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Paper 1 Answers

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PaperLevel 1 - Math I (Physics) Paper 1

Information

Section A: Functions and Trigonometry
Section B: Limits and Differentiation
Section C: Integration
Section D: Complex Numbers

Marking Guidance

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

Section A: Functions and Trigonometry

Answer 1. Sensor calibration functions

[15 marks]

A sensor arm rotates through an angle \(\theta\), where \(-\frac{\pi}{2}\le \theta\le \frac{\pi}{2}\). Its height above a reference rail is modelled by \(h(\theta)=1.20+0.50\sin\theta\), measured in metres. A calibration circuit converts height to voltage by \(V(h)=6h-4\).

a) Find the range of \(h\), and find an expression for \(h^{-1}(y)\).

[5 marks]

On the given interval, \(\sin\theta\) runs from \(-1\) to \(1\). Hence \[ h_{\min}=1.20-0.50=0.70,\qquad h_{\max}=1.20+0.50=1.70. \] The range is \(0.70\le h\le1.70\). For the inverse, write \(y=1.20+0.50\sin\theta\). Then \[ \sin\theta=\frac{y-1.20}{0.50}. \] Because the domain is the principal sine interval, \[ h^{-1}(y)=\sin^{-1}\left(\frac{y-1.20}{0.50}\right),\qquad 0.70\le y\le1.70. \]

Answer 1. Sensor calibration functions

[15 marks]

b) Find the voltage as a function of \(\theta\), written as \(V(h(\theta))\). Hence find the value of \(\theta\) for which the voltage is \(4.70\,\mathrm{V}\).

[5 marks]

Substitute the height model into the voltage model: \[ V(h(\theta))=6(1.20+0.50\sin\theta)-4. \] Therefore \[ V(h(\theta))=3.20+3\sin\theta. \] Set this equal to \(4.70\): \[ 3.20+3\sin\theta=4.70. \] Then \[ 3\sin\theta=1.50,\qquad \sin\theta=\frac{1}{2}. \] On \(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\), the solution is \[ \theta=\frac{\pi}{6}. \]

Answer 1. Sensor calibration functions

[15 marks]

c) During a sweep, \(\theta(t)=-\frac{\pi}{6}+\frac{\pi t}{12}\) for \(0\le t\le8\). Find the time at which the voltage is \(4.70\,\mathrm{V}\). A power model is \(P(V)=(V-3.20)^2+2\); find \(P(V(h(\theta)))\) in simplest form.

[5 marks]

From part (b), the voltage is \(4.70\,\mathrm{V}\) when \(\theta=\frac{\pi}{6}\). Substitute into the sweep model: \[ -\frac{\pi}{6}+\frac{\pi t}{12}=\frac{\pi}{6}. \] Thus \[ \frac{\pi t}{12}=\frac{\pi}{3}. \] Dividing by \(\pi\) gives \(\frac{t}{12}=\frac{1}{3}\), so \[ t=4\,\mathrm{s}. \] For the power model, use \(V(h(\theta))=3.20+3\sin\theta\): \[ P(V(h(\theta)))=(3\sin\theta)^2+2=9\sin^2\theta+2. \]

Section B: Limits and Differentiation

Answer 2. Rates from limits and derivatives

[15 marks]

A test cart has displacement \(x(t)=\frac{t^2}{t+1}\), where \(x\) is in metres and \(t\) is in seconds. A separate sensor response uses square-root behaviour for small changes.

a) Using first principles, show that if \(f(t)=t^2+3t\), then \(f'(t)=2t+3\).

[5 marks]

By first principles, \[ f'(t)=\lim_{\Delta t\to0}\frac{f(t+\Delta t)-f(t)}{\Delta t}. \] Now \[ f(t+\Delta t)=(t+\Delta t)^2+3(t+\Delta t). \] Therefore \[ f(t+\Delta t)-f(t)=2t\Delta t+(\Delta t)^2+3\Delta t. \] Divide by \(\Delta t\): \[ \frac{f(t+\Delta t)-f(t)}{\Delta t}=2t+\Delta t+3. \] Taking the limit gives \[ f'(t)=2t+3. \]

Answer 2. Rates from limits and derivatives

[15 marks]

b) Differentiate \(x(t)=\frac{t^2}{t+1}\) and find the cart velocity at \(t=2\).

[5 marks]

Use the quotient rule with numerator \(t^2\) and denominator \(t+1\): \[ x'(t)=\frac{(t+1)(2t)-t^2}{(t+1)^2}. \] Simplify the numerator: \[ x'(t)=\frac{2t^2+2t-t^2}{(t+1)^2}=\frac{t^2+2t}{(t+1)^2}. \] Thus \[ x'(t)=\frac{t(t+2)}{(t+1)^2}. \] At \(t=2\), \[ x'(2)=\frac{2(4)}{3^2}=\frac{8}{9}. \] The velocity is \(\frac{8}{9}\,\mathrm{m\,s^{-1}}\).

Answer 2. Rates from limits and derivatives

[15 marks]

c) Evaluate \(\lim_{u\to0}\frac{\sqrt{1+4u}-1}{u}\) and state what type of rate this limit represents.

[5 marks]

Rationalise the numerator: \[ \frac{\sqrt{1+4u}-1}{u}\cdot\frac{\sqrt{1+4u}+1}{\sqrt{1+4u}+1}=\frac{(1+4u)-1}{u(\sqrt{1+4u}+1)}. \] This simplifies to \[ \frac{4}{\sqrt{1+4u}+1}. \] Therefore \[ \lim_{u\to0}\frac{\sqrt{1+4u}-1}{u}=\frac{4}{1+1}=2. \] This is the instantaneous rate of change at zero of the response function \(g(u)=\sqrt{1+4u}\).

Section C: Integration

Answer 3. Accumulation by integration

[15 marks]

In a straight-track experiment, accumulated quantities are found by integrating simple rate or force models.

a) A variable force is \(F(x)=2x+3\) for \(0\le x\le4\). Find the work done and state the definite-integral interpretation.

[4 marks]

Work is the signed area under the force-displacement graph: \[ W=\int_0^4(2x+3)\,dx. \] An antiderivative is \(x^2+3x\). Hence \[ W=[x^2+3x]_0^4=(16+12)-0=28. \] The work done is \(28\,\mathrm{J}\).

Answer 3. Accumulation by integration

[15 marks]

b) Use substitution to evaluate \(\int_0^1 4t(1+t^2)^3\,dt\).

[5 marks]

Let \(u=1+t^2\). Then \(du=2t\,dt\), so \(4t\,dt=2\,du\). The limits become \(u=1\) when \(t=0\), and \(u=2\) when \(t=1\). Therefore \[ \int_0^1 4t(1+t^2)^3\,dt=\int_1^2 2u^3\,du. \] Now \[ \int_1^2 2u^3\,du=\left[\frac{u^4}{2}\right]_1^2=\frac{16}{2}-\frac{1}{2}=\frac{15}{2}. \]

Answer 3. Accumulation by integration

[15 marks]

c) A control input is \(I(t)=t e^{-t}\). Use integration by parts to find \(\int_0^2 t e^{-t}\,dt\).

[6 marks]

Use integration by parts with \(u=t\) and \(dv=e^{-t}\,dt\). Then \(du=dt\) and \(v=-e^{-t}\). Thus \[ \int t e^{-t}\,dt=-te^{-t}+\int e^{-t}\,dt=-te^{-t}-e^{-t}+C. \] Therefore \[ \int_0^2 t e^{-t}\,dt=[-te^{-t}-e^{-t}]_0^2. \] Substitute the limits: \[ (-2e^{-2}-e^{-2})-(-1)=1-3e^{-2}. \] Hence the accumulated control input is \[ 1-3e^{-2}\approx0.594. \]

Section D: Complex Numbers

Answer 4. Complex phasor arithmetic

[15 marks]

Complex numbers are used to represent amplitudes and phases of oscillatory signals. Write \(i^2=-1\).

a) Write \(z=\sqrt{3}+i\) in polar form \(r(\cos\theta+i\sin\theta)\).

[4 marks]

The modulus is \[ r=\sqrt{(\sqrt{3})^2+1^2}=2. \] The point is in the first quadrant and \[ \tan\theta=\frac{1}{\sqrt{3}}, \] so \(\theta=\frac{\pi}{6}\). Therefore \[ z=2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right). \]

Answer 4. Complex phasor arithmetic

[15 marks]

b) Let \(w=3\left(\cos\left(-\frac{\pi}{3}\right)+i\sin\left(-\frac{\pi}{3}\right)\right)\). Find \(zw\) in polar form and in the form \(a+bi\).

[4 marks]

From part (a), \(z\) has modulus \(2\) and argument \(\frac{\pi}{6}\). The number \(w\) has modulus \(3\) and argument \(-\frac{\pi}{3}\). Multiplying in polar form multiplies moduli and adds arguments, so \[ zw=6\left(\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right). \] Using \(\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\) and \(\sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2}\), \[ zw=6\left(\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)=3\sqrt{3}-3i. \]

Answer 4. Complex phasor arithmetic

[15 marks]

c) A signal is represented by \(\operatorname{Re}\left(6e^{i(3t+\frac{\pi}{3})}\right)\). Write it as a real trigonometric function and find its value at \(t=\frac{\pi}{18}\).

[4 marks]

Euler's formula gives \(e^{i\alpha}=\cos\alpha+i\sin\alpha\). Therefore \[ 6e^{i(3t+\frac{\pi}{3})}=6\cos\left(3t+\frac{\pi}{3}\right)+6i\sin\left(3t+\frac{\pi}{3}\right). \] The real displacement is \[ x(t)=6\cos\left(3t+\frac{\pi}{3}\right). \] At \(t=\frac{\pi}{18}\), \[ 3t+\frac{\pi}{3}=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}. \] Thus \[ x\left(\frac{\pi}{18}\right)=6\cos\frac{\pi}{2}=0. \]

Answer 4. Complex phasor arithmetic

[15 marks]

d) Write \(5-5i\) in the form \(Re^{i\phi}\), and interpret \(R\) and \(\phi\) for an oscillatory signal.

[3 marks]

The modulus is \[ R=|5-5i|=\sqrt{5^2+(-5)^2}=5\sqrt{2}. \] The argument is in the fourth quadrant and satisfies \(\tan\phi=-1\), so \[ \phi=-\frac{\pi}{4}. \] Therefore \[ 5-5i=5\sqrt{2}e^{-i\frac{\pi}{4}}. \] The modulus \(5\sqrt{2}\) is the resultant amplitude and \(-\frac{\pi}{4}\) is the phase angle.