Practise › Questions › Tangents, turning points and curve behaviour
Tangents, turning points and curve behaviour questions
The derivative now gets used. It gives tangent and normal equations, locates the stationary points where curves turn, distinguishes maxima from minima by a second derivative, and finds the largest volume obtainable from a sheet of card, which is the kind of question these methods were developed for.
8 original questions · 35 marks · the tangents, turning points and curve behaviour notes · Differentiation
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Find an equation of the tangent to the curve y = x2 at the point (2, 4).
Worked 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. The derivative supplies the slope and ordinary straight-line work does the rest.Find an equation of the normal to the curve y = x2 at the point (2, 4).
Worked 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. One derivative feeds both lines, and forgetting to invert as well as negate is the standard slip.Find the coordinates of the stationary points of the curve y = x3 − 3x2 − 9x + 5, and determine the nature of each.
Worked 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, and substituting into dy/dx to find y is a favourite way of losing two marks at the last moment.The curve C has equation y = x3 − 3x2 − 9x + 5. Find the interval on which C is decreasing.
Worked 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. Between the hilltop at x = −1 and the valley at x = 3 the only way is down.Show that the curve y = x3 has a stationary point at the origin which is neither a maximum nor a minimum.
Worked 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. When f″ = 0 the test is silent rather than negative, and the gradient's behaviour on both sides is what settles the classification.A rectangle has perimeter 40 cm. Use calculus to find the dimensions that give the greatest area, and state that area.
Worked 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. Write the quantity as a function of one variable first, then let the calculus do the choosing. Skipping the justification loses a mark even when the answer is right.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.
Worked 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. Writing 2000/r as 2000r−1 before differentiating avoids the most common error, which is differentiating the fraction to −2000/r and losing the power altogether. The minimum area is about 554 cm2.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.
Worked 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. Local extrema are local, and comparing values across different branches of a curve is meaningless.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise tangents, turning points and curve behaviour one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.