2\lambda\). The constraint: \(\dfrac{9\lambda}2+8\lambda=25\), so \(\dfrac{25\lambda}2=25\) and \(\lambda=2\). The router goes at \((3,4)\).
    3. The distance is \(\sqrt{9+16}=\) 5 m. The formula gives \(\dfrac{|25|}{\sqrt{3^2+4^2}}=\dfrac{25}5=5\), the same.
    4. Maximise \(x^2+y^2\) subject to \(x^2+4y^2=25\): \((2x,2y)=\mu(2x,8y)\). Either \(x=0\) (then \(y=\pm2.5\), \(f=6.25\)) or \(y=0\) (then \(x=\pm5\), \(f=25\)). The farthest points are \((\pm5,0)\), 5 m away.

17. Answer to Question 17

    1. \(T=\{(x,y):0\le y\le x\le1\}\): the triangle below the line \(y=x\) and left of \(x=1\).
    2. The inner integral \(\int e^{x^2}dx\) has no elementary antiderivative.
    3. In the other order, \(x\) goes from 0 to 1 and \(y\) from 0 to \(x\): \(I=\int_0^1\!\!\int_0^xe^{x^2}dy\,dx=\int_0^1xe^{x^2}dx=\left[\tfrac12e^{x^2}\right]_0^1=\) \(\dfrac{e-1}2\approx0.859\).
    4. Area \(=\dfrac12\). Average \(=\dfrac{I}{1/2}=e-1\approx\) 1.718.

18. Answer to Question 18

    1. \(J^2=\int e^{-x^2}dx\cdot\int e^{-y^2}dy=\iint e^{-x^2}e^{-y^2}dx\,dy=\iint e^{-(x^2+y^2)}dA\).
    2. In polar coordinates \(x^2+y^2=r^2\) and \(dA=r\,dr\,d\theta\): \(\int_0^{2\pi}\!\!\int_0^\infty e^{-r^2}r\,dr\,d\theta=2\pi\left[-\tfrac12e^{-r^2}\right]_0^\infty=2\pi\cdot\tfrac12=\pi\).
    3. \(J^2=\pi\) and \(J>0\), so \(J=\) \(\sqrt\pi\).
    4. Substituting \(x=u/\sqrt2\) gives \(\int e^{-u^2/2}du=\sqrt2\int e^{-x^2}dx=\sqrt2\sqrt\pi=\sqrt{2\pi}\). The density must integrate to 1, so we divide by \(\sqrt{2\pi}\).

19. Answer to Question 19

    1. \(T_x=2xy\) and \(T_y=x^2\).
    2. \(\dfrac{dT}{dt}=T_x\dfrac{dx}{dt}+T_y\dfrac{dy}{dt}=2xy(-2\sin t)+x^2(2\cos t)\). With \(x=2\cos t\), \(y=2\sin t\) this is \(-16\cos t\sin^2t+8\cos^3t\).
    3. At \(t=\frac\pi4\) both \(\cos t\) and \(\sin t\) equal \(\frac1{\sqrt2}\). The first term is \(-16\cdot\frac1{2\sqrt2}=-4\sqrt2\) and the second is \(8\cdot\frac1{2\sqrt2}=2\sqrt2\). The total is \(-2\sqrt2\approx\) \(-2.83\) \(^\circ\)C per minute: the drone is cooling.
    4. \(\dfrac{d}{dt}\left(8\cos^2t\sin t\right)=-16\cos t\sin^2t+8\cos^3t\), the same expression.

20. Answer to Question 20

    1. \(u_{xx}+u_{yy}=2+(-2)=0\). For \(e^x\cos y\): \(u_{xx}=e^x\cos y\) and \(u_{yy}=-e^x\cos y\); the sum is 0.
    2. \(u_x=\dfrac{2x}{x^2+y^2}\), \(u_{xx}=\dfrac{2(x^2+y^2)-4x^2}{(x^2+y^2)^2}=\dfrac{2y^2-2x^2}{(x^2+y^2)^2}\). By symmetry \(u_{yy}=\dfrac{2x^2-2y^2}{(x^2+y^2)^2}\). The sum is 0.
    3. By the chain rule \(v_x=f'(x-ct)\), \(v_{xx}=f''\), \(v_t=-cf'\), \(v_{tt}=c^2f''\). So \(v_{tt}=c^2v_{xx}\).
    4. The wave repeats every \(2\pi\) in \(x\) and every \(\pi\) in \(t\), and moves to the right with speed \(c=2\) (the pattern at time \(t\) is the pattern at time 0 shifted by \(2t\)).