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

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1. Answer to Question 1 (4 marks)

   1. \(f_x=2xe^y\), \(f_y=x^2e^y\), \(f_{xy}=2xe^y\) (2).
   2. At \((2,0)\): \(f_x=4\), \(f_y=4\) (1). The plane is \(z-4=4(x-2)+4y\), i.e. \(z=4x+4y-4\) (1).

   In (a), to find \(f_x\) treat \(y\) as a constant, so \(e^y\) is a constant factor; to find \(f_y\) treat \(x\) as a constant, so \(x^2\) is a constant factor. Differentiating \(f_x\) with respect to \(y\) gives \(f_{xy}\), and \(f_{yx}=\partial_x(x^2e^y)=2xe^y\) is the same, as Clairaut's theorem predicts. In (b), the tangent plane is \(z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)\); check that \((2,0,4)\) lies on it: \(8+0-4=4\).

2. Answer to Question 2 (4 marks)

   \(f_x=3x^2-12=0\) gives \(x=\pm2\); \(f_y=2y-4=0\) gives \(y=2\) (2). \(f_{xx}=6x\), \(f_{yy}=2\), \(f_{xy}=0\), so \(D=12x\) (1). At \((2,2)\): \(D=24>0\), \(f_{xx}>0\): minimum. At \((-2,2)\): \(D<0\): saddle (1).

   The critical points are where both first partial derivatives vanish, so solve the two equations together, giving \((2,2)\) and \((-2,2)\). The discriminant is \(D=f_{xx}f_{yy}-f_{xy}^2=6x\times2-0=12x\). At \((2,2)\), \(D\gt 0\) and \(f_{xx}=12\gt 0\), so it is a local minimum with value \(8-24+4-8=-20\). At \((-2,2)\), \(D=-24\lt 0\), so it is a saddle point: the function increases in some directions and decreases in others.

3. Answer to Question 3 (4 marks)

   \((y^2,2xy)=\lambda(1,1)\) so \(y^2=2xy\) (1), giving \(y=0\) or \(y=2x\) (1). With \(y=2x\): \(3x=3\), \(x=1\), \(y=2\) (1). Maximum value \(f=1\times4=4\); the other candidate \(y=0\) gives \(f=0\) (1).

   The method sets \(\nabla f=\lambda\nabla g\) with constraint \(g(x,y)=x+y=3\), whose gradient is \((1,1)\). Equating the two components gives \(y^2=\lambda\) and \(2xy=\lambda\). The endpoints of the constraint segment, \((3,0)\) and \((0,3)\), both give \(f=0\), so the maximum on the closed segment is 4. Check by substitution: \(f=(3-y)y^2\) has derivative \(6y-3y^2=0\) at \(y=2\).

4. Answer to Question 4 (4 marks)

   1. Inner: \(\int_0^1(x+2y)dx=\tfrac12+2y\) (1). Outer: \(\int_0^2(\tfrac12+2y)dy=1+4=5\) (1).
   2. \(\int_0^{2\pi}\!\!\int_0^1\dfrac{r\,dr}{1+r^2}\,d\theta=2\pi\cdot\tfrac12\ln2\) (1) \(=\pi\ln2\) (1).

   In (a), integrate first with respect to \(x\), treating \(y\) as a constant: \(\left[\tfrac{x^2}2+2xy\right]_0^1=\tfrac12+2y\). Then integrate with respect to \(y\): \(\left[\tfrac y2+y^2\right]_0^2=1+4=5\). In (b), change to polar coordinates with \(x^2+y^2=r^2\) and \(dA=r\,dr\,d\theta\); a common error is to forget the Jacobian factor \(r\). The disc corresponds to \(0\le r\le1\) and \(0\le\theta\le2\pi\). Substituting \(u=1+r^2\) gives \(\int_0^1\tfrac{r}{1+r^2}dr=\tfrac12\ln2\).

5. Answer to Question 5 (4 marks)

   1. \(V=24\) m\(^3\); \(V_x=yz=6\), \(V_y=xz=8\), \(V_z=xy=12\) (2).
   2. \(|dV|\le(6+8+12)(0.05)=1.3\) m\(^3\) (2), about 5.4% of \(V\).

   In (a), each partial derivative is the product of the other two sides, because \(V\) is linear in each variable. In (b), the total differential is \(dV=V_x\,dx+V_y\,dy+V_z\,dz\), and the worst case takes every error with the same sign, \(|dx|=|dy|=|dz|=0.05\). Check against the exact worst case: \(4.05\times3.05\times2.05\approx25.32\), which is about 1.32 m³ more than 24, close to the estimate. The relative error is \(1.3\div24\approx5.4\%\).

Total : 20 marks