Questions on tangents, turning points and curve behaviour

Question 1

Find an equation of the tangent to the curve y = x² at the point (2, 4).

Accepted Answer

dy/dx = 2x, which is 4 at x = 2, so the tangent has gradient 4. Using y − y₁ = m(x − x₁) gives y − 4 = 4(x − 2), that is y = 4x − 4. M1 for differentiating, A1 for the gradient, A1 for the equation.

Question 2

Find an equation of the normal to the curve y = x² at the point (2, 4).

Accepted Answer

The tangent gradient at x = 2 is 2x = 4, and the normal is perpendicular to it, so its gradient is the negative reciprocal, −¼. Then y − 4 = −¼(x − 2), which tidies to y = −x/4 + 9/2, or x + 4y = 18. M1 for the tangent gradient, M1 for the negative reciprocal, A1 for the equation. Check the perpendicularity by multiplying: 4 × (−¼) = −1.

Question 3

A curve has second derivative d²y/dx² = 6x − 12 for all real x. State the interval on which the curve is convex, the interval on which it is concave, and the value of x at which the curve has a point of inflection, giving a reason for that value.

Accepted Answer

A curve is convex where its second derivative is positive, so 6x − 12 > 0 gives convex for x > 2. B1. It is concave where the second derivative is negative, so 6x − 12 < 0 gives concave for x < 2. B1. A point of inflection is a point at which the curve changes between convex and concave, and 6x − 12 is negative on one side of 2 and positive on the other, so the change does happen there: x = 2. B1 for the value with the change of sign as the reason. The value of the second derivative at a single point is not by itself the reason. What settles a point of inflection is the change of sign, and here it is the linear form of 6x − 12 that guarantees the sign changes rather than merely touching zero.

Question 4

The curve C has equation y = x² − 6x + 1. Find the coordinates of the stationary point of C, and find an equation of the normal to C at that point.

Accepted Answer

dy/dx = 2x − 6, which is zero at x = 3. M1 for setting the derivative to zero. Then y = 9 − 18 + 1 = −8, so the stationary point is (3, −8). A1. The tangent there is horizontal, with gradient 0, and the normal is perpendicular to it. The negative reciprocal of 0 does not exist, so the normal is not written in the form y = mx + c: it is the vertical line x = 3. B1 for the vertical normal with its reason.

Question 5

Find the coordinates of the stationary points of the curve y = x³ − 3x² − 9x + 5, and determine the nature of each.

Accepted Answer

dy/dx = 3x² − 6x − 9 = 3(x − 3)(x + 1), which is zero at x = −1 and x = 3. Substituting into the original equation gives y(−1) = −1 − 3 + 9 + 5 = 10 and y(3) = 27 − 27 − 27 + 5 = −22, so the stationary points are (−1, 10) and (3, −22). The second derivative is 6x − 6, which is −12 at x = −1, a maximum, and +12 at x = 3, a minimum. M1 A1 for the derivative, M1 for solving, A1 for both x values, A1 for both y values, A1 for the classification. The heights come from the original curve, never from the derivative, so substitute the x values into the curve, not into dy/dx.

Question 6

The curve C has equation y = x³ − 3x² − 9x + 5. Find the interval on which C is decreasing.

Accepted Answer

A curve falls where its gradient is negative, so solve dy/dx < 0. Here dy/dx = 3x² − 6x − 9 = 3(x − 3)(x + 1), an upward parabola in x, which is negative strictly between its roots. So C is decreasing for −1 < x < 3. M1 for setting the derivative below zero, M1 for factorising or solving, A1 for the interval.

Question 7

Show that the curve y = x³ has a stationary point at the origin which is neither a maximum nor a minimum.

Accepted Answer

dy/dx = 3x², which is zero at x = 0, so the origin is a stationary point. The second derivative 6x is also zero there, so that test decides nothing. Instead look at the sign of dy/dx on either side: 3x² is positive for x = −1 and positive for x = 1, so the curve climbs into the point and climbs away from it. It is therefore a point of inflection with a horizontal tangent. M1 for the stationary point, M1 for testing the gradient either side, A1 for the conclusion.

Question 8

A rectangle has perimeter 40 cm. Use calculus to find the dimensions that give the greatest area, and state that area.

Accepted Answer

If one side is x then the other is 20 − x, so A = x(20 − x) = 20x − x². Then dA/dx = 20 − 2x, which is zero at x = 10, and d²A/dx² = −2, negative, confirming a maximum. The rectangle is a 10 cm by 10 cm square with area 100 cm². M1 for the second side in terms of x, A1 for the area function, M1 for differentiating and solving, A1 for x = 10, A1 for the area with its justification.

Question 9

The curve C has equation y = 3x⁵ − 5x⁴. Find the values of x for which d²y/dx² = 0, and determine for each of them whether C has a point of inflection there, giving your reasons.

Accepted Answer

dy/dx = 15x⁴ − 20x³ and d²y/dx² = 60x³ − 60x². M1 for differentiating twice, A1 for the second derivative, which factorises as 60x²(x − 1). It is zero at x = 0 and x = 1. A1 for both values. Now test the sign on each side of each. The factor 60x² is positive except at x = 0, so the sign is the sign of (x − 1): negative at x = −1, negative at x = ½, and positive at x = 1½. M1 for testing the sign either side of both values. Across x = 0 the second derivative is negative on both sides, so the curve is concave on both sides and there is no point of inflection at x = 0; across x = 1 it changes from negative to positive, so the curve changes from concave to convex and there is a point of inflection at (1, −2). A1 for both conclusions with the reasons.

Question 10

The curve C has equation y = x³ − 3x. The tangent to C at the point P(2, 2) meets C again at the point Q. Find the coordinates of Q.

Accepted Answer

First check P: 8 − 6 = 2, so P does lie on C. dy/dx = 3x² − 3, which is 12 − 3 = 9 at x = 2. M1 for the derivative and the gradient at P. The tangent is y − 2 = 9(x − 2), that is y = 9x − 16. A1. Where the tangent meets C, x³ − 3x = 9x − 16, so x³ − 12x + 16 = 0. M1 for equating and collecting. The tangent touches C at x = 2, so (x − 2) is a repeated factor, and x³ − 12x + 16 = (x − 2)²(x + 4). dM1 for the factorisation using the repeated root. The remaining root is x = −4, where y = −64 + 12 = −52, so Q is (−4, −52). A1. Substituting x = −4 into the tangent gives −36 − 16 = −52 as well, which checks the answer.

Question 11

The curve C has equation y = x³ − 6x² + 5x + 12. Find the coordinates of the two points on C at which the tangent is parallel to the line y = 5x − 1.

Accepted Answer

Parallel lines have equal gradients, so the tangent gradient must be 5: solve dy/dx = 5. M1 for setting the derivative equal to the line's gradient. Here dy/dx = 3x² − 12x + 5, so 3x² − 12x + 5 = 5 gives 3x² − 12x = 0. A1. Factorising, 3x(x − 4) = 0, so x = 0 or x = 4. A1 for both values. The heights come from C itself: y(0) = 12 and y(4) = 64 − 96 + 20 + 12 = 0, so the points are (0, 12) and (4, 0). A1. Setting the derivative to zero instead would find the stationary points, which is a different question.

Question 12

A closed cylinder has volume 1000 cm³. Show that its total surface area S cm² satisfies S = 2πr² + 2000/r, where r cm is the radius. Hence find the radius that minimises S, giving your answer to 3 significant figures, and justify that your value gives a minimum.

Accepted Answer

With height h, the volume gives πr²h = 1000, so h = 1000/(πr²). The closed cylinder has two circular ends and a curved surface, so S = 2πr² + 2πrh = 2πr² + 2πr × 1000/(πr²) = 2πr² + 2000/r, as required. Differentiating, dS/dr = 4πr − 2000/r², and setting this to zero gives 4πr³ = 2000, so r³ = 500/π and r = 5.42 cm. For the justification, d²S/dr² = 4π + 4000/r³, which is positive for every r > 0, so the stationary point is a minimum. M1 for h from the volume, M1 for substituting into S, A1 for the printed result, M1 for differentiating, A1 for dS/dr, A1 for r = 5.42, B1 for the second derivative justification. Write 2000/r as 2000r⁻¹ before differentiating, so that the power rule gives −2000r⁻². The minimum area is about 554 cm².

Question 13

The curve C has equation y = x + 4/x for x > 0. Find the minimum value of y, and justify that your value is a minimum.

Accepted Answer

Write y = x + 4x⁻¹, so dy/dx = 1 − 4x⁻² = 1 − 4/x². Setting this to zero gives x² = 4, so x = 2, taking the positive root because x > 0. Then y = 2 + 2 = 4. For the justification, d²y/dx² = 8x⁻³ = 8/x³, which is 1 at x = 2 and therefore positive, confirming a minimum. M1 for rewriting with a negative index, A1 for the derivative, M1 for solving, A1 for x = 2, A1 for y = 4 with justification. The domain restriction is doing real work here, since x = −2 is also stationary and gives a local maximum of −4, which sits below the minimum you have just found.

Question 14

A curve has equation y = ax³ + bx² + cx + d, where a, b, c and d are constants and a ≠ 0. Prove that the curve has exactly one point of inflection, and find its x-coordinate in terms of a and b. Show also that, when the curve has two stationary points, the point of inflection lies midway between them.

Accepted Answer

dy/dx = 3ax² + 2bx + c and d²y/dx² = 6ax + 2b. M1 for differentiating twice, A1 for 6ax + 2b. That second derivative is a linear function of x whose gradient 6a is not zero, since a ≠ 0, so it takes the value zero at exactly one value of x and changes sign as x passes through it. M1 for the argument from a non-zero gradient to one crossing with a change of sign. Solving 6ax + 2b = 0 gives x = −b/(3a), and since the sign changes there the curve does change between convex and concave, so there is exactly one point of inflection. A1. For the stationary points, 3ax² + 2bx + c = 0. When this quadratic has two roots, their sum is −2b/(3a) by the sum of roots. M1 for the sum of the roots. Their mean is therefore half that, −b/(3a), which is the x-coordinate already found, so the point of inflection lies midway between the two stationary points. A1. Nothing here needs the roots themselves, only their sum, and c and d never enter the answer at all: c moves the stationary points apart or together without moving their midpoint, and d raises the whole curve.

Question 15

The curve C has equation y = √x for x ≥ 0. The normal to C at the point P(4, 2) meets the x-axis at R and the y-axis at S, and O is the origin. Find an equation of the normal, and find the exact area of triangle ORS. Explain why the same method cannot be used to find the normal to C at the origin, and write down an equation of that normal.

Accepted Answer

P lies on C since √4 = 2. Writing y = x^1/2 gives dy/dx = ½x^−1/2 = 1/(2√x), which at x = 4 is 1/4. M1 for the derivative and the gradient at P. The normal gradient is the negative reciprocal, −4, so y − 2 = −4(x − 4), that is 4x + y = 18. A1. Putting y = 0 gives R(4.5, 0) and putting x = 0 gives S(0, 18). M1 for both intercepts. The triangle has its right angle at O, with legs of length 4.5 and 18, so its area is ½ × 4.5 × 18 = 81/2. A1. At the origin the method stops at the first step: 1/(2√x) is undefined at x = 0, and as x approaches 0 the gradient increases without bound, so C has no finite gradient there and there is no number to take the negative reciprocal of. M1 for identifying that the derivative does not exist at x = 0. The tangent at the origin is the vertical line x = 0, so the normal is perpendicular to it and horizontal: y = 0. A1.

Question 16

The graph shows the gradient function f′(x) of a curve y = f(x), for −2 ≤ x ≤ 4. The curve y = f(x) itself is not drawn. Use the graph to write down the x-coordinates of the stationary points of y = f(x) in this interval, and determine the nature of each, giving your reason from the graph. State the interval on which f is decreasing. Write down the value of x at which y = f(x) has a point of inflection, and explain how the graph of f′ shows it.

Accepted Answer

A stationary point of y = f(x) is a place where f′(x) = 0, that is where the drawn graph meets the x-axis. It does so at x = −1 and x = 3. B1 for x = −1, B1 for x = 3. Just to the left of x = −1 the graph is above the axis and just to the right it is below, so f′ changes from positive to negative there and f has a maximum at x = −1. B1 for the maximum with the change of sign as the reason. Just to the left of x = 3 the graph is below the axis and just to the right it is above, so f′ changes from negative to positive and f has a minimum at x = 3. B1 for the minimum with the change of sign as the reason. A curve is decreasing where its gradient is negative, and the graph lies below the axis between the two crossings, so f is decreasing for −1 < x < 3. B1 for the interval. The lowest point of the drawn graph is at x = 1. B1 for the value. Now f″ is the gradient of the graph of f′, and at that lowest point the graph of f′ is momentarily flat, so f″(1) = 0; the graph of f′ is falling before x = 1 and rising after it, so f″ changes from negative to positive there. That change of sign, not the zero on its own, is what makes x = 1 a point of inflection. B1 for the reason.