1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

7. (a) Given that \(x = e ^ { t }\), show that

1. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - t } \frac { \mathrm {~d} y } { \mathrm {~d} t }$$
2. $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \left( \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - \frac { \mathrm { d } y } { \mathrm {~d} t } \right)$$ (b) Use you answers to part (a) to show that the substitution \(x = \mathrm { e } ^ { t }\) transforms the differential equation $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = x ^ { 3 }$$ into $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } - 3 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 2 y = \mathrm { e } ^ { 3 t }$$ (c) Hence find the general solution of $$x ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 2 x \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = x ^ { 3 }$$

3. Given that $$y = x - \operatorname { artanh } \left( \frac { 2 x } { 1 + x ^ { 2 } } \right)$$

1. show that $$1 - \frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { 1 - x ^ { 2 } }$$ where \(k\) is a constant to be found.
2. Hence, or otherwise, show that $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \left( 1 - \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = 0$$

3. Given that $$y = \arctan \left( \frac { \sin x } { \cos x - 1 } \right) \quad x \neq 2 n \pi , \quad n \in \mathbb { Z }$$ Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = k$$ where \(k\) is a constant to be found. \(\_\_\_\_\) "

8. \begin{figure}[h] \end{figure} The curve \(C\) shown in Figure 1 has polar equation $$r = 1 + \sin \theta \quad - \frac { \pi } { 2 } < \theta \leqslant \frac { \pi } { 2 }$$ The point \(P\) lies on \(C\) such that the tangent to \(C\) at \(P\) is perpendicular to the initial line.

1. Use calculus to determine the polar coordinates of \(P\). The tangent to \(C\) at the point \(Q\) where \(\theta = \frac { \pi } { 2 }\) is parallel to the initial line.
   The tangent to \(C\) at \(Q\) meets the tangent to \(C\) at \(P\) at the point \(S\), as shown in Figure 1.
   The finite region \(R\), shown shaded in Figure 1, is bounded by the line segments \(Q S , S P\) and the curve \(C\).
2. Use algebraic integration to show that the area of \(R\) is $$\frac { 1 } { 32 } ( a \sqrt { 3 } + b \pi )$$ where \(a\) and \(b\) are integers to be determined.
   (6)

1. (a) Show that the substitution \(z = y ^ { - 2 }\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } + 2 x y = x \mathrm { e } ^ { - x ^ { 2 } } y ^ { 3 }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 4 x z = - 2 x \mathrm { e } ^ { - x ^ { 2 } }$$ (b) Solve differential equation (II) to find \(z\) as a function of \(x\).
(c) Hence find the general solution of differential equation (I), giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

1. (a) Show that the transformation \(z = y ^ { \frac { 1 } { 2 } }\) transforms the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } - 4 y \tan x = 2 y ^ { \frac { 1 } { 2 } }$$ into the differential equation $$\frac { \mathrm { d } z } { \mathrm {~d} x } - 2 z \tan x = 1$$ (b) Solve the differential equation (II) to find \(z\) as a function of \(x\).
(c) Hence obtain the general solution of the differential equation (I). $$\left[ \begin{array} { l } \text { Leave } \\ \text { blank } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \\ \text { " } \end{array} \right.$$

4. \(y = \arctan ( \sqrt { } x ) , \quad x > 0,0 < y < \frac { \pi } { 2 }\).

1. Find the value of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(\mathrm { x } = \frac { 1 } { 4 }\).
2. Show that \(2 x ( 1 + x ) \frac { d ^ { 2 } y } { d x ^ { 2 } } + ( 1 + 3 x ) \frac { d y } { d x } = 0\).

Differentiate, with respect to \(x\),

1. \(\mathrm { e } ^ { 3 x } + \ln 2 x\),
2. \(\left( 5 + x ^ { 2 } \right) ^ { \frac { 3 } { 2 } }\).

$$f ( x ) = \frac { 2 x + 3 } { x + 2 } - \frac { 9 + 2 x } { 2 x ^ { 2 } + 3 x - 2 } , \quad x > \frac { 1 } { 2 }$$

1. Show that \(\mathrm { f } ( x ) = \frac { 4 x - 6 } { 2 x - 1 }\).
2. Hence, or otherwise, find \(\mathrm { f } ^ { \prime } ( x )\) in its simplest form.

4. $$y = \operatorname { artanh } \left( \frac { \cos x + a } { \cos x - a } \right)$$ where \(a\) is a non-zero constant.
Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = k \tan x$$ where \(k\) is a constant to be determined.

1. The ellipse \(E\) has equation \(\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1\)

The point \(P ( 4 \cos \theta , 3 \sin \theta )\) lies on \(E\).

1. Use calculus to show that an equation of the tangent to \(E\) at \(P\) is $$3 x \cos \theta + 4 y \sin \theta = 12$$
2. Determine an equation for the normal to \(E\) at \(P\). The tangent to \(E\) at \(P\) meets the \(x\)-axis at the point \(A\).
   The normal to \(E\) at \(P\) meets the \(y\)-axis at the point \(B\).
3. Show that the locus of the midpoint of \(A\) and \(B\) as \(\theta\) varies has equation $$x ^ { 2 } \left( p - q y ^ { 2 } \right) = r$$ where \(p , q\) and \(r\) are integers to be determined.

1. The ellipse \(E\) has equation

$$\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 9 } = 1$$ The line \(l\) is the normal to \(E\) at the point \(P ( 5 \cos \theta , 3 \sin \theta )\) where \(0 < \theta < \frac { \pi } { 2 }\)

1. Using calculus, show that an equation for \(l\) is $$5 x \sin \theta - 3 y \cos \theta = 16 \sin \theta \cos \theta$$ Given that
   • \(\quad l\) intersects the \(y\)-axis at the point \(Q\)
   • the midpoint of the line segment \(P Q\) is \(M\)
   • determine the exact maximum area of triangle \(O M P\) as \(\theta\) varies, where \(O\) is the origin.
   You must justify your answer.

8. The hyperbola \(H\) has equation \(\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 4 } = 1\). The line \(l _ { 1 }\) is the tangent to \(H\) at the point \(P ( 4 \sec t , 2 \tan t )\).

1. Use calculus to show that an equation of \(l _ { 1 }\) is $$2 y \sin t = x - 4 \cos t$$ The line \(l _ { 2 }\) passes through the origin and is perpendicular to \(l _ { 1 }\).
   The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(Q\).
2. Show that, as \(t\) varies, an equation of the locus of \(Q\) is $$\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 16 x ^ { 2 } - 4 y ^ { 2 }$$

2 In this question, \(L\) is the straight line with equation \(\mathbf { r } = \left( \begin{array} { r } 2 \\ 1 \\ - 1 \end{array} \right) + \lambda \left( \begin{array} { r } - 2 \\ 2 \\ 1 \end{array} \right)\), and \(\mathrm { g } ( x , y , z ) = \left( x y + z ^ { 2 } \right) \mathrm { e } ^ { x - 2 y }\).

1. Find \(\frac { \partial \mathrm { g } } { \partial x } , \frac { \partial \mathrm {~g} } { \partial y }\) and \(\frac { \partial \mathrm { g } } { \partial z }\).
2. Show that the normal to the surface \(\mathrm { g } ( x , y , z ) = 3\) at the point \(( 2,1 , - 1 )\) is the line \(L\). On the line \(L\), there are two points at which \(\mathrm { g } ( x , y , z ) = 0\).
3. Show that one of these points is \(\mathrm { P } ( 0,3,0 )\), and find the coordinates of the other point Q .
4. Show that, if \(x = - 2 \mu , y = 3 + 2 \mu , z = \mu\), and \(\mu\) is small, then $$\mathrm { g } ( x , y , z ) \approx - 6 \mu \mathrm { e } ^ { - 6 }$$ You are given that \(h\) is a small number.
5. There is a point on \(L\), close to P , at which \(\mathrm { g } ( x , y , z ) = h\). Show that this point is approximately $$\left( \frac { 1 } { 3 } \mathrm { e } ^ { 6 } h , 3 - \frac { 1 } { 3 } \mathrm { e } ^ { 6 } h , - \frac { 1 } { 6 } \mathrm { e } ^ { 6 } h \right)$$
6. Find the approximate coordinates of the point on \(L\), close to Q , at which \(\mathrm { g } ( x , y , z ) = h\).

1 Differentiate \(\sqrt [ 3 ] { 1 + 6 x ^ { 2 } }\).

4 When the gas in a balloon is kept at a constant temperature, the pressure \(P\) in atmospheres and the volume \(V \mathrm {~m} ^ { 3 }\) are related by the equation $$P = \frac { k } { V }$$ where \(k\) is a constant. [This is known as Boyle's Law.]
When the volume is \(100 \mathrm {~m} ^ { 3 }\), the pressure is 5 atmospheres, and the volume is increasing at a rate of \(10 \mathrm {~m} ^ { 3 }\) per second.

1. Show that \(k = 500\).
2. Find \(\frac { \mathrm { d } P } { \mathrm {~d} V }\) in terms of \(V\).
3. Find the rate at which the pressure is decreasing when \(V = 100\).

6 A curve has equation \(y = \frac { x } { 2 + 3 \ln x }\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence find the exact coordinates of the stationary point of the curve.

3. A curve has the equation \(y = ( 3 x - 5 ) ^ { 3 }\).

1. Find an equation for the tangent to the curve at the point \(P ( 2,1 )\). The tangent to the curve at the point \(Q\) is parallel to the tangent at \(P\).
2. Find the coordinates of \(Q\).

6. \includegraphics[max width=\textwidth, alt={}, center]{14a2477a-c40e-4b4b-bc39-7100d1df9b4d-2_397_488_1299_632} The diagram shows a vertical cross-section through a vase.
The inside of the vase is in the shape of a right-circular cone with the angle between the sides in the cross-section being \(60 ^ { \circ }\). When the depth of water in the vase is \(h \mathrm {~cm}\), the volume of water in the vase is \(V \mathrm {~cm} ^ { 3 }\).

1. Show that \(V = \frac { 1 } { 9 } \pi h ^ { 3 }\). The vase is initially empty and water is poured in at a constant rate of \(120 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
2. Find, to 2 decimal places, the rate at which \(h\) is increasing
   1. when \(h = 6\),
   2. after water has been poured in for 8 seconds.

3. Differentiate each of the following with respect to \(x\) and simplify your answers.

1. \(\quad \ln ( 3 x - 2 )\)
2. \(\frac { 2 x + 1 } { 1 - x }\)
3. \(x ^ { \frac { 3 } { 2 } } \mathrm { e } ^ { 2 x }\)

9. \includegraphics[max width=\textwidth, alt={}, center]{d1cf3850-964a-4ff1-ae25-f1bc60a6aded-3_501_1111_877_413} The diagram shows a graph of the temperature of a room, \(T ^ { \circ } \mathrm { C }\), at time \(t\) minutes.
The temperature is controlled by a thermostat such that when the temperature falls to \(12 ^ { \circ } \mathrm { C }\), a heater is turned on until the temperature reaches \(18 ^ { \circ } \mathrm { C }\). The room then cools until the temperature again falls to \(12 ^ { \circ } \mathrm { C }\). For \(t\) in the interval \(10 \leq t \leq 60 , T\) is given by $$T = 5 + A \mathrm { e } ^ { - k t } ,$$ where \(A\) and \(k\) are constants.
Given that \(T = 18\) when \(t = 10\) and that \(T = 12\) when \(t = 60\),

1. show that \(k = 0.0124\) to 3 significant figures and find the value of \(A\),
2. find the rate at which the temperature of the room is decreasing when \(t = 20\). The temperature again reaches \(18 ^ { \circ } \mathrm { C }\) when \(t = 70\) and the graph for \(70 \leq t \leq 120\) is a translation of the graph for \(10 \leq t \leq 60\).
3. Find the value of the constant \(B\) such that for \(70 \leq t \leq 120\) $$T = 5 + B \mathrm { e } ^ { - k t }$$

3. Find the coordinates of the stationary points of the curve with equation $$y = \frac { x - 1 } { x ^ { 2 } - 2 x + 5 }$$

5. A curve has the equation \(y = \sqrt { 3 x + 11 }\). The point \(P\) on the curve has \(x\)-coordinate 3 .

1. Show that the tangent to the curve at \(P\) has the equation $$3 x - 4 \sqrt { 5 } y + 31 = 0$$ The normal to the curve at \(P\) crosses the \(y\)-axis at \(Q\).
2. Find the \(y\)-coordinate of \(Q\) in the form \(k \sqrt { 5 }\).