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The temperature, $T$ (°C), of a fluid is given by the formula

\n

\\[T=\\simplify{{a}*exp({b}*t)+{cc}*t+{d}}\\]

\n

where $t$ is the time in minutes.

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Part a)

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To calculate the temperature of the fluid after {am} minutes, substitute $t=\\var{am}$ into the formula for $T$:

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\\[T=\\var{a}\\times e^{\\var{b}\\times\\var{am}}-\\var{-cc}\\times\\var{am}+\\var{d}=\\var{precround(a_ans,1)}\\]

\n

Part b)

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Differentiate $T$:

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\\[\\frac{dT}{dt}=\\var{a}\\times\\var{b}\\times e^{\\simplify{{b}*t}}-\\var{-cc}=\\simplify{{a*b}*exp({b}*t)+{cc}}\\]

\n

Part c) and d)

\n

Substitute $t=\\var{cm}$ into the derivative to calculate the rate of change of temperature at that time:

\n

\\[\\frac{dT}{dt}=\\var{a*b}\\times e^{\\var{b}\\times\\var{cm}}-\\var{-cc}=\\var{precround(cf',3)}\\]

\n

This is {c_posneg}, which means the temperature is {c_risefall}.

\n

Part e)

\n

Substitute $t=\\var{dm}$ into the derivative to calculate the rate of change of temperature at that time:

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\\[\\frac{dT}{dt}=\\var{a*b}\\times e^{\\var{b}\\times\\var{dm}}-\\var{-cc}=\\var{precround(f'(dm),1)}\\]

\n

Therefore the rate of change of temperature is {precround(f'(dm),1)} °C/min

\n

Part f)

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If the temperature is {e_rise} at $\\var{abs(eT')}$ °C per minute, then $\\frac{dT}{dt}=\\var{eT'}$ so we solve the equation:

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\\[\\simplify{{a*b}*exp({b}*t)+{cc}}=\\var{eT'}\\]

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Add {-cc}:

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\\[\\simplify{{a*b}*exp({b}*t)}=\\var{eT'-cc}\\]

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Divide by {a*b}:

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\\[\\simplify{exp({b}*t)}=\\var{precround((eT'-cc)/(a*b),3)}\\]

\n

Take natural log:

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\\[\\simplify{{b}*t}=\\var{precround(ln((eT'-cc)/(a*b)),3)}\\]

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Divide by {b}:

\n

\\[t=\\var{precround(e_ans,3)}\\]

\n

Rounded to the nearest integer, $t=\\var{precround(e_ans,0)}$ minutes.

\n

Part g)

\n

The temperature stops falling, when $\\frac{dT}{dt}=0$, so we solve the equation:

\n

\\[\\simplify{{a*b}*exp({b}*t)+{cc}}=0\\]

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Add {-cc} and divide by {a*b}:

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\\[\\simplify{exp({b}*t)}=\\var{precround(-cc/(a*b),3)}\\]

\n

Take natural log and divide by {b}:

\n

\\[t=\\var{precround(ft0,3)}\\]

\n

Rounded to the nearest integer, $t=\\var{precround(ft0,0)}$ minutes.

\n

Part h)

\n

The lowest temperature will occur at the time when it first stops falling, i.e. when $t=\\var{precround(ft0,3)}$. (For an accurate answer, do not use the rounded value.) Substitute this into the formula for $T$:

\n

\\[T=\\var{a}\\times e^{\\var{b}\\times\\var{precround(ft0,3)}}-\\var{-cc}\\times\\var{precround(ft0,3)}+\\var{d}=\\var{precround(gT,3)}\\]

\n

Rounded to 2 significant figures, $T=\\var{siground(gT,2)}\\;^\\circ\\text{C}$.

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Calculate the temperature of the fluid after {am} minutes.

\n

[[0]] $^\\circ\\text{C}$

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Write down a formula for rate of change of temperature at time $t$.

\n

$\\frac{dT}{dt}=\\;$[[0]]

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Is the temperature rising or falling after {cm} minutes?

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Give a reason for your answer to c)

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Determine the rate of change of temperature (in °C/min) after $\\var{dm}$ minutes.

\n

[[0]] $^\\circ\\text{C}/\\text{min}$

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At what time is the temperature {e_rise} at $\\var{abs(eT')}$ °C per minute?

\n

[[0]] mins

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At what time does the temperature stop falling?

\n

[[0]] mins

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Hence calculate the temperature at its lowest value.

\n

[[0]] $^\\circ\\text{C}$

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A metal box will be constructed. The base length must be three times the base width, and the volume must be {volume} m3. The metal used to build the top and bottom costs £{tb_cost} per m2 and the metal used to build the sides costs just £{side_cost} per m2.

\n

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Enter a formula for the volume of the box in terms of $w$ and $h$, simplifying where possible.

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The volume must be {volume} m3. Use this volume constraint to write $h$ in terms of $w$.

\n

$h=\\;$[[0]]

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Equate the volume to {volume} and rearrange to make $h$ the subject.

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Enter a formula (in terms of $w$ only) for the total area of the top and bottom of the box. Simplify where possible.

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Add together the area of the top and the area of the bottom.

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The metal used to build the top and bottom costs £{tb_cost} per m2. Enter a formula for the cost of the top and bottom.

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Multiply the area by the cost per square metre.

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Enter a formula (in terms of $w$ and $h$) for the total area of the four sides of the box.

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Add together the area of the 4 sides of the box. Don't include the top or bottom.

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The metal used to build the sides costs £{side_cost} per m2. Enter a formula for the cost of the four sides.

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Multiply the area by the cost per square metre.

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Using your answer from part (b), write the formula above in terms of $w$ only.

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Substitute your formula for $h$ in part (b) into the formula in part (f).

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Now write down a formula for the total cost $C$ of building the box in terms of $w$.

\n

$C=\\;$[[0]]

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This is the cost of the top, bottom and sides. So add together the costs for these.

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Differentiate the formula for $C$ to obtain $\\frac{dC}{dw}$.

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Calculate the value of $w$ for which $C$ has a stationary point.

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Solve the equation

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\\[\\frac{dC}{dw}=0\\]

\n

by collecting all the $w$ terms together, and taking the cube root.

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Verify that the stationary point is a minimum by calculating the second derivative $\\frac{d^2C}{dw^2}$ at this value of $w$. (A positive second derivative implies a minimum.)

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Hence enter the dimensions of the box which minimise the cost of materials.

\n

Width: [[0]] m

\n

Length: [[1]] m

\n

Height: [[2]] m

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