# Tricky Edexcel C4 Examination Question Solved (January 2008) Paper

Question (8):

Liquid is pouring into a large vertical circular cylinder at a constant rate of 1600 cm^3 per second and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is 4000 cm^2.

a) Show that at time seconds, the height h cm of liquid in the cylinder satisfies the differential equation: $\frac { dh }{ dt } =0.4-k\sqrt { h }$, where is a positive constant.

Solution:

Well firstly let’s draw a little diagram to depict what is actually happening: As you can see, dV/dt is the rate at which the volume of liquid in the cylinder is changing over time. You have to remember that liquid is being poured into  the large vertical cylinder at a constant rate of 1600 cm^3 per second. What you also have to note is that the vertical cylinder is losing liquid at a rate proportional to the square root of the height already in the cylinder. Therefore we have $\frac { dV }{ dt } =1600-C\sqrt { h }$.

Now the area of the circular cross section should be: $A=\pi { r }^{ 2 }$. However, we are told that A=4000. Knowing that the Volume of the cylinder is $V=\pi { r }^{ 2 }h$, we can transform the volume formula to V=4000h.

Ok, so how can we get dh/dt? This is the formula we can use to derive it: $\frac { dh }{ dt } =\frac { dh }{ dV } \cdot \frac { dV }{ dt }$

We know that: $V=4000h,\\ \\ \therefore \quad \frac { dV }{ dh } =4000,\quad and\quad as\quad \frac { dh }{ dV } =\frac { 1 }{ \frac { dV }{ dh } } ,\\ \frac { dh }{ dV } =\frac { 1 }{ 4000 }$.

We also know that: $\frac { dV }{ dt } =1600-C\sqrt { h }$

So what we get is: $\frac { dh }{ dt } =\frac { 1 }{ 4000 } \cdot \left( 1600-C\sqrt { h } \right) \\ \\ =\frac { 1600 }{ 4000 } -\frac { C }{ 4000 } \sqrt { h } \\ \\ =0.4-k\sqrt { h } ,\\ \\ \therefore \quad k=\frac { C }{ 4000 }$.

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b) When h=25, water is leaking out of the hole at 400 cm^3 per second. Show that k=0.02.

Solution:

Ok, with this information we can say that: $\frac { dV }{ dt } =1600-400$

Moving on, we can determine that: $\frac { dV }{ dt } =1600-400,\quad but\quad \frac { dV }{ dt } =1600-C\sqrt { h } .\\ \\ Since,\quad h=25,\quad and\quad C\sqrt { h } =400,\\ \\ 5C=400,\quad \therefore \quad C=80.\\ \\ But\quad k=\frac { C }{ 4000 } =\frac { 80 }{ 4000 } ,\\ \\ \therefore \quad k=0.02.\\ \\$

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c) Separate the variables of the differential equation: $\frac { dh }{ dt } =0.4-0.02\sqrt { h } \\ \\$,

to show that the time taken to fill the cylinder from empty to a height of 100cm is given by: $\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh\\ \\$

Solution: $\frac { dh }{ dt } =\frac { 2 }{ 5 } -\frac { \sqrt { h } }{ 50 } =\frac { 20 }{ 50 } -\frac { \sqrt { h } }{ 50 } \\ \\ =\frac { 20-\sqrt { h } }{ 50 } \\ \\ \frac { dh }{ dt } =\frac { 20-\sqrt { h } }{ 50 } \\ \\ dh=\frac { 20-\sqrt { h } }{ 50 } dt\\ \\ 50dh=20-\sqrt { h } dt\\ \\ \frac { 50 }{ 20-\sqrt { h } } dh=1dt\\ \\ \int { 1dt=\int { \frac { 50 }{ 20-\sqrt { h } } } } dh\\ \\ \therefore \quad t=\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh\\ \\$

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d) Using the substitution: $h={ \left( 20-x \right) }^{ 2 }\\ \\$, or otherwise, find the exact value of: $\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh\\ \\$

Solution: $If\quad h={ \left( 20-x \right) }^{ 2 },\quad \\ \\ { h }^{ \frac { 1 }{ 2 } }=\sqrt { h } ={ \left[ { \left( 20-x \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }=\left( 20-x \right) .\\ \\ If\quad h={ \left( 20-x \right) }^{ 2 }={ p }^{ 2 },\quad \frac { dh }{ dp } =2p,\quad p=20-x\quad \therefore \quad \frac { dp }{ dx } =-1\\ \\ So\quad \frac { dh }{ dx } =-2\left( 20-x \right) ,\quad \therefore \quad dh=-2\left( 20-x \right) dx\\ \\ When\quad h=100,\quad { \left( 20-x \right) }^{ 2 }=100,\quad \therefore \quad x=10.\\ When\quad h=0,\quad { \left( 20-x \right) }^{ 2 }=0,\quad \therefore \quad x=20.\$

Therefore, we need to integrate: $\int _{ 20 }^{ 10 }{ \frac { 50 }{ 20-\left( 20-x \right) } \cdot \frac { -2\left( 20-x \right) }{ 1 } } dx=\int _{ 20 }^{ 10 }{ \frac { -100\left( 20-x \right) }{ x } } dx\\ \\ =\int _{ 20 }^{ 10 }{ \frac { -2000+100x }{ x } } dx=\int _{ 20 }^{ 10 }{ -\frac { 2000 }{ x } } +\frac { 100x }{ x } dx\\ \\ =\int _{ 20 }^{ 10 }{ -\frac { 2000 }{ x } } +100dx.\\ \\ But\quad as\quad -\int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ b }^{ a }{ f\left( x \right) dx } ,\\ \\ \int _{ 20 }^{ 10 }{ -\frac { 2000 }{ x } } +100dx=\int _{ 10 }^{ 20 }{ \frac { 2000 }{ x } } -100dx\\ \\ ={ \left[ 2000\ln { x } -100x \right] }_{ 10 }^{ 20 }\\ \\ =2000\ln { 20 } -2000-\left( 2000\ln { 10-1000 } \right) \\ \\ =2000\ln { 20 } -2000-2000\ln { 10 } +1000\\ \\ =2000\left( \ln { 20-\ln { 10 } } \right) -2000+1000\\ \\ =2000\ln { 2-1000 }$

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e) Hence find the time taken to fill the cylinder from empty to a height of 100cm, giving your answer in minutes and seconds to the nearest second.

Solution:

Remember that: $t=\int _{ 0 }^{ 100 }{ \frac { 50 }{ 20-\sqrt { h } } } dh$.

As this is the case, $t=2000\ln { 2 } -1000=386\quad seconds=6\quad minutes\quad and\quad 26\quad seconds.$

# Coming Up With The Formula For Areas Underneath Curves $Area\quad PMTN<\delta A

# Proving The Product Rule $If\quad y=u\cdot v,\quad \left( y+\delta y \right) =\left( u+\delta u \right) \left( v+\delta v \right) \\ \\ y+\delta y=uv+u\delta v+v\delta u+\delta u\delta v\\ \\ but\quad y=uv,\\ \\ \delta y=u\delta v+v\delta u+\delta u\delta v\\ \\ \frac { \delta y }{ \delta x } =u\frac { \delta v }{ \delta x } +v\frac { \delta u }{ \delta x } +\frac { \delta u }{ \delta x } \delta v\\ \\ But\quad as\quad \delta x\rightarrow 0,\quad \frac { \delta y }{ \delta x } \rightarrow \frac { dy }{ dx } ,\quad \frac { \delta v }{ \delta x } \rightarrow \frac { dv }{ dx } ,\quad \frac { \delta u }{ \delta x } \rightarrow \frac { du }{ dx } \quad and\quad \delta v\rightarrow 0.\\ \\ \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx }$

# Proving The Quotient Rule… $y=\frac { u }{ v } ,\quad \therefore \quad \left( y+\delta y \right) =\frac { \left( u+\delta u \right) }{ \left( v+\delta v \right) } \\ \\ \left( y+\delta y \right) \left( v+\delta v \right) =u+\delta u\\ \\ yv+y\delta v+v\delta y+\delta y\delta v=u+\delta u\\ \\ but...\quad yv=u,\\ \\ so...\\ \\ y\delta v+v\delta y+\delta y\delta v=\delta u\\ \\ \frac { u }{ v } \delta v+v\delta y+\delta y\delta v=\delta u\\ \\ u\delta v+{ v }^{ 2 }\delta y+v\delta y\delta v=v\delta u\\ \\ { v }^{ 2 }\delta y=v\delta u-u\delta v-v\delta y\delta v\\ \\ { v }^{ 2 }\frac { \delta y }{ \delta x } =v\frac { \delta u }{ \delta x } -u\frac { \delta v }{ \delta x } -v\delta v\frac { \delta y }{ \delta x } \\ \\ But\quad as\quad \delta x\rightarrow 0,\quad \frac { \delta y }{ \delta x } \rightarrow \frac { dy }{ dx } ,\quad \frac { \delta u }{ \delta x } \rightarrow \frac { du }{ dx } ,\quad \frac { \delta v }{ \delta x } \rightarrow \frac { dv }{ dx } \quad and\quad \delta v\rightarrow 0.\\ \\ So\quad you\quad get:\\ \\ { v }^{ 2 }\frac { dy }{ dx } =v\frac { du }{ dx } -u\frac { dv }{ dx } \\ \\ \frac { dy }{ dx } =\frac { v\frac { du }{ dx } -u\frac { dv }{ dx } }{ { v }^{ 2 } }$