1.07a Derivative as gradient: of tangent to curve

3 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 x ^ { - \frac { 1 } { 2 } }\), and the curve passes through the point (4,5). Find the equation of the curve.

5 A is the point \(( 2,1 )\) on the curve \(y = \frac { 4 } { x ^ { 2 } }\).
B is the point on the same curve with \(x\)-coordinate 2.1.

1. Calculate the gradient of the chord AB of the curve. Give your answer correct to 2 decimal places.
2. Give the \(x\)-coordinate of a point C on the curve for which the gradient of chord AC is a better approximation to the gradient of the curve at A .
3. Use calculus to find the gradient of the curve at A .

12

1. Calculate the gradient of the chord joining the points on the curve \(y = x ^ { 2 } - 7\) for which \(x = 3\) and \(x = 3.1\).
2. Given that \(\mathrm { f } ( x ) = x ^ { 2 } - 7\), find and simplify \(\frac { \mathrm { f } ( 3 + h ) - \mathrm { f } ( 3 ) } { h }\).
3. Use your result in part (ii) to find the gradient of \(y = x ^ { 2 } - 7\) at the point where \(x = 3\), showing your reasoning.
4. Find the equation of the tangent to the curve \(y = x ^ { 2 } - 7\) at the point where \(x = 3\).
5. This tangent crosses the \(x\)-axis at the point P . The curve crosses the positive \(x\)-axis at the point Q . Find the distance PQ , giving your answer correct to 3 decimal places.

5 The gradient of a curve is given by the function \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 - x\).
The curve passes through the point \(( 1,2 )\).
Find the equation of the curve.

7 The gradient of a curve \(y = \mathrm { f } ( x )\) is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - 10 x + 6\). The curve passes through the point \(( 2,3 )\) Find the equation of the curve.

2 A particle moves along a straight line containing a point O . Its displacement, \(x \mathrm {~m}\), from O at time \(t\) seconds is given by $$x = 12 t - t ^ { 3 } , \text { where } - 10 \leqslant t \leqslant 10$$ Find the values of \(x\) for which the velocity of the particle is zero.

2 A particle moves along the \(x\)-axis with velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), at time \(t\) given by $$v = 24 t - 6 t ^ { 2 }$$ The positive direction is in the sense of \(x\) increasing.

1. Find an expression for the acceleration of the particle at time \(t\).
2. Find the times, \(t _ { 1 }\) and \(t _ { 2 }\), at which the particle has zero speed.
3. Find the distance travelled between the times \(t _ { 1 }\) and \(t _ { 2 }\).

5 The position vector of a particle at time \(t\) is given by $$\mathbf { r } = \frac { 1 } { 2 } t \mathbf { i } + \left( t ^ { 2 } - 1 \right) \mathbf { j } ,$$ referred to an origin \(\mathbf { O }\) where \(\mathbf { i }\) and \(\mathbf { j }\) are the standard unit vectors in the directions of the cartesian axes \(\mathrm { O } x\) and Oy respectively.

1. Write down the value of \(t\) for which the \(x\)-coordinate of the position of the particle is 2 . Find the \(y\)-coordinate at this time.
2. Show that the cartesian equation of the path of the particle is \(y = 4 x ^ { 2 } - 1\).
3. Find the coordinates of the point where the particle is moving at \(45 ^ { \circ }\) to both \(\mathrm { O } x\) and \(\mathrm { O } y\). Section B (36 marks)

4 Fig. 4 shows the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) in the directions of the cartesian axes \(\mathrm { O } x\) and \(\mathrm { O } y\), respectively. O is the origin of the axes and of position vectors. \begin{figure}[h] \end{figure} The position vector of a particle is given by \(\mathbf { r } = 3 t \mathbf { i } + \left( 18 t ^ { 2 } - 1 \right) \mathbf { j }\) for \(t \geqslant 0\), where \(t\) is time.

1. Show that the path of the particle cuts the \(x\)-axis just once.
2. Find an expression for the velocity of the particle at time \(t\). Deduce that the particle never travels in the j direction.
3. Find the cartesian equation of the path of the particle, simplifying your answer.

3 The points \(\mathrm { P } ( 2,3.6 )\) and \(\mathrm { Q } ( 2.2,2.4 )\) lie on the curve \(y = \mathrm { f } ( x )\). Use P and Q to estimate the gradient of the curve at the point where \(x = 2\).

5 The equation of a curve is \(y = \sqrt { 1 + 2 x }\).

1. Calculate the gradient of the chord joining the points on the curve where \(x = 4\) and \(x = 4\). Give your answer correct to 4 decimal places.
2. Showing the points you use, calculate the gradient of another chord of the curve which is a closer approximation to the gradient of the curve when \(x = 4\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{f540b962-ee6b-409a-a2a1-cd7ad4945514-2_1031_1113_273_499} \captionsetup{labelformat=empty} \caption{Fig. 5}
   \end{figure} Fig. 5 shows the graph of \(y = 2 ^ { x }\).

8 Fig. 6 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \arctan x\). \begin{figure}[h] \end{figure}

1. Find the range of the function \(\mathrm { f } ( x )\), giving your answer in terms of \(\pi\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). Find the gradient of the curve \(y = \mathrm { f } ^ { - 1 } ( x )\) at the origin.
3. Hence write down the gradient of \(y = \frac { 1 } { 2 } \arctan x\) at the origin.

5 Fig. 8 shows the curve \(y = \frac { x } { \sqrt { x - 2 } }\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \(( 11,11 )\). \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 3 .
2. Show that, for the curve, \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 4 } { 2 ( x - 2 ) ^ { \frac { 3 } { 2 } } }\). Hence find the gradient of the curve at P . Use the result to show that the curve is not symmetrical about \(y = x\).
3. Using the substitution \(u = x - 2\), show that \(\int _ { 3 } ^ { 11 } \frac { x } { \sqrt { x - 2 } } \mathrm {~d} x = 25 \frac { 1 } { 3 }\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\).

7.Figure 2 shows a rectangular section of marshland,\(O A B C\) ,which is \(a\) metres long by \(b\) metres wide,where \(a > b\) . \begin{figure}[h] \end{figure} Edgar intends to get from \(O\) to \(B\) in the shortest possible time.In order to do this,he runs along edge \(O A\) for a distance \(x\) metres \(( 0 \leqslant x < a )\) to the point \(D\) before wading through the marsh directly from \(D\) to \(B\) . Edgar can wade through the marsh at a constant speed of \(1 \mathrm {~ms} ^ { - 1 }\) ,and he can run along the edge of the marsh at a constant speed of \(\lambda \mathrm { ms } ^ { - 1 }\) ,where \(\lambda > 1\)

1. By finding an expression in terms of \(x\) for the time taken,\(t\) seconds,for Edgar to reach \(B\) from \(O\) ,show that $$\frac { \mathrm { d } t } { \mathrm {~d} x } = \frac { 1 } { \lambda } - \frac { a - x } { \sqrt { ( a - x ) ^ { 2 } + b ^ { 2 } } }$$
2. 1. Find,in terms of \(a , b\) and \(\lambda\) ,the value of \(x\) for which \(\frac { \mathrm { d } t } { \mathrm {~d} x } = 0\)
   2. Show that this value of \(x\) lies in the interval \(0 \leqslant x < a\) provided \(\lambda \geqslant \sqrt { 1 + \frac { b ^ { 2 } } { a ^ { 2 } } }\)
   3. For \(\lambda\) in this range,show that the value of \(x\) found in(b)(i)gives a minimum value of \(t\) .
3. Find the minimum time taken for Edgar to get from \(O\) to \(B\) if
   1. \(\lambda \geqslant \sqrt { 1 + \frac { b ^ { 2 } } { a ^ { 2 } } }\)
   2. \(1 < \lambda < \sqrt { 1 + \frac { b ^ { 2 } } { a ^ { 2 } } }\) Edgar's friend,Frankie,also runs at a constant speed of \(\lambda \mathrm { m } \mathrm { s } ^ { - 1 }\) .Frankie runs along the edges \(O A\) and \(A B\) .Given that \(\lambda \geqslant \sqrt { 1 + \frac { b ^ { 2 } } { a ^ { 2 } } }\)
4. find the range of values of \(\lambda\) for which Frankie gets to \(B\) from \(O\) in a shorter time than Edgar's minimum time.

7. \begin{figure}[h] \end{figure} Figure 3 shows part of the curve \(C\) with equation \(y = x ^ { 4 } - 10 x ^ { 3 } + 33 x ^ { 2 } - 34 x\) and the line \(L\) with equation \(y = m x + c\) . The line \(L\) touches \(C\) at the points \(P\) and \(Q\) with \(x\) coordinates \(p\) and \(q\) respectively.

1. Explain why $$x ^ { 4 } - 10 x ^ { 3 } + 33 x ^ { 2 } - ( 34 + m ) x - c = ( x - p ) ^ { 2 } ( x - q ) ^ { 2 }$$ The finite region \(R\) ,shown shaded in Figure 3,is bounded by \(C\) and \(L\) .
2. Use integration by parts to show that the area of \(R\) is \(\frac { ( q - p ) ^ { 5 } } { 30 }\)
3. Show that $$( x - p ) ^ { 2 } ( x - q ) ^ { 2 } = x ^ { 4 } - 2 ( p + q ) x ^ { 3 } + S x ^ { 2 } - T x + U$$ where \(S , T\) and \(U\) are expressions to be found in terms of \(p\) and \(q\) .
4. Using part(a)and part(c)find the value of \(p\) ,the value of \(q\) and the equation of \(L\) .

4. $$\mathrm { g } ( x ) = \frac { x ^ { 4 } + x ^ { 3 } - 7 x ^ { 2 } + 8 x - 48 } { x ^ { 2 } + x - 12 } , \quad x > 3 , \quad x \in \mathbb { R }$$

1. Given that $$\frac { x ^ { 4 } + x ^ { 3 } - 7 x ^ { 2 } + 8 x - 48 } { x ^ { 2 } + x - 12 } \equiv x ^ { 2 } + A + \frac { B } { x - 3 }$$ find the values of the constants \(A\) and \(B\).
2. Hence, or otherwise, find the equation of the tangent to the curve with equation \(y = \mathrm { g } ( x )\) at the point where \(x = 4\). Give your answer in the form \(y = m x + c\), where \(m\) and \(c\) are constants to be determined.
   (5)
   

6 \includegraphics[max width=\textwidth, alt={}, center]{918d83c3-1608-4482-9d3d-8af05e65f353-2_394_846_1868_648} The diagram shows part of the curve \(y = x ^ { 2 } + 5\). The point \(A\) has coordinates ( 1,6 ). The point \(B\) has coordinates ( \(a , a ^ { 2 } + 5\) ), where \(a\) is a constant greater than 1 . The point \(C\) is on the curve between \(A\) and \(B\).

1. Find by differentiation the value of the gradient of the curve at the point \(A\).
2. The line segment joining the points \(A\) and \(B\) has gradient 2.3. Find the value of \(a\).
3. State a possible value for the gradient of the line segment joining the points \(A\) and \(C\).

8 A curve has equation \(y = 2 x ^ { 2 }\). The points \(A\) and \(B\) lie on the curve and have \(x\)-coordinates 5 and \(5 + h\) respectively, where \(h > 0\).

1. Show that the gradient of the line \(A B\) is \(20 + 2 h\).
2. Explain how the answer to part (i) relates to the gradient of the curve at \(A\).
3. The normal to the curve at \(A\) meets the \(y\)-axis at the point \(C\). Find the \(y\)-coordinate of \(C\).

2 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x - 4\). The curve passes through the distinct points ( 2,5 ) and ( \(p , 5\) ).

1. Find the equation of the curve.
2. Find the value of \(p\).

3 The equation of a curve is \(y = \sqrt { 1 + 2 x }\).

1. Calculate the gradient of the chord joining the points on the curve where \(x = 4\) and \(x = 4.1\). Give your answer correct to 4 decimal places.
2. Showing the points you use, calculate the gradient of another chord of the curve which is a closer approximation to the gradient of the curve when \(x = 4\).

5

1. The graph of the function \(y = \mathrm { f } ( x )\) passes through the point \(P\) with coordinates (2, 6), and is a one-one function. State the coordinates of the point corresponding to \(P\) on each of the following curves.
   1. \(\quad y = \mathrm { f } ( x ) + 3\)
   2. \(\quad y = 2 \mathrm { f } ( 3 x - 1 )\)
   3. \(y = \mathrm { f } ^ { - 1 } ( x )\)
2. \includegraphics[max width=\textwidth, alt={}, center]{6b766f5c-8533-4e0c-bb10-0d9949dc777b-5_494_739_806_333} The diagram shows part of the graph of \(y = \mathrm { g } ^ { \prime } ( x )\). This is the graph of the gradient function of \(y = \mathrm { g } ( x )\). The graph intersects the \(x\)-axis at \(x = - 2\) and \(x = 4\).
   1. State the \(x\)-coordinate of any stationary points on the graph of \(y = \mathrm { g } ( x )\).
   2. State the set of values of \(x\) for which \(y = \mathrm { g } ( x )\) is a decreasing function.
   3. State the \(x\)-coordinate of any points of inflection on the graph of \(y = \mathrm { g } ( x )\).

5. \begin{figure}[h] \end{figure} Figure 1 shows part of the curve with equation \(y = 3 x ^ { 2 } - 2\) The point \(P ( 2,10 )\) lies on the curve.

1. Find the gradient of the tangent to the curve at \(P\). The point \(Q\) with \(x\) coordinate \(2 + h\) also lies on the curve.
2. Find the gradient of the line \(P Q\), giving your answer in terms of \(h\) in simplest form.
3. Explain briefly the relationship between part (b) and the answer to part (a).

10 A particle \(P\) is moving in a straight line. At time \(t\) seconds \(P\) has velocity \(v \mathrm {~ms} ^ { - 1 }\) where \(v = ( 2 t + 1 ) ( 3 - t )\).

1. Find the deceleration of \(P\) when \(t = 4\).
2. State the positive value of \(t\) for which \(P\) is instantaneously at rest.
3. Find the total distance that \(P\) travels between times \(t = 0\) and \(t = 4\).

5 \includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-04_700_727_260_242} The diagram shows a curve \(C\) for which \(y\) is inversely proportional to \(x\). The curve passes through the point \(\left( 1 , - \frac { 1 } { 2 } \right)\).

1. 1. Determine the equation of the gradient function for the curve \(C\).
   2. Sketch this gradient function on the axes in the Printed Answer Booklet.
2. The diagram indicates that the curve \(C\) has no stationary points. State what feature of your sketch in part (a)(ii) corresponds to this.
3. The curve \(C\) is translated by the vector \(\binom { - 2 } { 0 }\). Find the equation of the curve after it has been translated.

11 \includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-08_586_672_1231_242} A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds, where \(t \geqslant 0\), the velocity of \(P\) in the positive \(x\)-direction is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). It is given that \(v = t ( t - 3 ) ( 8 - t )\). \(P\) attains its maximum velocity at time \(T\) seconds. The diagram shows part of the velocity-time graph for the motion of \(P\).

1. State the acceleration of \(P\) at time \(T\).
2. In this question you must show detailed reasoning. Determine the value of \(T\).
3. Find the total distance that \(P\) travels between times \(t = 0\) and \(t = T\). \includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-09_524_410_251_242} Particles \(P\) and \(Q\), of masses 4 kg and 6 kg respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley. The system is in equilibrium with \(P\) hanging 1.75 m above a horizontal plane and \(Q\) resting on the plane. Both parts of the string below the pulley are vertical (see diagram).
   1. Find the magnitude of the normal reaction force acting on \(Q\). The mass of \(P\) is doubled, and the system is released from rest. You may assume that in the subsequent motion \(Q\) does not reach the pulley.
   2. Determine the magnitude of the force exerted on the pulley by the string before \(P\) strikes the plane.
   3. Determine the total distance travelled by \(Q\) between the instant when the system is released and the instant when \(Q\) first comes momentarily to rest. When this motion is observed in practice, it is found that the total distance travelled by \(Q\) between the instant when the system is released and the instant when \(Q\) first comes momentarily to rest is less than the answer calculated in part (c).
   4. State one factor that could account for this difference.