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