Related rates problems

10 The equation of a curve is \(\mathrm { y } = ( 5 - 2 \mathrm { x } ) ^ { \frac { 3 } { 2 } } + 5\) for \(x < \frac { 5 } { 2 }\).

1. A point \(P\) is moving along the curve in such a way that the \(y\)-coordinate of point \(P\) is decreasing at 5 units per second. Find the rate at which the \(x\)-coordinate of point \(P\) is increasing when \(y = 32\).
2. Point \(A\) on the curve has \(y\)-coordinate 32. Point \(B\) on the curve is such that the gradient of the curve at \(B\) is - 3 . Find the equation of the perpendicular bisector of \(A B\). Give your answer in the form \(\mathrm { ax } + \mathrm { by } + \mathrm { c } = 0\), where \(a , b\) and \(c\) are integers.
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4 A curve has equation \(y = x ^ { 2 } - 2 x - 3\). A point is moving along the curve in such a way that at \(P\) the \(y\)-coordinate is increasing at 4 units per second and the \(x\)-coordinate is increasing at 6 units per second. Find the \(x\)-coordinate of \(P\).

3 An oil pipeline under the sea is leaking oil and a circular patch of oil has formed on the surface of the sea. At midday the radius of the patch of oil is 50 m and is increasing at a rate of 3 metres per hour. Find the rate at which the area of the oil is increasing at midday.

1. \hspace{0pt} [In this question you may assume the formula for the area of a circle and the following formulae:
   a sphere of radius \(r\) has volume \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\) and surface area \(S = 4 \pi r ^ { 2 }\)
   a cylinder of radius \(r\) and height \(h\) has volume \(V = \pi r ^ { 2 } h\) and curved surface area \(S = 2 \pi r h ]\)

\begin{figure}[h] \end{figure} Figure 5 shows the model for a building. The model is made up of three parts. The roof is modelled by the curved surface of a hemisphere of radius \(R \mathrm {~cm}\). The walls are modelled by the curved surface of a circular cylinder of radius \(R \mathrm {~cm}\) and height \(H \mathrm {~cm}\). The floor is modelled by a circular disc of radius \(R \mathrm {~cm}\). The model is made of material of negligible thickness, and the walls are perpendicular to the base. It is given that the volume of the model is \(800 \pi \mathrm {~cm} ^ { 3 }\) and that \(0 < R < 10.6\)

1. Show that $$H = \frac { 800 } { R ^ { 2 } } - \frac { 2 } { 3 } R$$
2. Show that the surface area, \(A \mathrm {~cm} ^ { 2 }\), of the model is given by $$A = \frac { 5 \pi R ^ { 2 } } { 3 } + \frac { 1600 \pi } { R }$$
3. Use calculus to find the value of \(R\), to 3 significant figures, for which \(A\) is a minimum.
4. Prove that this value of \(R\) gives a minimum value for \(A\).
5. Find, to 3 significant figures, the value of \(H\) which corresponds to this value for \(R\).

5. At time \(t\) seconds the radius of a sphere is \(r \mathrm {~cm}\), its volume is \(V \mathrm {~cm} ^ { 3 }\) and its surface area is \(S \mathrm {~cm} ^ { 2 }\). [You are given that \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\) and that \(S = 4 \pi r ^ { 2 }\) ] The volume of the sphere is increasing uniformly at a constant rate of \(3 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).

1. Find \(\frac { \mathrm { d } r } { \mathrm {~d} t }\) when the radius of the sphere is 4 cm , giving your answer to 3 significant figures.
2. Find the rate at which the surface area of the sphere is increasing when the radius is 4 cm .

6. Some ink is poured onto a piece of cloth forming a stain that then spreads. The area of the stain, \(A \mathrm {~cm} ^ { 2 }\), after \(t\) seconds is given by $$A = ( p + q t ) ^ { 2 }$$ where \(p\) and \(q\) are positive constants.
Given that when \(t = 0 , A = 4\) and that when \(t = 5 , A = 9\),

1. find the value of \(p\) and show that \(q = \frac { 1 } { 5 }\),
2. find \(\frac { \mathrm { d } A } { \mathrm {~d} t }\) in terms of \(t\),
3. find the rate at which the area of the stain is increasing when \(t = 15\).

3 The volume, \(V \mathrm {~m} ^ { 3 }\), of water in a tank after time \(t\) seconds is given by $$V = \frac { t ^ { 3 } } { 4 } - 3 t + 5$$

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} t }\).
2. 1. Find the rate of change of volume, in \(\mathrm { m } ^ { 3 } \mathrm {~s} ^ { - 1 }\), when \(t = 1\).
   2. Hence determine, with a reason, whether the volume is increasing or decreasing when \(t = 1\).
3. 1. Find the positive value of \(t\) for which \(V\) has a stationary value.
   2. Find \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} t ^ { 2 } }\), and hence determine whether this stationary value is a maximum value or a minimum value.
      (3 marks)

8. Some ink is poured onto a piece of cloth forming a stain that then spreads. The area of the stain, \(A \mathrm {~cm} ^ { 2 }\), after \(t\) seconds is given by $$A = ( p + q t ) ^ { 2 } ,$$ where \(p\) and \(q\) are positive constants.
Given that when \(t = 0 , A = 4\) and that when \(t = 5 , A = 9\),

1. find the value of \(p\) and show that \(q = \frac { 1 } { 5 }\),
2. find \(\frac { \mathrm { d } A } { \mathrm {~d} t }\) in terms of \(t\),
3. find the rate at which the area of the stain is increasing when \(t = 15\).

10. Water is being emptied out of a sink. The depth of water, \(y \mathrm {~cm}\), at time \(t\) seconds, may be modelled by $$y = t ^ { 2 } - 14 t + 49 \quad 0 \leqslant t \leqslant 7$$

1. Find the value of \(t\) when the depth of water is 25 cm .
2. Find the rate of decrease of the depth of water when \(t = 3\).

9 Prove that the sum of a rational number and an irrational number is always irrational.
Jodie is attempting to use differentiation from first principles to prove that the gradient of \(y = \sin x\) is zero when \(x = \frac { \pi } { 2 }\) Jodie's teacher tells her that she has made mistakes starting in Step 4 of her working. Her working is shown below.
\includegraphics[max width=\textwidth, alt={}, center]{6b1312f4-9a5c-4465-8129-7d37e99efefe-14_645_1298_584_370} Step 1 Gradient of chord \(A B = \frac { \sin \left( \frac { \pi } { 2 } + h \right) - \sin \left( \frac { \pi } { 2 } \right) } { h }\) Step 2 \(= \frac { \sin \left( \frac { \pi } { 2 } \right) \cos ( h ) + \cos \left( \frac { \pi } { 2 } \right) \sin ( h ) - \sin \left( \frac { \pi } { 2 } \right) } { h }\) Step 3 $$= \sin \left( \frac { \pi } { 2 } \right) \left( \frac { \cos ( h ) - 1 } { h } \right) + \cos \left( \frac { \pi } { 2 } \right) \frac { \sin ( h ) } { h }$$ Step 4
For gradient of curve at \(A\),
let \(h = 0\) then
\(\frac { \cos ( h ) - 1 } { h } = 0\) and \(\frac { \sin ( h ) } { h } = 0\)
Step 5
Hence the gradient of the curve at \(A\) is given by \(\sin \left( \frac { \pi } { 2 } \right) \times 0 + \cos \left( \frac { \pi } { 2 } \right) \times 0 = 0\) Complete Steps 4 and 5 of Jodie's working below, to correct her proof. Step 4 For gradient of curve at \(A\), Step 5 Hence the gradient of the curve at \(A\) is given by