SPM.2.AHL.TZ0.11

pestleMathematicsAAHLPaper 2SPM· sl-5-5-integration-introduction-areas-between-curve-and-x-axissource ↗

A large tank initially contains pure water. Water containing salt begins to flow into the tank The solution is kept uniform by stirring and leaves the tank through an outlet at its base. Let $x$ grams represent the amount of salt in the tank and let $t$ minutes represent the time since the salt water began flowing into the tank.

The rate of change of the amount of salt in the tank, $\frac{{{\text{d}}x}}{{{\text{d}}t}}$ , is described by the differential equation $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 10{{\text{e}}^{- \frac{t}{4}}} - \frac{x}{{t + 1}}$ .

Show that $t$ + 1 is an integrating factor for this differential equation.

[2]

Hence, by solving this differential equation, show that $x\left( t \right) = \frac{{200 - 40{{\text{e}}^{ - \frac{t}{4}}}\left( {t + 5} \right)}}{{t + 1}}$ .

[8]

Sketch the graph of $x$ versus $t$ for 0 ≤ $t$ ≤ 60 and hence find the maximum amount of salt in the tank and the value of $t$ at which this occurs.

[5]

Find the value of $t$ at which the amount of salt in the tank is decreasing most rapidly.

[2]

The rate of change of the amount of salt leaving the tank is equal to $\frac{x}{{t + 1}}$ .

Find the amount of salt that left the tank during the first 60 minutes.

[4]

Markscheme / solution

METHOD 1

$I(t) = {{\text{e}}^{\int {P\left( t \right)} \,{\text{d}}t}}$ M1

${{\text{e}}^{\int {\frac{1}{{t + 1}}} \,{\text{d}}t}}$

= ${{\text{e}}^{{\text{ln}}\left( {t + 1} \right)}}$ A1

$= t + 1$ AG

METHOD 2

attempting product rule differentiation on $\frac{{\text{d}}}{{{\text{d}}t}}\left( {x\left( {t + 1} \right)} \right)$ M1

$\frac{{\text{d}}}{{{\text{d}}t}}\left( {x\left( {t + 1} \right)} \right) = \frac{{{\text{d}}x}}{{{\text{d}}t}}\left( {t + 1} \right) + x$

$= \left( {t + 1} \right)\left( {\frac{{{\text{d}}x}}{{{\text{d}}t}} + \frac{x}{{t + 1}}} \right)$ A1

so $t + 1$ is an integrating factor for this differential equation AG

[2 marks]

attempting to multiply through by $\left( {t + 1} \right)$ and rearrange to give (M1)

$\left( {t + 1} \right)\frac{{{\text{d}}x}}{{{\text{d}}t}} + x = 10\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}}$ A1

$\frac{{\text{d}}}{{{\text{d}}t}}\left( {x\left( {t + 1} \right)} \right) = 10\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}}$

$x\left( {t + 1} \right) = \int {10\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}}} {\text{d}}t$ A1

attempting to integrate the RHS by parts M1

$= - 40\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}} + 40\int {{{\text{e}}^{ - \frac{t}{4}}}} \,{\text{d}}t$

$= - 40\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}} - 160{{\text{e}}^{ - \frac{t}{4}}} + C$ A1

Note: Condone the absence of C.

EITHER

substituting $t = 0,\,\,x = 0 \Rightarrow C = 200$ M1

$x = \frac{{- 40\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}} - 160{{\text{e}}^{ - \frac{t}{4}}} + 200}}{{t + 1}}$ A1

using $- 40{{\text{e}}^{ - \frac{t}{4}}}$ as the highest common factor of $- 40\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}}$ and $- 160{{\text{e}}^{ - \frac{t}{4}}}$ M1

using $- 40{{\text{e}}^{ - \frac{t}{4}}}$ as the highest common factor of $- 40\left( {t + 1} \right){{\text{e}}^{ - \frac{t}{4}}}$ and $- 160{{\text{e}}^{ - \frac{t}{4}}}$ giving

$x\left( {t + 1} \right) = - 40{{\text{e}}^{ - \frac{t}{4}}}\left( {t + 5} \right) + C$ (or equivalent) M1A1

substituting $t = 0,\,\,x = 0 \Rightarrow C = 200$ M1

THEN

$x\left( t \right) = \frac{{200 - 40{{\text{e}}^{ - \frac{t}{4}}}\left( {t + 5} \right)}}{{t + 1}}$ AG

[8 marks]

graph starts at the origin and has a local maximum (coordinates not required) A1

sketched for 0 ≤ $t$ ≤ 60 A1

correct concavity for 0 ≤ $t$ ≤ 60 A1

maximum amount of salt is 14.6 (grams) at $t$ = 6.60 (minutes) A1A1

[5 marks]

using an appropriate graph or equation (first or second derivative) M1

amount of salt is decreasing most rapidly at $t$ = 12.9 (minutes) A1

[2 marks]

EITHER

attempting to form an integral representing the amount of salt that left the tank M1

$\int\limits_0^{60} {\frac{{x\left( t \right)}}{{t + 1}}{\text{d}}t}$

$\int\limits_0^{60} {\frac{{200 - 40{{\text{e}}^{ - \frac{t}{4}}}\left( {t + 5} \right)}}{{{{\left( {t + 1} \right)}^2}}}{\text{d}}t}$ A1

attempting to form an integral representing the amount of salt that entered the tank minus the amount of salt in the tank at $t$ = 60(minutes)

amount of salt that left the tank is $\int\limits_0^{60} {10} {{\text{e}}^{ - \frac{t}{4}}}\,{\text{d}}t - x\left( {60} \right)$ A1

THEN

= 36.7 (grams) A2

[4 marks]

Examiners’ report

[N/A]