1. Problem statement: Solve the cooling, hoop fluctuation, optimization and differentiation problems as given. 1. (a) Coffee model: The temperature of the coffee is modeled by $T(t)=25+40e^{kt}$, where $t$ is minutes after placement and $k$ is a constant to determine. 1. (a) (i) Find the initial temperature $T(0)$. 1. Evaluate at $t=0$ to get $T(0)=25+40e^{0}$. 1. Since $e^{0}=1$ we have $T(0)=25+40=65$. 1. Answer: The initial temperature is $65$ degrees Celsius. 1. (a) (ii) Determine $k$ given $T(5)=45$. 1. Substitute $t=5$ and $T(5)=45$ into the model to get $$45=25+40e^{5k}.$$ 1. Isolate the exponential to obtain $$40e^{5k}=20$$ and hence $$e^{5k}=\frac{1}{2}.$$ 1. Take natural logarithms to get $$5k=\ln\left(\frac{1}{2}\right)$$ and therefore $$k=\frac{1}{5}\ln\left(\frac{1}{2}\right).$$ 1. Numerically $k\approx\frac{-0.693147}{5}\approx-0.13863$. 1. (a) (iii) When does $T(t)=30$ degrees Celsius? 1. Set $30=25+40e^{kt}$ and solve for $t$ to get $$40e^{kt}=5$$ and $$e^{kt}=\frac{1}{8}.$$ 1. Taking logs gives $$kt=\ln\left(\frac{1}{8}\right)$$ and $$t=\frac{\ln\left(\frac{1}{8}\right)}{k}.$$ 1. Using $k=\frac{1}{5}\ln\left(\frac{1}{2}\right)$ we can simplify since $\ln\left(\frac{1}{8}\right)=\ln\left(\frac{1}{2^3}\right)=-3\ln 2$ and $k=\frac{-\ln 2}{5}$ to find $t=15$ minutes. 1. Answer: The coffee reaches $30$ degrees after $15$ minutes. 1. (a) (iv) How fast is the coffee cooling at $t=8$ minutes? 1. Differentiate $T(t)$ with respect to $t$ to get $$\frac{dT}{dt}=40k e^{kt}.$$ 1. Evaluate at $t=8$ using $k\approx-0.138629$ to obtain $$\frac{dT}{dt}\Big|_{t=8}=40\cdot k\cdot e^{8k}\approx40\cdot(-0.138629)\cdot e^{-1.109035}\n\approx-1.836.$$ 1. The negative sign indicates the coffee is cooling at about $1.836$ degrees Celsius per minute at $t=8$. 1. (b) Hoop diameter model: $d(t)=-\tfrac{1}{2}\cos(At)+B$, with $d(0)=5/2$ and $d(4/3)=11/4$. 1. (b) (i) Determine constants $A$ and $B$. 1. Use $t=0$: $d(0)=-\tfrac{1}{2}\cos(0)+B=-\tfrac{1}{2}+B=\tfrac{5}{2}$, so $B=3$. 1. Use $t=\tfrac{4}{3}$: $d(4/3)=-\tfrac{1}{2}\cos\left(\tfrac{4A}{3}\right)+3=\tfrac{11}{4}$, so $-\tfrac{1}{2}\cos\left(\tfrac{4A}{3}\right)= -\tfrac{1}{4}$. 1. Multiply by $-2$ to get $\cos\left(\tfrac{4A}{3}\right)=\tfrac{1}{2}$. 1. The principal solution is $\tfrac{4A}{3}=\tfrac{\pi}{3}$, hence $A=\tfrac{3\pi}{12}=\tfrac{\pi}{4}$. 1. Answer: $A=\tfrac{\pi}{4}$ and $B=3$, noting other solutions differ by integer multiples of $2\pi$ in the cosine argument. 1. (b) (ii) Sketch description over two periods starting at $t=0$. 1. For $A=\tfrac{\pi}{4}$ the period is $T_p=\frac{2\pi}{A}=\frac{2\pi}{\pi/4}=8$ minutes, so two periods span $t\in[0,16]$. 1. The graph is a cosine curve with amplitude $0.5$ about the midline $y=B=3$, so it oscillates between $3.5$ and $2.5$. 1. At $t=0$ the value is $3.5$ (a maximum), then it decreases to $2.5$ at $t=4$, returns to $3.5$ at $t=8$, and repeats once more to $t=16$. 1. This description is sufficient to sketch a smooth cosine curve with these features and period $8$ and amplitude $0.5$. 2. (a) Minimise $f=(y-2)(x+4)$ subject to $2y-x=40$. 2. Express $x$ from the constraint as $x=2y-40$. 2. Substitute into $f$ to obtain $$f(y)=(y-2)(2y-36)=2y^{2}-40y+72.$$ 2. This is a quadratic in $y$ with $a=2,b=-40$, so the minimum occurs at $y=\frac{-b}{2a}=\frac{40}{4}=10$. 2. Then $x=2y-40=20-40=-20$. 2. The minimum value is $f(10)=2(100)-400+72=-128$. 2. Answer: $(x,y)=(-20,10)$ gives the minimum with $f=-128$. 2. (b) Fence a rectangle divided by a horizontal internal fence with total fencing $120$ metres; width is horizontal $W$ and length vertical $L$. 2. The total fence length counts the two vertical sides ($2L$), the top and bottom ($2W$) and the internal dividing horizontal fence of length $W$, but the top or bottom is not duplicated because the internal divider replaces one, so the configuration given yields total fence $3W+2L=120$. 2. Solve the constraint for $W$: $$W=\frac{120-2L}{3}.$$ 2. Area $A=W\cdot L= L\frac{120-2L}{3}=\frac{120L-2L^{2}}{3}.$ 2. Differentiate with respect to $L$: $$\frac{dA}{dL}=\frac{120-4L}{3}$$ and set to zero to find $L=30$. 2. Then $W=\frac{120-2\cdot30}{3}=\frac{60}{3}=20$ and the maximum area is $600$ square metres. 2. Answer: Width $20$ m and length $30$ m maximize the area. 2. (c) Cylinder with constraint $r+h=90$ cm and volume $V=\pi r^{2}h$. 2. Express $h=90-r$ and substitute to get $$V(r)=\pi r^{2}(90-r)=\pi(90r^{2}-r^{3}).$$ 2. Differentiate: $$\frac{dV}{dr}=\pi(180r-3r^{2})=\pi r(180-3r).$$ 2. Set derivative zero to get $r=0$ or $180-3r=0$ which gives $r=60$ cm as the positive maximiser. 2. Then $h=90-60=30$ cm and this yields a maximum volume. 2. Answer: Radius $60$ cm and height $30$ cm maximize the volume. 3. Differentiate the following functions with respect to $x$. 3. (a) $f(x)=\dfrac{\tan(2x)}{\sqrt{x}}$. 3. Write $f(x)=\tan(2x)x^{-1/2}$ and apply product and chain rules to obtain $$f'(x)=2\sec^{2}(2x)x^{-1/2}-\tfrac{1}{2}\tan(2x)x^{-3/2}.$$ 3. This can be written as $$f'(x)=\frac{2\sec^{2}(2x)}{\sqrt{x}}-\frac{\tan(2x)}{2x^{3/2}}.$$ 3. (b) $y=\ln\bigl(3x^{6}-5x+7\bigr)^{2}$. 3. Use the log power rule to write $y=2\ln\bigl(3x^{6}-5x+7\bigr)$, then differentiate to get $$y'=2\cdot\frac{1}{3x^{6}-5x+7}\cdot(18x^{5}-5).$$ 3. Simplify to $$y'=\frac{36x^{5}-10}{3x^{6}-5x+7}.$$ 3. (c) $g(x)=\cos(2x)e^{-4x}$. 3. Use the product rule and chain rule to get $$g'(x)=-2\sin(2x)e^{-4x}-4\cos(2x)e^{-4x}.$$ 3. Factor to a compact form $$g'(x)=-2e^{-4x}\bigl(\sin(2x)+2\cos(2x)\bigr).$$ 3. (d) $h(x)=\cot(3x)$. 3. Use the derivative $\dfrac{d}{dx}\cot(u)=-\csc^{2}(u)u'$ with $u=3x$ to obtain $$h'(x)=-3\csc^{2}(3x).$$ 3. Each differentiation step used the chain rule, product rule or standard trig/log derivatives and intermediate algebra is shown above.