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Paper 1 Answers

Time3 hours

Marks100

SetSet 1

PaperLevel 1 - Math II (Physics) Paper 1

Information

Section A: Vectors and Kinematics
Section B: Ordinary Differential Equations
Section C: Fourier Analysis
Section D: Multivariable Calculus
Section E: Critical Points and Vector Calculus
Section F: Multiple Integrals and Vector Identities
Section G: Probability and Statistics

Marking Guidance

These worked answers show the expected method and final result. Equivalent correct reasoning should receive credit.

Section A: Vectors and Kinematics

Answer 1. Lines, Angles, and Volume

[14 marks]

Four points in \(\mathbb R^3\) are given by \(A=(1,0,2)\), \(B=(3,1,0)\), \(C=(0,2,1)\), and \(D=(2,0,3)\).

a) Find vector equations for the line \(\ell_1\) through \(A\) and \(B\), and the line \(\ell_2\) through \(C\) and \(D\).

[4 marks]

The direction vector for \(\ell_1\) is \(B-A=(2,1,-2)\), so \[\ell_1:\ \mathbf r=(1,0,2)+s(2,1,-2),\qquad s\in\mathbb R.\] The direction vector for \(\ell_2\) is \(D-C=(2,-2,2)\), so \[\ell_2:\ \mathbf r=(0,2,1)+t(2,-2,2),\qquad t\in\mathbb R.\]

Answer 1. Lines, Angles, and Volume

[14 marks]

b) Determine the shortest distance between \(\ell_1\) and \(\ell_2\).

[4 marks]

Let \(\mathbf u=(2,1,-2)\) and \(\mathbf v=(2,-2,2)\). Then \(\mathbf u\times\mathbf v=(-2,-8,-6)\), with magnitude \(2\sqrt{26}\). Also \(C-A=(-1,2,-1)\). The distance between the skew lines is \[\frac{|(C-A)\cdot(\mathbf u\times\mathbf v)|}{|\mathbf u\times\mathbf v|}=\frac{|(-1,2,-1)\cdot(-2,-8,-6)|}{2\sqrt{26}}=\frac{8}{2\sqrt{26}}=\frac{4}{\sqrt{26}}.\]

Answer 1. Lines, Angles, and Volume

[14 marks]

c) Find the cosine of the angle between \(\overrightarrow{BA}\) and \(\overrightarrow{AC}\).

[3 marks]

We have \(\overrightarrow{BA}=A-B=(-2,-1,2)\) and \(\overrightarrow{AC}=C-A=(-1,2,-1)\). Their dot product is \(2-2-2=-2\). The magnitudes are \(|\overrightarrow{BA}|=3\) and \(|\overrightarrow{AC}|=\sqrt6\). Therefore \[\cos\theta=\frac{\overrightarrow{BA}\cdot\overrightarrow{AC}}{|\overrightarrow{BA}|\,|\overrightarrow{AC}|}=-\frac{2}{3\sqrt6}.\]

Answer 1. Lines, Angles, and Volume

[14 marks]

d) Compute the volume of the tetrahedron with vertices \(A\), \(B\), \(C\), and \(D\).

[3 marks]

The tetrahedron volume is one sixth of the scalar triple product based at \(A\). Here \(B-A=(2,1,-2)\), \(C-A=(-1,2,-1)\), and \(D-A=(1,0,1)\). Since \((C-A)\times(D-A)=(2,0,-2)\), \[(B-A)\cdot((C-A)\times(D-A))=(2,1,-2)\cdot(2,0,-2)=8.\] Hence the volume is \(8/6=4/3\).

Section B: Ordinary Differential Equations

Answer 2. Linear ODEs and Coupled Populations

[14 marks]

Answer both independent modelling questions.

a) Solve \(\frac{dy}{dx}+2xy=4x\), subject to \(y(0)=3\).

[6 marks]

The integrating factor is \(\exp\int 2x\,dx=e^{x^2}\). Multiplying the equation by this factor gives \[\frac{d}{dx}(ye^{x^2})=4xe^{x^2}.\] Integrating, \[ye^{x^2}=2e^{x^2}+C,\] so \(y=2+Ce^{-x^2}\). The condition \(y(0)=3\) gives \(3=2+C\), hence \(C=1\). Therefore \[y(x)=2+e^{-x^2}.\]

Answer 2. Linear ODEs and Coupled Populations

[14 marks]

b) Two populations satisfy \(S'=3S-2W\) and \(W'=S-W\). Deduce a second-order differential equation for \(S(t)\).

[5 marks]

From \(S'=3S-2W\), rearrange to \(W=(3S-S')/2\). Differentiate the first equation: \[S''=3S'-2W'.\] Using \(W'=S-W\), this becomes \[S''=3S'-2S+2W.\] Substitute \(2W=3S-S'\), giving \[S''=3S'-2S+3S-S'=2S'+S.\] Therefore \[S''-2S'-S=0.\]

Answer 2. Linear ODEs and Coupled Populations

[14 marks]

c) Solve the second-order equation from part (b) and describe the long-term behaviour of a non-zero solution with a positive coefficient of the growing mode.

[3 marks]

The auxiliary equation is \(r^2-2r-1=0\), so \(r=1\pm\sqrt2\). Thus \[S(t)=Ae^{(1+\sqrt2)t}+Be^{(1-\sqrt2)t}.\] Since \(1+\sqrt2>0\) and \(1-\sqrt2<0\), the second mode decays while the first mode grows. If \(A>0\), the population grows like \(Ae^{(1+\sqrt2)t}\) for large \(t\).

Section C: Fourier Analysis

Answer 3. Fourier Series and Parseval

[14 marks]

Let \(h\) be the \(2\pi\)-periodic function with \(h(x)=1\) for \(0<x<\pi\), \(h(x)=-1\) for \(-\pi<x<0\), and \(h(0)=h(\pi)=0\).

a) Describe the graph of \(h\) on \((-2\pi,2\pi)\).

[3 marks]

The function is an odd square wave with jumps at integer multiples of \(\pi\). On \((-2\pi,2\pi)\), periodicity gives \(h=1\) on \((-2\pi,-\pi)\) and \((0,\pi)\), while \(h=-1\) on \((-\pi,0)\) and \((\pi,2\pi)\). At the jump points the assigned values do not affect the Fourier coefficients.

Answer 3. Fourier Series and Parseval

[14 marks]

b) Find the Fourier series of \(h\), giving the first six non-zero terms explicitly.

[6 marks]

Since \(h\) is odd, only sine terms occur. The coefficients are \[b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}h(x)\sin nx\,dx=\frac{2}{\pi}\int_0^{\pi}\sin nx\,dx=\frac{2}{\pi}\frac{1-(-1)^n}{n}.\] Hence \(b_n=0\) for even \(n\), and \(b_n=4/(\pi n)\) for odd \(n\). Therefore \[h(x)\sim\frac{4}{\pi}\left(\sin x+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\frac{\sin 7x}{7}+\frac{\sin 9x}{9}+\frac{\sin 11x}{11}+\cdots\right).\]

Answer 3. Fourier Series and Parseval

[14 marks]

c) Use Parseval's theorem to evaluate \(\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}\).

[5 marks]

Parseval's theorem gives \[\frac{1}{\pi}\int_{-\pi}^{\pi}|h(x)|^2\,dx=\sum_{n=1}^{\infty}b_n^2.\] Since \(|h(x)|=1\) except at jump points, the left side is \(2\). The right side is \[\sum_{k=0}^{\infty}\left(\frac{4}{\pi(2k+1)}\right)^2=\frac{16}{\pi^2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}.\] Thus \[2=\frac{16}{\pi^2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2},\] and the required sum is \(\pi^2/8\).

Section D: Multivariable Calculus

Answer 4. Chain Rule and Directional Derivatives

[14 marks]

Let \(F(u,v)=f(x,y)\), where \(x=u+v\) and \(y=uv\). Assume the mixed partial derivatives of \(f\) are equal.

a) Find \(F_u\) and \(F_v\) in terms of partial derivatives of \(f\).

[5 marks]

Using the chain rule, \(x_u=1\), \(x_v=1\), \(y_u=v\), and \(y_v=u\). Therefore \[F_u=f_xx_u+f_yy_u=f_x+vf_y,\] and similarly \[F_v=f_xx_v+f_yy_v=f_x+uf_y.\] The derivatives \(f_x\) and \(f_y\) are evaluated at \((x,y)=(u+v,uv)\).

Answer 4. Chain Rule and Directional Derivatives

[14 marks]

b) Show that \(F_{uv}=f_{xx}+(u+v)f_{xy}+uvf_{yy}+f_y\).

[5 marks]

Differentiate \(F_u=f_x+vf_y\) with respect to \(v\). For the first term, \[\frac{\partial}{\partial v}f_x=f_{xx}x_v+f_{xy}y_v=f_{xx}+uf_{xy}.\] For the second term, use the product rule: \[\frac{\partial}{\partial v}(vf_y)=f_y+v(f_{yx}x_v+f_{yy}y_v)=f_y+v(f_{xy}+uf_{yy}).\] Adding these terms gives \[F_{uv}=f_{xx}+(u+v)f_{xy}+uvf_{yy}+f_y.\]

Answer 4. Chain Rule and Directional Derivatives

[14 marks]

c) The temperature in a tank is \(T(x,y,z)=20-x^2-2y^2-z^2\). At \((1,-1,2)\), find the unit direction of greatest increase and the corresponding rate of increase.

[4 marks]

The direction of greatest increase is the gradient direction. Here \[\nabla T=(-2x,-4y,-2z),\] so at \((1,-1,2)\) we have \(\nabla T=(-2,4,-4)\). Its magnitude is \(\sqrt{4+16+16}=6\). Hence the required unit vector is \[\frac{\nabla T}{|\nabla T|}=\left(-\frac13,\frac23,-\frac23\right),\] and the maximum rate of increase is \(|\nabla T|=6\).

Section E: Critical Points and Vector Calculus

Answer 5. Exact Differentials, Divergence, and Curl

[14 marks]

Answer both parts.

a) Determine all constants \(a\) and \(b\) for which \(df=(2x+ay)\,dx+(bx+6y)\,dy\) is exact. For those values, find a potential function.

[6 marks]

Write \(M=2x+ay\) and \(N=bx+6y\). Exactness requires \(M_y=N_x\). Since \(M_y=a\) and \(N_x=b\), the differential is exact exactly when \(a=b\). When this holds, integrate \(M\) with respect to \(x\): \[f(x,y)=x^2+axy+g(y).\] Differentiating with respect to \(y\) gives \(f_y=ax+g'(y)\), and this must equal \(ax+6y\). Hence \(g'(y)=6y\), so \(g(y)=3y^2+C\). Thus \[f(x,y)=x^2+axy+3y^2+C.\]

Answer 5. Exact Differentials, Divergence, and Curl

[14 marks]

b) For \(\mathbf V=(x^2y,xz,yz^2)\), compute \(\nabla\cdot\mathbf V\) and \(\nabla\times\mathbf V\).

[8 marks]

The divergence is \[\nabla\cdot\mathbf V=\frac{\partial}{\partial x}(x^2y)+\frac{\partial}{\partial y}(xz)+\frac{\partial}{\partial z}(yz^2)=2xy+0+2yz.\] For the curl, use \(\nabla\times\mathbf V=(R_y-Q_z, P_z-R_x, Q_x-P_y)\) with \(P=x^2y\), \(Q=xz\), and \(R=yz^2\). This gives \[\nabla\times\mathbf V=(z^2-x,0,z-x^2).\]

Section F: Multiple Integrals and Vector Identities

Answer 6. Double Integrals and Volumes

[15 marks]

Evaluate the following integrals and state the geometry used.

a) Evaluate \(\iint_R (x+y)\,dA\), where \(R\) is the triangle bounded by \(x=0\), \(y=0\), and \(x+y=2\).

[7 marks]

The region may be written as \(0\le x\le2\), \(0\le y\le2-x\). Hence \[\iint_R(x+y)\,dA=\int_0^2\int_0^{2-x}(x+y)\,dy\,dx.\] The inner integral is \[x(2-x)+\frac{(2-x)^2}{2}.\] Integrating from \(0\) to \(2\) gives \[\int_0^2\left(x(2-x)+\frac{(2-x)^2}{2}\right)dx=\frac{8}{3}.\] This agrees with symmetry, since \(\int_Rx\,dA=\int_Ry\,dA=4/3\).

Answer 6. Double Integrals and Volumes

[15 marks]

b) Find the volume enclosed by the ellipsoid \(\frac{x^2}{4}+\frac{y^2}{9}+z^2=1\).

[8 marks]

Use the scaling \(x=2u\), \(y=3v\), and \(z=w\). The ellipsoid becomes the unit sphere \(u^2+v^2+w^2\le1\). The Jacobian determinant of the scaling is \(2\cdot3\cdot1=6\). Therefore the ellipsoid volume is \[6\times\frac{4\pi}{3}=8\pi.\] Equivalently, this is the standard ellipsoid formula \(\frac{4\pi}{3}abc\) with semi-axes \(2\), \(3\), and \(1\).

Section G: Probability and Statistics

Answer 7. Conditional Probability and Approximations

[15 marks]

A field laboratory monitors three independent warning systems during a week of observations.

a) Let \(A\), \(B\), and \(C\) be independent events with probabilities \(p_A\), \(p_B\), and \(p_C\). An alert is triggered if \(A\) occurs or if both \(B\) and \(C\) occur. Show that the alert probability is \(p_A+p_Bp_C-p_Ap_Bp_C\).

[5 marks]

The alert event is \(A\cup(B\cap C)\). Therefore \[P(A\cup(B\cap C))=P(A)+P(B\cap C)-P(A\cap B\cap C).\] Independence gives \(P(B\cap C)=p_Bp_C\) and \(P(A\cap B\cap C)=p_Ap_Bp_C\). Hence the alert probability is \[p_A+p_Bp_C-p_Ap_Bp_C.\]

Answer 7. Conditional Probability and Approximations

[15 marks]

b) Find \(P(A\mid\hbox{alert})\) in terms of \(p_A\), \(p_B\), and \(p_C\).

[4 marks]

Since \(A\) always triggers an alert, \(A\cap\hbox{alert}=A\). Therefore \[P(A\mid\hbox{alert})=\frac{P(A)}{P(\hbox{alert})}=\frac{p_A}{p_A+p_Bp_C-p_Ap_Bp_C}.\] This is a conditional probability: the denominator is the total chance that an alert occurs, while the numerator is the part of those alerts caused by event \(A\).

Answer 7. Conditional Probability and Approximations

[15 marks]

c) A rare marker appears independently in each of \(800\) samples with probability \(0.005\). Use a Poisson approximation to estimate the probability of at most two appearances.

[3 marks]

Let \(X\sim\operatorname{Bin}(800,0.005)\). Since \(n\) is large and \(p\) is small, use \(X\approx\operatorname{Poisson}(\lambda)\) with \(\lambda=np=4\). Thus \[P(X\le2)\approx e^{-4}\left(1+4+\frac{4^2}{2}\right)=13e^{-4}\approx0.238.\]

Answer 7. Conditional Probability and Approximations

[15 marks]

d) Over \(40\) independent weeks the weekly count is approximated by \(\operatorname{Poisson}(4)\). Approximate the probability that the sample mean is less than \(3.5\).

[3 marks]

For one week, the mean and variance are both \(4\). For \(40\) weeks, the sample mean \(\bar X\) is approximately normal with mean \(4\) and variance \(4/40=0.1\). Therefore \[P(\bar X<3.5)\approx P\left(Z<\frac{3.5-4}{\sqrt{0.1}}\right)=P(Z<-1.58).\] From standard normal tables this is about \(0.057\).