# Derivative of y=cotx

$y=cotx=\frac { cosx }{ sinx } =\frac { u }{ v } ,\\ \\ If\quad y=\frac { u }{ v } ,\quad \frac { dy }{ dx } =\frac { v\frac { du }{ dx } -u\frac { dv }{ dx } }{ { v }^{ 2 } } .\\ \\ u=cosx,\quad \therefore \quad \frac { du }{ dx } =-sinx\\ v=sinx,\quad \therefore \quad \frac { dv }{ dx } =cosx,\quad { v }^{ 2 }=\sin ^{ 2 }{ x } \\ \\ So:\\ \\ \frac { dy }{ dx } =\frac { sinx\cdot \left( -sinx \right) -cosxcosx }{ \sin ^{ 2 }{ x } } \\ \\ =\frac { -\sin ^{ 2 }{ x } -\cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } =\frac { -\sin ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } -\frac { \cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } \\ \\ =-1-\cot ^{ 2 }{ x } =-{ cosec }^{ 2 }x\\ \\ \therefore \quad If\quad y=cotx,\quad \frac { dy }{ dx } =-{ cosec }^{ 2 }x\\ \\ \\$

# tan(A-B)=(tanA-tanB)/(1+tanAtanB)

Prove that:

$tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB }$

Firstly:

$tan\left( \alpha +\beta \right) =\frac { tan\alpha +tan\beta }{ 1-tan\alpha tan\beta }$

Also, remember that:

$tan\left( -\theta \right) =-tan\theta$

So:

$\beta =-\phi \\ \\ tan\left( \alpha +\left( -\phi \right) \right) =\frac { tan\alpha +tan\left( -\phi \right) }{ 1-tan\alpha tan\left( -\phi \right) } \\ \\ =\frac { tan\alpha -tan\phi }{ 1+tan\alpha tan\phi } =tan\left( \alpha -\phi \right) \\ \\ \therefore \quad tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB }$

# How To Derive The Quadratic Formula

$a{ x }^{ 2 }+bx+c=0\\ \\ a{ x }^{ 2 }+bx=-c\\ \\ { x }^{ 2 }+\frac { b }{ a } x=-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }-{ \left( \frac { b }{ 2a } \right) }^{ 2 }=-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }-\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } =-\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } -\frac { c }{ a } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=\frac { { b }^{ 2 } }{ 4{ a }^{ 2 } } -\frac { 4ac }{ 4{ a }^{ 2 } } \\ \\ { \left( x+\frac { b }{ 2a } \right) }^{ 2 }=\frac { { b }^{ 2 }-4ac }{ 4{ a }^{ 2 } } \\ \\ x+\frac { b }{ 2a } =\pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ \sqrt { 4 } \sqrt { { a }^{ 2 } } } =\pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \\ x=-\frac { b }{ 2a } \pm \frac { \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \\ x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

— USEFUL FORMULAS:

$\frac { a }{ c } \pm \frac { b }{ c } =\frac { a\pm b }{ c } \\ \\ \frac { a }{ b } \pm \frac { c }{ d } =\frac { ad }{ bd } \pm \frac { bc }{ bd } =\frac { ad\pm bc }{ bd } \\ \\ \sqrt { { a }^{ 2 } } ={ a }^{ \frac { 2 }{ 2 } }=a\\ \\ \sqrt { a } \sqrt { b } =\sqrt { ab } \\ \\ \frac { \sqrt { a } }{ \sqrt { b } } =\sqrt { \frac { a }{ b } }$

# Quick Way To Derive The Integration By Parts Formula

Firstly, you need to know what the product rule is:

$If\quad y=u\cdot v,\quad \frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } .$

Then…

$\frac { dy }{ dx } =u\frac { dv }{ dx } +v\frac { du }{ dx } \\ \\ u\frac { dv }{ dx } =\frac { dy }{ dx } -v\frac { du }{ dx }$

Now integrate each term with respect to x:

$\int { u\frac { dv }{ dx } } dx=\int { \frac { dy }{ dx } } dx-\int { v\frac { du }{ dx } } dx$

Leaving:

$\int { u\frac { dv }{ dx } } dx=y-\int { v\frac { du }{ dx } } dx\\ \\ As\quad y=u\cdot v,\\ \\ \int { u\frac { dv }{ dx } } dx=uv-\int { v\frac { du }{ dx } } dx$

# Derivate Of y=cosx Proof

Prove that:

$If\quad f(x)=cosx,\quad f$

$\lim _{ \delta x\rightarrow 0 }{ \frac { f\left( x+\delta x \right) -f\left( x \right) }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cos\left( x+\delta x \right) -cosx }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosxcos\delta x-sinxsin\delta x-cosx }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosx\left( cos\delta x-1 \right) -sinxsin\delta x }{ \delta x } } \\ \\ =\lim _{ \delta x\rightarrow 0 }{ \frac { cosx\left( cos\delta x-1 \right) }{ \delta x } } -\frac { sinxsin\delta x }{ \delta x } \\ \\ =0-sinx=-sinx\\ \\ As:\\ \\ \lim _{ \delta x\rightarrow 0 }{ \frac { \left( cos\delta x-1 \right) }{ \delta x } } =0\\ \\ \lim _{ \delta x\rightarrow 0 }{ \frac { sin\delta x }{ \delta x } } =1$

# sin(A-B)=sinAcosB-cosAsinB

Prove that:

$sin(A-B)=sinAcosB-cosAsinB$

$sin\left( \alpha +\beta \right) =sin\alpha cos\beta +cos\alpha sin\beta \\ \\ But:\quad \\ \\ sin(-\theta )=-sin\theta \\ cos(-\theta )=cos\theta \\ tan(-\theta )=-tan\theta \\ \\ Say:\quad \beta =-\phi \\ \\ \therefore \quad sin\left( \alpha +\left( -\phi \right) \right) =sin\alpha cos\left( -\phi \right) +cos\alpha sin\left( -\phi \right) \\ \\ =sin\alpha cos\phi -cos\alpha sin\phi =sin\left( \alpha -\phi \right) \\ \\ If\quad \alpha =A\quad and\quad \phi =B\\ \\ sin\left( A-B \right) =sinAcosB-cosAsinB.$

# tan(A+B)=(tanA+tanB)/(1-tanAtanB)

Prove that:

$tan(A+B)=\frac { tanA+tanB }{ 1-tanAtanB }$

$LHS=tan(A+B)=\frac { sin(A+B) }{ cos(A+B) } \\ \\ =\frac { sinAcosB+cosAsinB }{ cosAcosB-sinAsinB } =\frac { \frac { sinA }{ 1 } \cdot \frac { sinB }{ tanB } +\frac { sinA }{ tanA } \cdot \frac { sinB }{ 1 } }{ \frac { sinA }{ tanA } \cdot \frac { sinB }{ tanB } -\frac { sinAsinB }{ 1 } } \\ \\ =\frac { \frac { sinAsinB }{ tanB } +\frac { sinAsinB }{ tanA } }{ \frac { sinAsinB }{ tanAtanB } -\frac { sinAsinB }{ 1 } } =\frac { sinAsinB\left( \frac { 1 }{ tanB } +\frac { 1 }{ tanA } \right) }{ sinAsinB\left( \frac { 1 }{ tanAtanB } -\frac { 1 }{ 1 } \right) } \\ \\ =\frac { \frac { 1 }{ tanB } +\frac { 1 }{ tanA } }{ \frac { 1 }{ tanAtanB } -\frac { 1 }{ 1 } } =\frac { \frac { tanA+tanB }{ tanAtanB } }{ \frac { 1-tanAtanB }{ tanAtanB } } =\frac { tanA+tanB }{ tanAtanB } \cdot \frac { tanAtanB }{ 1-tanAtanB } \\ \\ =\frac { tanA+tanB }{ 1-tanAtanB } =RHS$

# 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:

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

—————————————————————

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:

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

———————————————————————

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:

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

——————————————————————–

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

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