1.07m Tangents and normals: gradient and equations

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

1. Use calculus to find the coordinates of the turning points on this curve. Determine the nature of these turning points. [5]
2. Show that the tangent to the curve at the point where \(x = -1\) has gradient 9. Find the coordinates of the other point, P, on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P. Show that the area of the triangle bounded by the normal at P and the \(x\)- and \(y\)-axes is 8 square units. [8]

\includegraphics{figure_11} Fig. 11 shows a sketch of the curve with equation \(y = x - \frac{4}{x^2}\).

1. Find \(\frac{dy}{dx}\) and show that \(\frac{d^2y}{dx^2} = -\frac{24}{x^4}\). [3]
2. Hence find the coordinates of the stationary point on the curve. Verify that the stationary point is a maximum. [5]
3. Find the equation of the normal to the curve when \(x = -1\). Give your answer in the form \(ax + by + c = 0\). [5]

\includegraphics{figure_2} Figure 2 shows the curve with equation \(y = 5 + x - x^2\) and the normal to the curve at the point \(P(1, 5)\).

1. Find an equation for the normal to the curve at \(P\) in the form \(y = mx + c\). [5]
2. Find the coordinates of the point \(Q\), where the normal to the curve at \(P\) intersects the curve again. [2]
3. Show that the area of the shaded region bounded by the curve and the straight line \(PQ\) is \(\frac{4}{3}\). [6]

\includegraphics{figure_2} Figure 2 shows the curve \(C\) with equation \(y = 3x - 4\sqrt{x} + 2\) and the tangent to \(C\) at the point \(A\). Given that \(A\) has \(x\)-coordinate 4,

1. show that the tangent to \(C\) at \(A\) has the equation \(y = 2x - 2\). [6]

The shaded region is bounded by \(C\), the tangent to \(C\) at \(A\) and the positive coordinate axes.

1. Find the area of the shaded region. [8]

The gradient of a curve is given by \(\frac{dy}{dx} = 4x + 3\). The curve passes through the point \((2, 9)\).

1. Find the equation of the tangent to the curve at the point \((2, 9)\). [3]
2. Find the equation of the curve and the coordinates of its points of intersection with the \(x\)-axis. Find also the coordinates of the minimum point of this curve. [7]
3. Find the equation of the curve after it has been stretched parallel to the \(x\)-axis with scale factor \(\frac{1}{2}\). Write down the coordinates of the minimum point of the transformed curve. [3]

Find the equation of the normal to the curve \(y = 8x^4 + 4\) at the point where \(x = \frac{1}{2}\). [5]

\includegraphics{figure_1} Fig. 9 shows a sketch of the graph of \(y = x^3 - 10x^2 + 12x + 72\).

1. Write down \(\frac{dy}{dx}\). [2]
2. Find the equation of the tangent to the curve at the point on the curve where \(x = 2\). [4]
3. Show that the curve crosses the \(x\)-axis at \(x = -2\). Show also that the curve touches the \(x\)-axis at \(x = 6\). [3]
4. Find the area of the finite region bounded by the curve and the \(x\)-axis, shown shaded in Fig. 9. [4]

Fig. 10 shows a sketch of the curve \(y = x^2 - 4x + 3\). The point A on the curve has \(x\)-coordinate 4. At point B the curve crosses the \(x\)-axis. \includegraphics{figure_2}

1. Use calculus to find the equation of the normal to the curve at A and show that this normal intersects the \(x\)-axis at C (16, 0). [6]
2. Find the area of the region ABC bounded by the curve, the normal at A and the \(x\)-axis. [5]

The point A has \(x\)-coordinate 5 and lies on the curve \(y = x^2 - 4x + 3\).

1. Sketch the curve. [2]
2. Use calculus to find the equation of the tangent to the curve at A. [4]
3. Show that the equation of the normal to the curve at A is \(x + 6y = 53\). Find also, using an algebraic method, the \(x\)-coordinate of the point at which this normal crosses the curve again. [6]

\includegraphics{figure_3} A is the point with coordinates (1, 4) on the curve \(y = 4x^2\). B is the point with coordinates (0, 1), as shown in Fig. 10.

1. The line through A and B intersects the curve again at the point C. Show that the coordinates of C are \(\left(-\frac{1}{4}, \frac{1}{4}\right)\). [4]
2. Use calculus to find the equation of the tangent to the curve at A and verify that the equation of the tangent at C is \(y = -2x - \frac{1}{4}\). [6]
3. The two tangents intersect at the point D. Find the \(y\)-coordinate of D. [2]

Find the equation of the tangent to the curve \(y = 6\sqrt{x}\) at the point where \(x = 16\). [5]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = \text{f}(x)\), where $$\text{f}(x) = 10 + \ln(3x) - \frac{1}{2}e^x, \quad 0.1 \leq x \leq 3.3.$$ Given that f(k) = 0,

1. show, by calculation, that \(3.1 < k < 3.2\). [2]
2. Find f'(x). [3]

The tangent to the graph at \(x = 1\) intersects the \(y\)-axis at the point \(P\).

1. 1. Find an equation of this tangent.
   2. Find the exact \(y\)-coordinate of \(P\), giving your answer in the form \(a + \ln b\). [5]

\includegraphics{figure_2} Figure 2 shows part of the curve \(C\) with equation \(y = \text{f}(x)\), where $$\text{f}(x) = 0.5e^x - x^2.$$ The curve \(C\) cuts the \(y\)-axis at \(A\) and there is a minimum at the point \(B\).

1. Find an equation of the tangent to \(C\) at \(A\). [4]

The \(x\)-coordinate of \(B\) is approximately 2.15. A more exact estimate is to be made of this coordinate using iterations \(x_{n+1} = \ln g(x_n)\).

1. Show that a possible form for \(g(x)\) is \(g(x) = 4x\). [3]
2. Using \(x_{n+1} = \ln 4x_n\), with \(x_0 = 2.15\), calculate \(x_1\), \(x_2\) and \(x_3\). Give the value of \(x_3\) to 4 decimal places. [2]

The curve \(C\) has equation \(y = \text{f}(x)\), where $$\text{f}(x) = 3 \ln x + \frac{1}{x}, \quad x > 0.$$ The point \(P\) is a stationary point on \(C\).

1. Calculate the \(x\)-coordinate of \(P\). [4]
2. Show that the \(y\)-coordinate of \(P\) may be expressed in the form \(k - k \ln k\), where \(k\) is a constant to be found. [2]

The point \(Q\) on \(C\) has \(x\)-coordinate 1.

1. Find an equation for the normal to \(C\) at \(Q\). [4]

The normal to \(C\) at \(Q\) meets \(C\) again at the point \(R\).

1. Show that the \(x\)-coordinate of \(R\)
   1. satisfies the equation \(6 \ln x + x + \frac{2}{x} - 3 = 0\),
   2. lies between 0.13 and 0.14. [4]

The curve \(C\) has equation \(y = 2e^x + 3x^2 + 2\). The point \(A\) with coordinates \((0, 4)\) lies on \(C\). Find the equation of the tangent to \(C\) at \(A\). [5]

f(x) = \(x + \frac{e^x}{5}\), \(x \in \mathbb{R}\).

1. Find f'(x). [2]

The curve \(C\), with equation \(y = \)f(x), crosses the \(y\)-axis at the point \(A\).

1. Find an equation for the tangent to \(C\) at \(A\). [3]
2. Complete the table, giving the values of \(\sqrt{x + \frac{e^x}{5}}\) to 2 decimal places.

\(x\)	0	0.5	1	1.5	2
\(\sqrt{x + \frac{e^x}{5}}\)	0.45	0.91

[2]

Given that \(y = \log_a x\), \(x > 0\), where \(a\) is a positive constant,

1. 1. express \(x\) in terms of \(a\) and \(y\), [1]
   2. deduce that \(\ln x = y \ln a\). [1]
2. Show that \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{x \ln a}\). [2]

The curve \(C\) has equation \(y = \log_{10} x\), \(x > 0\). The point \(A\) on \(C\) has \(x\)-coordinate 10. Using the result in part (b),

1. find an equation for the tangent to \(C\) at \(A\). [4]

The tangent to \(C\) at \(A\) crosses the \(x\)-axis at the point \(B\).

1. Find the exact \(x\)-coordinate of \(B\). [2]

The curve with equation \(y = \ln 3x\) crosses the \(x\)-axis at the point \(P(p, 0)\).

1. Sketch the graph of \(y = \ln 3x\), showing the exact value of \(p\). [2]

The normal to the curve at the point \(Q\), with \(x\)-coordinate \(q\), passes through the origin.

1. Show that \(x = q\) is a solution of the equation \(x^2 + \ln 3x = 0\). [4]
2. Show that the equation in part (b) can be rearranged in the form \(x = \frac{1}{3}e^{-x^2}\). [2]
3. Use the iteration formula \(x_{n + 1} = \frac{1}{3}e^{-x_n^2}\), with \(x_0 = \frac{1}{3}\), to find \(x_1, x_2, x_3\) and \(x_4\). Hence write down, to 3 decimal places, an approximation for \(q\). [3]

\includegraphics{figure_8} The diagram shows part of the curve \(y = \ln(5 - x^2)\) which meets the \(x\)-axis at the point \(P\) with coordinates \((2, 0)\). The tangent to the curve at \(P\) meets the \(y\)-axis at the point \(Q\). The region \(A\) is bounded by the curve and the lines \(x = 0\) and \(y = 0\). The region \(B\) is bounded by the curve and the lines \(PQ\) and \(x = 0\).

1. Find the equation of the tangent to the curve at \(P\). [5]
2. Use Simpson's Rule with four strips to find an approximation to the area of the region \(A\), giving your answer correct to 3 significant figures. [4]
3. Deduce an approximation to the area of the region \(B\). [2]

Find the equation of the tangent to the curve \(y = \sqrt{4x + 1}\) at the point \((2, 3)\). [5]