Introduction to Calculus · applications

Applications: Introduction to Calculus

10 question types · Pre-University · each worked step by step with a figure that follows the steps

01

A Car Pulling Away from Traffic Lights: Its Speed at One Instant as the Limit of Average Speeds

methodFind the Average Speed over Shorter and Shorter Intervals Starting at the Instant, Write It for an Interval of Length h, Then Let h Tend to Zero

A car pulls away from a set of traffic lights. After x seconds it has traveled s = 2x2 meters. (a) Find the average speed of the car from x = 3 to x = 4, from x = 3 to x = 3.5 and from x = 3 to x = 3.1. (b) Find the average speed from x = 3 to x = 3 + h, and use it to find the speed of the car at the instant x = 3.

01020304050012345seconds after the lights, xmeters traveled, s(3, 18)x = 3: s = 2 × 32= 18x = 4: s = 32, so (32 − 18)/1 = 14 m/s
At x = 3 the car has traveled 18 m, and at x = 4 it has traveled 32 m. The chord between the two points has gradient 32 − 181 = 14, the average speed in m/s.
At x = 3 the car has traveled s = 2 × 32 = 18 m. At x = 4 it has traveled 2 × 42 = 32 m, so the average speed from x = 3 to x = 4 is 32 − 184 − 3 = 14 m/s.
step 1 of 6

The graph of s = 2x2 is a curve, so the speed of the car keeps changing. The average speed over an interval is the gradient of a chord. As the interval shrinks toward the instant x = 3, the chords close in on the tangent at that point, and their gradients close in on the speed at that instant.

  1. At x = 3 the car has traveled s = 2 × 32 = 18 m. At x = 4 it has traveled 2 × 42 = 32 m, so the average speed from x = 3 to x = 4 is 32 − 184 − 3 = 14 m/s.
  2. At x = 3.5, s = 2 × 3.52 = 24.5, so the average speed is 24.5 − 180.5 = 13 m/s. At x = 3.1, s = 2 × 3.12 = 19.22, so the average speed is 19.22 − 180.1 = 12.2 m/s.
  3. (a) The average speeds are 14 m/s, 13 m/s and 12.2 m/s. The shorter the interval, the closer the chord lies to the tangent at (3, 18).
  4. Over an interval of length h the car travels 2(3 + h)2 − 18 = 18 + 12h + 2h2 − 18 = 12h + 2h2 meters. Divide by the time h: the average speed is 12h + 2h2h = 12 + 2h m/s.
  5. Check the formula against part (a): h = 1 gives 14, h = 0.5 gives 13 and h = 0.1 gives 12.2. As h tends to zero, 2h tends to zero, so limh → 0 (12 + 2h) = 12.
  6. (b) The average speed from x = 3 to x = 3 + h is 12 + 2h m/s, and the speed at the instant x = 3 is 12 m/s. It is the gradient of the tangent to the curve at (3, 18).

answer(a) 14 m/s, 13 m/s and 12.2 m/s; (b) 12 + 2h m/s, so the speed at x = 3 is 12 m/s

techniqueLimits · The Gradient of a Curve · From First Principles

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing the distance at x = 3 by the time, 183 = 6 m/s. That is the average speed over the first 3 seconds from a standing start, not the speed at the instant x = 3.
  • Putting h = 0 into 2(3 + h)2 − 18h before simplifying, which gives 00. Expand the top and cancel the h first; only then let h tend to zero.
02

A Metal Cube Expanding as It Is Heated: How Fast Its Volume Grows, from First Principles

methodExpand f(x + h) − f(x), Divide by h, Cancel, and Let h Tend to Zero; Then Multiply the Derivative by a Small Change to Estimate the Change It Causes

A metal cube is heated, and its edge grows. When the edge is x cm long, the volume is V = x3 cm3. (a) Use first principles to show that dVdx = 3x2, and find the rate at which the volume grows with the edge when the edge is 10 cm. (b) Use your answer to estimate the increase in volume when the edge grows from 10 cm to 10.1 cm, and compare the estimate with the exact increase.

0600120018006789101112edge of the cube (cm), xvolume in cubic cm, V(10, 1000)(x + h)3− x3= 3x2h + 3xh2+ h3
When the edge grows from x to x + h, the volume grows by (x + h)3 − x3 = 3x2h + 3xh2 + h3.
When the edge grows from x to x + h, the volume grows by (x + h)3 − x3 = x3 + 3x2h + 3xh2 + h3 − x3 = 3x2h + 3xh2 + h3.
step 1 of 5

The derivative is the limit of the gradient of a chord. Take the chord from the point where the edge is x to the point where it is x + h, simplify its gradient, and let h shrink to zero. Close to a point the curve and its tangent are close together, so the derivative times a small change in x estimates the change in V.

  1. When the edge grows from x to x + h, the volume grows by (x + h)3 − x3 = x3 + 3x2h + 3xh2 + h3 − x3 = 3x2h + 3xh2 + h3.
  2. The gradient of the chord is this increase divided by h: 3x2h + 3xh2 + h3h = 3x2 + 3xh + h2. At x = 10 it is 364 for h = 2, 331 for h = 1 and 303.01 for h = 0.1.
  3. As h tends to zero, 3xh and h2 both tend to zero, so dVdx = limh → 0 (3x2 + 3xh + h2) = 3x2.
  4. (a) At x = 10, dVdx = 3 × 102 = 300. The volume is growing at 300 cm3 for each centimeter of edge.
  5. (b) For a small change δ x = 0.1, the increase is about dVdx × δ x = 300 × 0.1 = 30 cm3. The exact increase is 10.13 − 103 = 1030.301 − 1000 = 30.301 cm3, so the estimate is short by only 0.301 cm3, about 1%.

answer(a) dVdx = 3x2, which is 300 cm3 per cm when x = 10; (b) about 30 cm3, against an exact increase of 30.301 cm3

techniqueFrom First Principles · Limits · The Gradient of a Curve

examsGCSE Higher

Common pitfalls

  • Writing (x + h)3 = x3 + h3. A bracket cubed has four terms, x3 + 3x2h + 3xh2 + h3, and the derivative comes from the middle two.
  • Putting h = 0 into (x + h)3 − x3h straight away, which gives 00. Divide each term of the top by h first, and only then let h tend to zero.
03

A Ball Thrown Straight Up: Its Velocity and Acceleration from Its Height

methodDifferentiate the Height Term by Term with the Power Rule for the Velocity, Differentiate Again for the Acceleration, and Set the Velocity to Zero for the Top of the Flight

A ball is thrown straight up. After x seconds its height above the ground is y = 2 + 20x − 5x2 meters. (a) Find the velocity of the ball after 1 second and after 3 seconds, and say what each one tells you. (b) Find the acceleration of the ball, and its height at the moment it stops rising.

051015202501234seconds after the throw, xheight (m), yy = 2 + 20x − 5x2v = dy/dx = 20 − 10x
Differentiate term by term: the velocity is v = dydx = 20 − 10x m/s.
Differentiate term by term. The constant 2 gives 0, 20x gives 20 and −5x2 gives −10x, so the velocity is v = dydx = 20 − 10x m/s.
step 1 of 6

The velocity is the rate of change of the height, dydx, and the acceleration is the rate of change of the velocity, d2ydx2. The power rule turns axn into naxn−1, and a constant into 0.

  1. Differentiate term by term. The constant 2 gives 0, 20x gives 20 and −5x2 gives −10x, so the velocity is v = dydx = 20 − 10x m/s.
  2. At x = 1, v = 20 − 10 = 10. At x = 3, v = 20 − 30 = −10.
  3. (a) After 1 second the ball is rising at 10 m/s. After 3 seconds the velocity is −10 m/s: the minus sign means that the height is decreasing, so the ball is falling at 10 m/s.
  4. Differentiate the velocity: d2ydx2 = −10. The acceleration is −10 m/s2 at every moment of the flight, directed downward. It is the pull of gravity.
  5. The ball stops rising when v = 0: 20 − 10x = 0, so x = 2. Its height then is y = 2 + 20 × 2 − 5 × 22 = 2 + 40 − 20 = 22 m.
  6. (b) The acceleration is −10 m/s2, and the ball stops rising at a height of 22 m. Check: the height is 17 m at x = 1 and again at x = 3, because the flight is symmetrical about x = 2.

answer(a) 10 m/s upward after 1 second, and −10 m/s after 3 seconds, so the ball is falling at 10 m/s; (b) −10 m/s2, and a height of 22 m

techniqueDifferentiating Powers · The Gradient of a Curve

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Treating the velocity −10 m/s as a slip in the working. A negative velocity is correct: it says that the height is decreasing, so the ball is on its way down.
  • Differentiating the constant 2 to 2, or dropping the power in −5x2 and writing −5. A constant does not change, so its derivative is 0, and x2 becomes 2x, so −5x2 becomes −10x.
04

A Skateboard Ramp: The Line a Rider Leaves It Along, and a Strut at Right Angles to Its Surface

methodDifferentiate for the Gradient at the Point, Write the Tangent Through the Point with That Gradient, and Use the Negative Reciprocal of the Gradient for the Normal

The side view of a skateboard ramp is the curve y = x24 for 0 ≤ x ≤ 2, where x and y are in meters and the ground is the line y = 0. A rider leaves the ramp at its top edge, (2, 1), moving along the tangent there. (a) Find the equation of the tangent at (2, 1) and the angle it makes with the ground. (b) A support strut is fixed to the ramp at (2, 1), at right angles to its surface, and runs in a straight line from there to the ground. Find the equation of the line of the strut, where it meets the ground, and its length.

012301234meters along the ground, xheight (m), y(2, 1)y = x2/4, so dy/dx = 2x/4 = x/2x = 2: gradient = 2/2 = 1
Differentiate: dydx = x2, so the gradient of the ramp at its top edge (2, 1) is 1.
Differentiate: dydx = 2x4 = x2. At x = 2 the gradient of the ramp is 22 = 1.
step 1 of 6

The tangent at a point has the gradient of the curve at that point, which is the value of dydx there. The normal is at right angles to the tangent, so the two gradients multiply to −1.

  1. Differentiate: dydx = 2x4 = x2. At x = 2 the gradient of the ramp is 22 = 1.
  2. The tangent passes through (2, 1) with gradient 1: y − 1 = 1(x − 2), so y = x − 1.
  3. (a) The tangent is y = x − 1. A gradient of 1 is a rise of 1 m for each 1 m across, so the angle with the ground is tan−1 1 = 45°.
  4. The strut lies along the normal. Its gradient is −11 = −1, so y − 1 = −1(x − 2), which gives y = 3 − x.
  5. At the ground y = 0, so x = 3. The strut runs from (2, 1) to (3, 0), and its length is √12 + 12 = √2 ≈ 1.41 m, to 2 decimal places.
  6. (b) The strut lies along y = 3 − x and meets the ground at (3, 0), 1 m beyond the foot of the top edge. It is 1.41 m long. Check: the gradients multiply to 1 × (−1) = −1, so the strut is at right angles to the tangent.

answer(a) y = x − 1, at 45° to the ground; (b) y = 3 − x, meeting the ground at (3, 0), and the strut is √2 ≈ 1.41 m long

techniqueTangents and Normals · Differentiating Powers

Common pitfalls

  • Using the gradient of the curve, 1, for the normal as well. That line is the tangent again. The gradient of the normal is the negative reciprocal, −1.
  • Finding the gradient from the coordinates of the point, 12, as if the ramp were a straight line from the origin. The ramp is curved, so its gradient at (2, 1) comes from dydx = x2, which is 1.
05

An Open Box Folded from a Square Sheet of Card: The Cut That Gives the Greatest Volume

methodWrite the Volume in Terms of the Cut, Set dV/dx = 0 for the Stationary Points, Reject Any Outside the Situation, and Use the Second Derivative to Show a Maximum

An open box is made from a square sheet of card 12 cm by 12 cm. A square of side x cm is cut from each corner, and the four sides are folded up. (a) Show that the volume of the box is V = 4x3 − 48x2 + 144x cm3, and state the values of x that make sense. (b) Find the value of x that gives the greatest volume, show that it is a maximum, and find that volume.

x12 − 2x12 cm0501001500123456side of the cut square (cm), xvolume in cubic cm, Vbase (12 − 2x) by (12 − 2x), height xV = x(12 − 2x)2= 4x3− 48x2+ 144x
The base is a square of side 12 − 2x and the height is x, so V = x(12 − 2x)2 = 4x3 − 48x2 + 144x.
Each side of the base loses x at both ends, so the base is (12 − 2x) cm square, and the height is x cm. The volume is V = x(12 − 2x)2 = x(144 − 48x + 4x2) = 4x3 − 48x2 + 144x.
step 1 of 5

The base of the box is a square of side 12 − 2x, and its height is x. The greatest volume is at a stationary point, where the graph of V against x is flat and dVdx = 0. A negative second derivative shows that the stationary point is a maximum.

  1. Each side of the base loses x at both ends, so the base is (12 − 2x) cm square, and the height is x cm. The volume is V = x(12 − 2x)2 = x(144 − 48x + 4x2) = 4x3 − 48x2 + 144x.
  2. (a) The volume is V = 4x3 − 48x2 + 144x cm3. The cut must be more than 0, and the base must be more than 0, so 12 − 2x > 0: the values that make sense are 0 < x < 6.
  3. Differentiate: dVdx = 12x2 − 96x + 144 = 12(x2 − 8x + 12) = 12(x − 2)(x − 6). It is zero at x = 2 and at x = 6.
  4. Reject x = 6: it cuts the whole edge away and leaves no base. Differentiate again: d2Vdx2 = 24x − 96, which is 48 − 96 = −48 at x = 2. It is negative, so x = 2 gives a maximum.
  5. (b) The greatest volume is at x = 2: V = 2 × (12 − 4)2 = 2 × 64 = 128 cm3. Check a cut on either side: x = 1 gives 1 × 102 = 100 and x = 3 gives 3 × 62 = 108, both less than 128.

answer(a) V = 4x3 − 48x2 + 144x, for 0 < x < 6; (b) x = 2 cm, a maximum because d2Vdx2 = −48 < 0, and the greatest volume is 128 cm3

techniqueStationary Points · Differentiating Powers

Common pitfalls

  • Keeping x = 6 because it solves dVdx = 0. At x = 6 the base has side 12 − 12 = 0 and the volume is 0: it is the least volume, not the greatest.
  • Taking the side of the base as 12 − x. A square is cut from both ends of every edge, so each side of the base loses 2x.
06

A Sheep Pen Against a Wall with 40 m of Fencing: The Width That Encloses the Greatest Area

methodUse the Fixed Length of Fence to Write the Third Side in Terms of x, Write the Area as a Function of x Alone, Then Set dA/dx = 0

A farmer has 40 m of fencing to make a rectangular sheep pen against a long straight wall. The wall forms one side of the pen, and the fence forms the other three. Each of the two sides at right angles to the wall is x m long. (a) Write the area A of the pen in terms of x. (b) Find the value of x that gives the greatest area, the length of the side along the wall, and the greatest area.

wallxx40 − 2x010020005101520width out from the wall (m), xarea in square m, Aalong the wall: 40 − 2xA = x(40 − 2x) = 40x − 2x2
(a) The side along the wall is 40 − 2x m, so A = x(40 − 2x) = 40x − 2x2 m2.
The side along the wall is 40 − 2x m, so A = x(40 − 2x) = 40x − 2x2. (a) The area is A = 40x − 2x2 m2, for 0 < x < 20.
step 1 of 4

The two sides at right angles to the wall use 2x m of the fencing, so the side along the wall is 40 − 2x m. Once the area is written in terms of x alone, it has one stationary point, and the second derivative shows that the area is greatest there.

  1. The side along the wall is 40 − 2x m, so A = x(40 − 2x) = 40x − 2x2. (a) The area is A = 40x − 2x2 m2, for 0 < x < 20.
  2. Differentiate: dAdx = 40 − 4x. It is zero when 4x = 40, so x = 10.
  3. Differentiate again: d2Adx2 = −4. It is negative, so x = 10 gives a maximum.
  4. (b) The pen is 10 m out from the wall and 40 − 20 = 20 m along it, and the greatest area is 10 × 20 = 200 m2. Check: x = 9 gives 9 × 22 = 198 and x = 11 gives 11 × 18 = 198, both less than 200.

answer(a) A = 40x − 2x2 m2; (b) x = 10 m, with 20 m along the wall, for the greatest area of 200 m2

techniqueStationary Points · Differentiating Powers

Common pitfalls

  • Fencing all four sides, 2x + 2y = 40, as for a pen in an open field. The wall is one side, so only three sides need fence: 2x + y = 40.
  • Expecting a square pen to give the greatest area. Three equal sides of 403 ≈ 13.3 m enclose only about 177.8 m2. With a wall along one side, the best pen is twice as long as it is wide.
07

A Workshop's Profit Curve: The Output That Makes the Most Profit, and the Output That Makes the Least

methodSolve dP/dx = 0 for the Stationary Points, Read the Sign of the Second Derivative at Each One, and Compare the Maximum with the Ends of the Range

A workshop makes x hundred chairs a month, where 0 ≤ x ≤ 7. Its monthly profit is P = −x3 + 9x2 − 15x − 10 thousand dollars. (a) Find the two stationary points of the profit curve, and use the second derivative to decide which is a maximum and which is a minimum. (b) How many chairs a month give the greatest profit, and what is that profit? Check that it is more than the profit at either end of the range.

−20−100102001234567hundreds of chairs a month, xprofit (thousand $), PdP/dx = −3x2+ 18x − 15= −3(x − 1)(x − 5): zero at x = 1 and x = 5
Differentiate: dPdx = −3x2 + 18x − 15 = −3(x − 1)(x − 5), which is zero at x = 1 and at x = 5.
Differentiate: dPdx = −3x2 + 18x − 15 = −3(x2 − 6x + 5) = −3(x − 1)(x − 5). It is zero at x = 1 and at x = 5.
step 1 of 6

At a stationary point the tangent is flat, so dPdx = 0. Where d2Pdx2 is negative the curve bends downward, which makes a maximum; where it is positive the curve bends upward, which makes a minimum. On a limited range the ends must be checked as well.

  1. Differentiate: dPdx = −3x2 + 18x − 15 = −3(x2 − 6x + 5) = −3(x − 1)(x − 5). It is zero at x = 1 and at x = 5.
  2. Find the profit at each: P(1) = −1 + 9 − 15 − 10 = −17 and P(5) = −125 + 225 − 75 − 10 = 15.
  3. Differentiate again: d2Pdx2 = −6x + 18. At x = 1 it is 12, which is positive, so (1, −17) is a minimum. At x = 5 it is −12, which is negative, so (5, 15) is a maximum.
  4. (a) The stationary points are a minimum at (1, −17) and a maximum at (5, 15).
  5. At the ends of the range, P(0) = −10 and P(7) = −343 + 441 − 105 − 10 = −17. Both are less than 15.
  6. (b) The greatest profit is at x = 5: 500 chairs a month give a profit of 15 thousand dollars, which is $15 000.

answer(a) a minimum at (1, −17) and a maximum at (5, 15); (b) 500 chairs a month, for a profit of $15 000

techniqueStationary Points · Differentiating Powers

Common pitfalls

  • Stopping at dPdx = 0 and taking both points as maxima. The first stationary point, at 100 chairs, is the lowest profit on the curve, a loss of $17 000; the second derivative tells the two points apart.
  • Assuming that the largest output makes the largest profit. At 700 chairs the profit is −17, a loss of $17 000, because the cost of the extra chairs grows faster than the money they bring in.
08

A Car Overtaking a Lorry: The Distance Traveled as the Area Under Its Speed-Time Curve

methodThe Distance Is the Area Under the Speed-Time Curve: Integrate the Speed Between the Two Times, and Compare It with the Rectangle for the Steady Speed

A car traveling at 12 m/s pulls out to overtake a lorry. For the next 6 seconds its speed is v = 12 + 6x − x2 m/s, where x is the time in seconds, and at x = 6 it is back at 12 m/s. (a) Find the greatest speed of the car while it overtakes. (b) Find the distance the car travels in the 6 seconds, and how much further that is than if it had stayed at 12 m/s.

061218240123456seconds after pulling out, xspeed (m/s), v21 m/sdv/dx = 6 − 2x = 0 at x = 3greatest speed 12 + 18 − 9 = 21 m/s
(a) The speed is greatest where dvdx = 6 − 2x = 0, at x = 3, and it is 21 m/s.
The speed is greatest where dvdx = 6 − 2x = 0, which is at x = 3. (a) The greatest speed is v = 12 + 18 − 9 = 21 m/s.
step 1 of 5

The distance traveled is the area under the speed-time graph, and that area is a definite integral of the speed. Integration reverses differentiation: raise each power of x by one and divide by the new power.

  1. The speed is greatest where dvdx = 6 − 2x = 0, which is at x = 3. (a) The greatest speed is v = 12 + 18 − 9 = 21 m/s.
  2. The distance is the area under the speed-time curve from x = 0 to x = 6, which is ∫06 (12 + 6x − x2) dx.
  3. Integrate term by term, raising each power by one and dividing by the new power: [12x + 3x2 − x33]06.
  4. Put in the limits: (72 + 108 − 72) − 0 = 108. The car travels 108 m in the 6 seconds.
  5. At a steady 12 m/s it would have traveled 12 × 6 = 72 m, the area of the rectangle under the line v = 12. (b) The car travels 108 m, which is 108 − 72 = 36 m further. Check: ∫06 (6x − x2) dx = 108 − 72 = 36.

answer(a) 21 m/s; (b) 108 m, which is 36 m further than at a steady 12 m/s

techniqueArea Under a Curve · Integration · Stationary Points

examsO-Level · GCSE Higher · H2

Common pitfalls

  • Multiplying the greatest speed by the time, 21 × 6 = 126 m. The car is at 21 m/s for only one instant; the area under the curve adds up the speed at every moment.
  • Differentiating the speed to find the distance. Differentiating the speed gives the acceleration; the distance comes from integrating, which reverses differentiation.
09

Water Running into a Tank at a Rising Rate: The Volume in the Tank from the Rate

methodIntegrate the Rate to Recover the Volume, Use the Starting Volume to Find the Constant, Then Solve for the Time

A tank holds 50 liters of water when a valve is opened. Water then flows in at a rate of dVdx = 4x + 6 liters per minute, where x is the time in minutes since the valve was opened. (a) Find the volume V in the tank after x minutes, and the volume after 5 minutes. (b) How long after the valve is opened does the tank hold 310 liters?

010203040500246810minutes since the valve opened, xrate (liters a minute)rate = 4x + 6V = 2x2+ 6x + ccheck: the derivative is 4x + 6
Integrate the rate: V = ∫ (4x + 6) dx = 2x2 + 6x + c.
Integrate the rate: V = ∫ (4x + 6) dx = 2x2 + 6x + c. Check by differentiating: the derivative of 2x2 + 6x + c is 4x + 6.
step 1 of 5

Integration reverses differentiation: a function whose derivative is 4x + 6 is 2x2 + 6x, plus a constant. The 50 liters in the tank at x = 0 fix the constant. The water added between two times is the area under the graph of the rate.

  1. Integrate the rate: V = ∫ (4x + 6) dx = 2x2 + 6x + c. Check by differentiating: the derivative of 2x2 + 6x + c is 4x + 6.
  2. At x = 0 the tank holds 50 liters, so 0 + 0 + c = 50 and c = 50. The volume is V = 2x2 + 6x + 50.
  3. (a) After 5 minutes, V = 2 × 25 + 30 + 50 = 130 liters. The 80 liters added is the area under the rate graph from x = 0 to x = 5: ∫05 (4x + 6) dx = 50 + 30 = 80.
  4. For 310 liters, 2x2 + 6x + 50 = 310, so 2x2 + 6x − 260 = 0. Divide both sides by 2: x2 + 3x − 130 = 0, and so (x − 10)(x + 13) = 0.
  5. (b) x = 10 or x = −13. A time after the valve is opened cannot be negative, so reject x = −13: the tank holds 310 liters after 10 minutes. Check: 2 × 100 + 60 + 50 = 310.

answer(a) V = 2x2 + 6x + 50 liters, and 130 liters after 5 minutes; (b) after 10 minutes

techniqueIntegration · Area Under a Curve

examsO-Level

Common pitfalls

  • Leaving out the constant and writing V = 2x2 + 6x, which says that the tank starts empty. Integrating always leaves a constant, and here it is the 50 liters already in the tank.
  • Multiplying the starting rate by the time, 6 × 5 = 30 liters added. The rate rises minute by minute, so the water added is the area under the rate graph, 80 liters.
10

A Bouncing Ball and a Leaning Stack of Books: Two Endless Sums

methodSum a Geometric Series with a Ratio Between −1 and 1 by a/(1 − r); for the Harmonic Series, Group Its Terms into Blocks and Compare Each Block with One Half

(a) A ball is dropped from a height of 8 m. Each time it lands it bounces back up to half the height it fell from, and it goes on bouncing without end. Find the total distance it travels. (b) Identical books are stacked at the edge of a table, each leaning out as far as it can without the stack toppling. With n books the top book reaches 12(1 + 12 + 13 + ⋯ + 1n) book lengths beyond the edge of the table. How many books are needed for the top book to lie wholly beyond the edge, and is there a limit to how far the stack can reach?

081624321234567landings, nmeters so far0121248163264books, noverhang (book lengths)8 + 2(4 + 2 + 1 + ...)
After the first drop of 8 m the ball rises and falls 4 m, then 2 m, then 1 m: the distance is 8 + 2(4 + 2 + 1 + ⋯).
After the first drop of 8 m, the ball rises and falls 4 m, then 2 m, then 1 m, and so on. The distance is 8 + 2(4 + 2 + 1 + ⋯).
step 1 of 6

An endless sum converges when its partial sums settle toward a limit, and diverges when they do not. A geometric series whose ratio lies between −1 and 1 converges to a1 − r. The harmonic series 1 + 12 + 13 + ⋯ has terms that shrink to zero, and still it diverges.

  1. After the first drop of 8 m, the ball rises and falls 4 m, then 2 m, then 1 m, and so on. The distance is 8 + 2(4 + 2 + 1 + ⋯).
  2. The partial sums of the distance are 8, 16, 20, 22, 23, 23.5, … meters. Each one closes half of the gap that is left to 24 m, and none of them passes it.
  3. The bracket is a geometric series with first term 4 and ratio 12. Since −1 < 12 < 1 it converges, to 41 − 12 = 8. (a) The total distance is 8 + 2 × 8 = 24 m.
  4. For the books, the overhang with n = 1, 2, 3, 4 is 12, 34, 1112 and 2524 book lengths. The first of these to pass 1 is 2524, with 4 books.
  5. Group the terms of the harmonic series: 13 + 14 > 12, 15 + 16 + 17 + 18 > 12, and each next group of twice as many terms is also more than 12. So the sum passes any number you name: ∑n=1∞ 1n diverges.
  6. (b) The top book lies wholly beyond the edge with 4 books. There is no limit to the overhang. Each doubling of the number of books adds at least 14 of a book length, so the overhang passes 2 book lengths with 31 books and keeps on growing, more and more slowly.

answer(a) 24 m; (b) 4 books, for an overhang of 2524 book lengths; there is no limit, because the harmonic series diverges

techniqueTwo Infinite Sums, Opposite Fates · Limits

examsH2

Common pitfalls

  • Adding 8 + 4 + 2 + 1 + ⋯ = 16 m, which counts each bounce once. After every landing the ball goes up and then comes back down, so each bounce height is traveled twice.
  • Deciding that the overhang must settle because the terms 1n shrink to zero. A sum whose terms do not shrink to zero cannot converge, but terms that do shrink are not enough: the harmonic series still grows without limit.
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