a) height = diameter

Therefore, \(\large r=\frac{h}{2}\)

In this case, the height of water = x

b) The rate at which the depth of the water is increasing is \(​\large \frac{dx}{dt}\)

To find it, use the chain rule

\(​\large \frac{dV}{dt}\)=\(\large \frac{dV}{dx}\)\(\large \times\)\(\large \frac{dx}{dt}\)

a) The volume of a cone, \(\large V = \frac{1}{3}\pi r^2h\)

In this case the height = diameter

Therefore, \(\large r=\frac{h}{2}\)

The volume of a cone, \(\large V = \frac{1}{3}\pi (\frac{h}{2})^2h\\ \large V = \frac{1}{3}\pi (\frac{h^2}{4})h\\ \large V = \pi \frac{h^3}{12}\)

When the depth (height) of water= x, then

\(\large V=\frac{1}{12} \pi x^3\)

b) \(\large V=\frac{1}{12} \pi x^3\)

Differentiate with respect to x , \(\large \frac{dV}{dx}=\frac{1}{4}\pi r^2\)

When x = 12 cm , \(\large \frac{dV}{dx}=\frac{1}{4}\pi \times12^2=36\pi\)

Water is added at a rate of \(\large 18\pi\) cm³ per minute, \(\large \frac{dV}{dt}=18 \pi\)

Using the chain rule

\(​\large \frac{dV}{dt}\)=\(\large \frac{dV}{dx}\)\(\large \times\)\(\large \frac{dx}{dt}\)

\(\large 18 \pi\) =\(\large36\pi\)\(\large \times\)\(\large \frac{dx}{dt}\)

\(\large \frac{dx}{dt}=\frac{18\pi}{36\pi}\)

\(\large \frac{dx}{dt}=0.5\) cm per minute

If we let x be the distance along the wall, then the speed of the spotlight is\(​\large \frac{dx}{dt}\)

To find it, use the chain rule

\(​\large \frac{dx}{dt}\)=\(\large \frac{dx}{d\theta}\)\(\large \times\)\(\large \frac{d\theta}{dt}\)

To find \(\large \frac{dx}{d\theta}\) use right-angled triangle trigonometry to find x in terms of \(\large \theta\)

The searchlight is rotating at 2 revolutions per minute

1 revolution is \(\large 2\pi \) radians

Therefore, \(\large \frac{d\theta}{dt}=4\pi\) radians per minute

Let the distance fom from the perpendicular be x

\(\large \sin \theta=\frac{x}{30}\)

\(\large x=30 \sin\theta\)

Differentiate with respect to \(\large \theta\)

\(\large \frac{dx}{d\theta}=30 \cos \theta\)

When \(\theta\)=45°, \(\large \frac{dx}{d\theta}=30 \cos45°=15\sqrt{2}\)

Use the chain rule

\(​\large \frac{dx}{dt}\)=\(\large \frac{dx}{d\theta}\)\(\large \times\)\(\large \frac{d\theta}{dt}\)

\(​\large \frac{dx}{dt}\)=\(\large 15\sqrt{2}\)\(\large \times\)\(\large 4\pi\)

\(​\large \frac{dx}{dt}=60\sqrt{2}\pi\approx267\)m per minute

If we let x be the distance along the ground, then the speed end A slips away from the wall is \(​\large \frac{dx}{dt}\)

If w let y be the distance up the wall, then the speed end B slips down the wall is \(​\large \frac{dy}{dt}\)

Use Pythagoras' Theorem

The speed end A slips away from the wall is \(​\large \frac{dx}{dt}\)

\(​\large \frac{dx}{dt}=0.5\) ms^-1

The speed end B slips down the wall is \(​\large \frac{dy}{dt}\)

Using Pythagoras' Theorem

\(\large x^2+y^2=8^2\)

\(\large x^2+y^2=64\)

Differentiate with respect to t (implicit differentiation)

\(\large 2x \frac{dx}{dt}+2y\frac{dy}{dt}=0\)

When y =5 ,

\(\large x^2+5^2=64\\ \large x^2=39\\ \large x=\sqrt {39}\)

Substitute these values in the differential equation

\(\large 2x \frac{dx}{dt}+2y\frac{dy}{dt}=0\)

\(\large 2\sqrt{39}\times 0.5+2\times 5\frac{dy}{dt}=0\\ \large 10\frac{dy}{dt}=-\sqrt{39}\\ \large \frac{dy}{dt}=-\frac{\sqrt{39}}{10}\\ \large \frac{dy}{dt}=-0.624\)

Notice that the value is negative. This represents the velocity and shows that it is slipping down.

The speed is 0.624 ms^-1

We can find the volume of the solid to be \(\large V=\frac{2}{3}\pi r^3+\pi r^2h\)

We know \(​\large \frac{dr}{dt}\) and \(​\large \frac{dV}{dt}\). We are required to find \(​\large \frac{dh}{dt}\).

Use implicit differentiation and differentiate the formula for V with respect to t

a) The volume of a hemisphere is \(\large \frac{2}{3}\pi r^3\)

The volume of a cylinder is \(\large \pi r^2h\)

The volume of the solid, \(\large V=\frac{2}{3}\pi r^3+\pi r^2h\)

b) We can find the value of h when r = 6 and V = \(\large 684\pi\)

\(\large 684\pi=\frac{2}{3}\pi \times 6^3+\pi \times 6^2\times h\\ \large 684\pi=144\pi+36\pi h\\ \large 540\pi=36\pi h\\ \large h = 15\)

The radius is changing at a rate of 1.5 cm/minute, \(​\large \frac{dr}{dt}=1.5\)

The volume is changing at a rate of \(\large 1800\pi\) cm³ /min, \(​\large \frac{dV}{dt}=1800\pi\)

We are required to find \(​\large \frac{dh}{dt}\).

Use implicit differentiation and differentiate the formula for V with respect to t

\(\large V=\frac{2}{3}\pi r^3+\pi r^2h\)

Notice that the second part of the expression is a product (both r and h are variables). Therefore we need to use the Product Rule for differentiation

\(\large \frac{dV}{dt}=2\pi r^2\frac{dr}{dt}+\pi r^2\frac{dh}{dt}+2\pi r^2\frac{dr}{dt}\cdot h\)

\(\large 1800\pi=2\pi \times 6^2\times 1.5+\pi \times 6^2\frac{dh}{dt}+2\pi \times 6^2\times 1.5\times 15\\ \large 1800\pi=108\pi+36\pi \cdot \frac{dh}{dt}+1620\pi\\ \large 72\pi=36\pi \cdot \frac{dh}{dt}\\ \large \frac{dh}{dt}=2\)

\(\large \frac{dh}{dt}=2\) cm/min

When you apply l'Hôpital's Rule, simplify your answer.

\(\large \lim\limits_{x\rightarrow 0}\frac{e^{x^2}-1}{\sin x^2}=\frac{1-1}{0}=\frac{0}{0}\)

This is in the indeterminate form

Let \(\large f(x) = e^{x^2}-1\) and \(\large g(x) =\sin x^2\)

Then, \(\large f'(x) = 2xe^{ x^2}\) and \(\large g'(x) = 2x\cos x^2\)

Therefore,

\(\large \lim\limits_{x\rightarrow 0}\frac{e^{x^2}-1}{\sin x^2}= \lim\limits_{x\rightarrow 0}\frac{2xe^{x^2}}{2x\cos x^2}\)

We can simplify this

\(\large \lim\limits_{x\rightarrow 0}\frac{e^{x^2}}{\cos x^2}=\frac{1}{1}=1\)

You will need to apply l'Hôpital's rule twice.

The second application will need simplifying

\(\large \lim\limits_{x\rightarrow 0}\frac{1-\cos x^2}{x^4}=\frac{1-1}{0}=\frac{0}{0}\)

This is in the indeterminate form

Let \(\large f(x) = 1-\cos x^2\) and \(\large g(x) = x^4\)

Then, \(\large f'(x) = 2x\sin x^2\) and \(\large g'(x) = 4x^3\)

Therefore,

\(\large \lim\limits_{x\rightarrow 0}\frac{1-\cos x^2}{x^4}= \lim\limits_{x\rightarrow 0}\frac{2x\sin x^2}{4x^3}\)

We can simplify this

\(\large \lim\limits_{x\rightarrow 0}\frac{\sin x^2}{2x^2}=\frac{0}{0}\)

We can apply l'Hôpital's rule again

Let \(\large f(x) = \sin x^2\) and \(\large g(x) = 2x^2\)

Then, \(\large f'(x) = 2x\cos x^2\) and \(\large g'(x) = 4x\)

Therefore, \(\large \lim\limits_{x\rightarrow 0}\frac{\sin x^2}{2x^2}= \lim\limits_{x\rightarrow 0}\frac{2x\cos x^2}{4x}\)

We can simplify this

\(\large \lim\limits_{x\rightarrow 0}\frac{\cos x^2}{2}=\frac{1}{2}\)

Therefore, \(\large \lim\limits_{x\rightarrow 0}\frac{1-\cos x^2}{x^4}=\frac{1}{2}\)

You will need to apply l'Hôpital's rule three times.

Ensure that you simplify the limit if it is possible.

\(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}=\frac{0-0}{0}=\frac{0}{0}\)

This is in the indeterminate form

Let \(\large f(x) = 6\tan x-6x\) and \(\large g(x) = x^3\)

Then, \(\large f'(x) = 6 \sec ^2x-6\) and \(\large g'(x) = 3x^2\)

Therefore,

\(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}= \lim\limits_{x\rightarrow 0}\frac{\frac{6}{(\cos x)^2}-6}{3x^2}=\frac{0}{0}\)

We can apply l'Hôpital's rule again

Let

\(\large f(x) = \frac{6}{(\cos x)^2}-6\\ \large f(x) =6(\cos x)^{-2}-6\) and \(\large g(x) = 3x^2\)

Then,

\(\large f'(x) =6(-2)(-\sin x)(\cos x)^{-3}\\ \large f'(x)=\frac{12\sin x}{(\cos x)^3}\) and \(\large g'(x) = 6x\)

Therefore, \(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}= \lim\limits_{x\rightarrow 0}\frac{\frac{12\sin x}{(\cos x)^3}}{6x}\)

We can simplify this

\(\large \lim\limits_{x\rightarrow 0}\frac{2\sin x}{x(\cos x)^3}\)

We can apply l'Hôpital's rule again

Let

\(\large f(x) = 2\sin x\) and \(\large g(x) = x(\cos x)^3\)

Then,

\(\large f'(x) =2\cos x\) and \(\large g'(x) = 1(\cos x)^3+x\cdot3(-\sin x)(\cos x)^2\\ \large g'(x) = (\cos x)^3-3x\sin x(\cos x)^2\)

Therefore, \(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}=\lim\limits_{x\rightarrow 0}\frac{2\cos x}{ (\cos x)^3-3x\sin x(\cos x)^2}=\frac{2}{1-0}=2\)

\(\large \lim\limits_{x\rightarrow 0}\frac{6\tan x-6x}{x^3}=2\)

The volume of the solid is found using the formula

\(\large V= \int_{0}^{3} y^2 \,dx \\ \large V= \int_{0}^{3} (4-(x-1)^2 \,)dx \)

You can use your calculator for this question.

Ensure that your calculator is in radian mode.

You can find the volume generated by rotating the trapezium around the y axis and subtract the volume generated by rotating the curve around the y axis

To perfom the integration, write

\(\large \cos^3y=\cos y\cdot\cos^2y=\cos y\ (1-\sin^2y)\)

It is the integration that makes this question difficult.

You will need to use both integration by parts and integration by substitution/recognition.

For the integration by substitution, let

\(\large u=x \qquad \frac{dv}{dx}=\sec^2x\)

a i) \(\large f(x)=\frac{x^2-7x+12}{x-1} \quad ,x\neq1\)

x intercepts when f(x) = 0

\(\large \frac{x^2-7x+12}{x-1}=0\\ \large {x^2-7x+12}=0\\ \large (x-3)(x-4)=0\\ \large x = 3, \quad x = 4\)

a ii) y intercepts when x = 0

\(\large f(0)=\frac{12}{-1} =-12\)

b) \(\large f(x)=\frac{x^2-7x+12}{x-1} \quad ,x\neq1\)

Vertical asymptote when \(\large x - 1 = 0\\ \large x= 1\)

c) To find oblique asymptote, divide

Hence, \(\large f(x)=x-6+\frac{18}{x-1}\)

Hence, oblique asymptote at y = x - 6

d)

e) \(\large \frac{1}{f(x)}=\frac{x-1}{x^2-7x+12} \quad ,x\neq3, x\neq4\)

f)

This question requires you to how to find the area under the graph: Area = \(\int _{ a }^{ b }{ y }\ dx\)

The integral requires integration by substitution

u = x² +1

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a) This requires you to use the Quotient Rule

b) Hence means that you should use the result from part a)

Use Integration by substitution u = tanx

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To complete this question, we need to split the integral into two parts

\(\int { \frac { arcsinx+x }{ \sqrt { 1-x^{ 2 } } } } dx=\int { \frac { arcsinx }{ \sqrt { 1-x^{ 2 } } } } dx+\int { \frac { x }{ \sqrt { 1-x^{ 2 } } } } dx\)

Both these integrals can be completed using Integration by substitution

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This might help

\(sin^5x = sin^4x\cdot sinx = (sin^2x)^2\cdot sinx\)

You will need to use Integration by substitution

Can you think of a substitution that will work?

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This is a tricky question that requires very careful manipulation

The substitution gives \(\frac{dx}{d\theta}=\frac{4}{3}cos\theta\)

The integral becomes \(\int { \sqrt { 16-16sin^{ 2 }\theta } \cdot \frac { 4 }{ 3 } cos\theta \quad d\theta } \)

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The solution requires you to use Integration by Parts

Think of the question as \(\int { 1\cdot arctanxdx } \)

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This question requires you to use Integration by Parts

In the second part you will need to integrate a rational function. Since the degree of the numerator is equal to the degree of the denominator, you need to divide the polynomials

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