Substitution method

7 \includegraphics[max width=\textwidth, alt={}, center]{e26f21c5-3776-4c86-8440-6959c5e37486-12_444_382_258_886} A water tank has vertical sides and a horizontal rectangular base, as shown in the diagram. The area of the base is \(2 \mathrm {~m} ^ { 2 }\). At time \(t = 0\) the tank is empty and water begins to flow into it at a rate of \(1 \mathrm {~m} ^ { 3 }\) per hour. At the same time water begins to flow out from the base at a rate of \(0.2 \sqrt { } h \mathrm {~m} ^ { 3 }\) per hour, where \(h \mathrm {~m}\) is the depth of water in the tank at time \(t\) hours.

1. Form a differential equation satisfied by \(h\) and \(t\), and show that the time \(T\) hours taken for the depth of water to reach 4 m is given by $$T = \int _ { 0 } ^ { 4 } \frac { 10 } { 5 - \sqrt { } h } \mathrm {~d} h$$
2. Using the substitution \(u = 5 - \sqrt { } h\), find the value of \(T\).

9. (a) Use the substitution \(u = 4 - \sqrt { } x\) to find $$\int \frac { \mathrm { d } x } { 4 - \sqrt { } x }$$ A team of scientists is studying a species of slow growing tree.
The rate of change in height of a tree in this species is modelled by the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 4 - \sqrt { } h } { 20 }$$ where \(h\) is the height in metres and \(t\) is the time measured in years after the tree is planted.
(b) Find the range in values of \(h\) for which the height of a tree in this species is increasing.
(c) Given that one of these trees is 1 metre high when it is planted, calculate the time it would take to reach a height of 10 metres. Write your answer to 3 significant figures. \includegraphics[max width=\textwidth, alt={}, center]{5b698944-41ac-4072-b5e1-c580b7752c39-31_154_145_2599_1804}

1. (a) Given \(0 \leqslant h < 25\), use the substitution \(u = 5 - \sqrt { h }\) to show that

$$\int \frac { \mathrm { d } h } { 5 - \sqrt { h } } = - 10 \ln ( 5 - \sqrt { h } ) - 2 \sqrt { h } + k$$ where \(k\) is a constant.
(6) A team of scientists is studying a species of tree.
The rate of change in height of a tree of this species is modelled by the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { t ^ { 0.2 } ( 5 - \sqrt { h } ) } { 5 }$$ where \(h\) is the height of the tree in metres and \(t\) is the time in years after the tree is planted.
One of these trees is 2 metres high when it is planted.
(b) Use integration to calculate the time it would take for this tree to reach a height of 15 metres, giving your answer to one decimal place.
(c) Hence calculate the rate of change in height of this tree when its height is 15 metres. Write your answer in centimetres per year to the nearest centimetre.

8. Liquid is pouring into a large vertical circular cylinder at a constant rate of \(1600 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) 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 \(4000 \mathrm {~cm} ^ { 2 }\).

1. Show that at time \(t\) seconds, the height \(h \mathrm {~cm}\) of liquid in the cylinder satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - k \sqrt { } h \text {, where } k \text { is a positive constant. }$$ When \(h = 25\), water is leaking out of the hole at \(400 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Show that \(k = 0.02\)
3. Separate the variables of the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - 0.02 \sqrt { } h$$ to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by $$\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { } h } \mathrm {~d} h$$ Using the substitution \(h = ( 20 - x ) ^ { 2 }\), or otherwise,
4. find the exact value of \(\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h\).
5. Hence find the time taken to fill the cylinder from empty to a height of 100 cm , giving your answer in minutes and seconds to the nearest second.

A balloon in the shape of a sphere has volume \(V\) and radius \(r\). Air is pumped into the balloon at a constant rate of \(40\pi\) starting when time \(t = 0\) and \(r = 0\). At the same time, air begins to flow out of the balloon at a rate of \(0.8\pi r\). The balloon remains a sphere at all times.

1. Show that \(r\) and \(t\) satisfy the differential equation $$\frac{dr}{dt} = \frac{50 - r}{5r^2}.$$ [3]
2. Find the quotient and remainder when \(5r^2\) is divided by \(50 - r\). [3]
3. Solve the differential equation in part (a), obtaining an expression for \(t\) in terms of \(r\). [6]
4. Find the value of \(t\) when the radius of the balloon is 12. [1]