1.07a Derivative as gradient: of tangent to curve

Given that \(y = \arcsin x\) prove that

1. \(\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\) [3]
2. \((1-x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0\) [4]

(Total 7 marks)

A particle \(P\) moves on the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v\) m s\(^{-1}\) in the direction of \(x\) increasing, where $$v = (t - 2)(3t - 10), \quad t \geq 0$$ When \(t = 0\), \(P\) is at the origin \(O\).

1. Find the acceleration of \(P\) when \(t = 3\) [3]
2. Find the total distance travelled by \(P\) in the first 3 seconds of its motion. [6]
3. Show that \(P\) never returns to \(O\). [2]

At time \(t\) seconds, a particle \(P\) has position vector \(r\) metres relative to a fixed origin \(O\), where $$r = (t^2 + 2t)\mathbf{i} + (t - 2t^2)\mathbf{j}.$$ Show that the acceleration of \(P\) is constant and find its magnitude. [5]

A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v \text{ ms}^{-1}\) in the positive \(x\)-direction, where \(v = 3t^2 - 4t + 3\). When \(t = 0\), \(P\) is at the origin \(O\). Find the distance of \(P\) from \(O\) when \(P\) is moving with minimum velocity. [8]

A particle \(P\) is moving in a plane. At time \(t\) seconds, \(P\) is moving with velocity \(\mathbf{v}\) m s\(^{-1}\), where \(\mathbf{v} = 2t\mathbf{i} - 3t^2\mathbf{j}\). Find

1. the speed of \(P\) when \(t = 4\) [2]
2. the acceleration of \(P\) when \(t = 4\) [3]

Given that \(P\) is at the point with position vector \((-4\mathbf{i} + \mathbf{j})\) m when \(t = 1\),

1. find the position vector of \(P\) when \(t = 4\) [5]

The curve \(C\) has equation \(y = x^3 - 2x^2 - x + 3\) The point \(P\), which lies on \(C\), has coordinates \((2, 1)\).

1. Show that an equation of the tangent to \(C\) at the point \(P\) is \(y = 3x - 5\) [5]

The point \(Q\) also lies on \(C\). Given that the tangent to \(C\) at \(Q\) is parallel to the tangent to \(C\) at \(P\),

1. find the coordinates of the point \(Q\). [5]

A curve has the equation \(y = x + \frac{3}{x}\), \(x \neq 0\). The point \(P\) on the curve has \(x\)-coordinate \(1\).

1. Show that the gradient of the curve at \(P\) is \(-2\). [3]
2. Find an equation for the normal to the curve at \(P\), giving your answer in the form \(y = mx + c\). [4]
3. Find the coordinates of the point where the normal to the curve at \(P\) intersects the curve again. [4]

A and B are points on the curve \(y = 4\sqrt{x}\). Point A has coordinates \((9, 12)\) and point B has \(x\)-coordinate \(9.5\). Find the gradient of the chord AB. The gradient of AB is an approximation to the gradient of the curve at A. State the \(x\)-coordinate of a point C on the curve such that the gradient of AC is a closer approximation. [3]

In Fig. 5, A and B are the points on the curve \(y = 2^x\) with \(x\)-coordinates 3 and 3.1 respectively. \includegraphics{figure_5}

1. Find the gradient of the chord AB. Give your answer correct to 2 decimal places. [2]
2. Stating the points you use, find the gradient of another chord which will give a closer approximation to the gradient of the tangent to \(y = 2^x\) at A. [2]

\includegraphics{figure_5} Fig. 5 shows the graph of \(y = 2^x\).

1. On the copy of Fig. 5, draw by eye a tangent to the curve at the point where \(x = 2\). Hence find an estimate of the gradient of \(y = 2^x\) when \(x = 2\). [3]
2. Calculate the \(y\)-values on the curve when \(x = 1.8\) and \(x = 2.2\). Hence calculate another approximation to the gradient of \(y = 2^x\) when \(x = 2\). [2]

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

A and B are points on the curve \(y = 4\sqrt{x}\). Point A has coordinates \((9, 12)\) and point B has \(x\)-coordinate \(9.5\). Find the gradient of the chord AB. The gradient of AB is an approximation to the gradient of the curve at A. State the \(x\)-coordinate of a point C on the curve such that the gradient of AC is a closer approximation. [3]

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]
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. [1]
3. Use calculus to find the gradient of the curve at A. [2]

Fig. 9 shows the line \(y = x\) and the curve \(y = f(x)\), where \(f(x) = \frac{1}{2}(e^x - 1)\). The line and the curve intersect at the origin and at the point P\((a, a)\). \includegraphics{figure_9}

1. Show that \(e^a = 1 + 2a\). [1]
2. Show that the area of the region enclosed by the curve, the \(x\)-axis and the line \(x = a\) is \(\frac{1}{2}a\). Hence find, in terms of \(a\), the area enclosed by the curve and the line \(y = x\). [6]
3. Show that the inverse function of f\((x)\) is g\((x)\), where g\((x) = \ln(1 + 2x)\). Add a sketch of \(y = g(x)\) to the copy of Fig. 9. [5]
4. Find the derivatives of f\((x)\) and g\((x)\). Hence verify that \(g'(a) = \frac{1}{f'(a)}\). Give a geometrical interpretation of this result. [7]

A curve \(C\) has equation \(y = x(x + 3)\).

1. Find the gradient of the line passing through the point \((-5, 10)\) and the point on \(C\) with \(x\)-coordinate \(-5 + h\). Give your answer in its simplest form. [3 marks]
2. Show how the answer to part (a) can be used to find the gradient of the curve \(C\) at the point \((-5, 10)\). State the value of this gradient. [2 marks]

A curve \(C\) has equation \(y = \frac{1}{x(x + 2)}\).

1. Write down the equations of all the asymptotes of \(C\). [2 marks]
2. The curve \(C\) has exactly one stationary point. The \(x\)-coordinate of the stationary point is \(-1\).
   1. Find the \(y\)-coordinate of the stationary point. [1 mark]
   2. Sketch the curve \(C\). [2 marks]
3. Solve the inequality $$\frac{1}{x(x + 2)} \leqslant \frac{1}{8}$$ [5 marks]

A curve \(C\) has equation \(y = (2 - x)(1 + x) + 3\).

1. A line passes through the point \((2, 3)\) and the point on \(C\) with \(x\)-coordinate \(2 + h\). Find the gradient of the line, giving your answer in its simplest form. [3 marks]
2. Show how your answer to part (a) can be used to find the gradient of the curve \(C\) at the point \((2, 3)\). State the value of this gradient. [2 marks]

\(f(x) = 3x^2\) Obtain \(\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\) Circle your answer. [1 mark] \(\frac{3h^2}{h}\) \quad \(x^3\) \quad \(\frac{3(x + h)^2 - 3x^2}{h}\) \quad \(6x\)

The diagram shows the graph of \(y = f(x)\), where \(f(x)\) is a quadratic function of \(x\). A copy of the diagram is given in the Printed Answer Booklet. \includegraphics{figure_2}

1. On the copy of the diagram in the Printed Answer Booklet, draw a possible graph of the gradient function \(y = f'(x)\). [3]
2. State the gradient of the graph of \(y = f''(x)\). [1]

A particle P has position vector \(\mathbf{r}\) m at time \(t\) s given by \(\mathbf{r} = (t^3 - 3t^2)\mathbf{i} - (4t^2 + 1)\mathbf{j}\) for \(t \geq 0\). Find the magnitude of the acceleration of P when \(t = 2\). [4]

The curve \(C\) has equation \(y = 2x^2 + 5x - 12\) and the line \(L\) has equation \(y = mx - 14\), where \(m\) is a real constant.

1. Given that \(L\) is a tangent to \(C\),
   1. show that \(m\) satisfies the equation $$m^2 - 10m + 9 = 0,$$ [5]
   2. find the coordinates of the two possible points of contact of \(C\) and \(L\). [6]
2. Given instead that \(L\) intersects \(C\) at two distinct points, find the range of values of \(m\). [2]

In this question you should show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. \includegraphics{figure_2} Figure 2 Figure 2 shows a sketch of part of the curve \(C\) with equation $$y = x^3 - 10x^2 + 27x - 23$$ The point \(P(5, -13)\) lies on \(C\) The line \(l\) is the tangent to \(C\) at \(P\)

1. Use differentiation to find the equation of \(l\), giving your answer in the form \(y = mx + c\) where \(m\) and \(c\) are integers to be found. [4]
2. Hence verify that \(l\) meets \(C\) again on the \(y\)-axis. [1]

The finite region \(R\), shown shaded in Figure 2, is bounded by the curve \(C\) and the line \(l\).

1. Use algebraic integration to find the exact area of \(R\). [4]