IB Mathematics AI HL 5.1 Calculus Question Bank

[Maximum number: 3]

Conrad is investigating the motion of a particle. The velocity of the particle, in $\mathrm{ms}^{-1}$ , is given by $v(t)=2 \cos t+\sin 2 t-0.2$ where t is the time, in seconds, after the investigation begins.

Figure for Question 5.1 Calculus - SL content question 1 — IB Maths AI HL

On the axes,

label, with coordinates, the point(s) where the particle has zero acceleration.

[Maximum number: 2]

This question uses differential equations to model the maximum velocity of a skydiver in free fall.
In 2012, Felix Baumgartner jumped from a height of 40000 m . He was attempting to travel at the speed of sound, $330 \mathrm{~m} \mathrm{~s}^{-1}$ , whilst free-falling to the Earth.
Before making his attempt, Felix used mathematical models to check how realistic his attempt would be. The simplest model he used suggests that

\frac{\mathrm{d} v}{\mathrm{~d} t}=g

where $v \mathrm{~m} \mathrm{~s}^{-1}$ is Felix's velocity and $g \mathrm{~ms}^{-2}$ is the acceleration due to gravity. The time since he began to free-fall is t seconds and the displacement from his initial position is s metres.

Throughout this question, the direction towards the centre of the Earth is taken to be positive and v is a positive quantity.

When s=0, it is given that Felix jumps with an initial velocity v=10.

To test the model

\frac{\mathrm{d} v}{\mathrm{~d} t}=g

Felix conducted a trial jump from a lower height, and data for v against t was found.

If the model is correct, describe the shape of the graph of v against t.

Felix's data are plotted on the following graph.

Figure for Question 5.1 Calculus - SL content question 2 — IB Maths AI HL

[Maximum number: 31]

Linda owns a field, represented by the shaded region R. The plan view of the field is shown in the following diagram, where both axes represent distance and are measured in metres.

Figure for Question 5.1 Calculus - SL content question 3 — IB Maths AI HL

The segments [AB], [CD] and [AD] respectively represent the western, eastern and southern boundaries of the field. The function, f(x), models the northern boundary of the field between points B and C and is given by

f(x)=\frac{-x^{2}}{50}+2 x+30, \text { for } 0 \leq x \leq 70

(a)

Find $f^{\prime}(x)$ .

[ 5 ]

$f^{\prime}(x)=\frac{-2 x}{50}+2\left(=\frac{-x}{25}+2,-0.04 x+2\right) \quad$ A1A1

Note: Award A1 for each correct term. Award at most A0A1 if extra terms are seen.

(b)

Hence find the coordinates of the point on the field that is furthest north.

Point A has coordinates (0,0), point B has coordinates (0,30), point C has coordinates (70,72) and point D has coordinates (70,0).

[ 5 ]

$0=\frac{-x}{25}+2$ OR sketch of $f^{\prime}(x)$ with x-intercept indicated M1
$x=50 \quad$ A1
$y=80 \quad$ A1
(50,80)
Note: Award M0A0A1 for the coordinate (50,80) seen either with no working or found from a graph of f(x).

(c)

Write down the integral which can be used to find the area of the shaded region R.

[ 4 ]

$\quad \int_{0}^{70} \frac{-x^{2}}{50}+2 x+30 \mathrm{~d} x \quad$ A1A1

Note: Award A1 for a correct integral, A1 for correct limits in the correct location. Award at most A0A1 if d x is omitted.

(d)

Find the area of Linda's field.

Linda used the trapezoidal rule with ten intervals to estimate the area. This calculation underestimated the area by $11.4 \mathrm{~m}^{2}$ .

[ 4 ]

( Area $=) 4710 \mathrm{~m}^{2}\left(4713.33 \ldots, \frac{14140}{3}\right) \quad$ A2

(e)

Suggest how Linda might be able to reduce the error whilst still using the trapezoidal rule.

Linda would like to construct a building on her field. The square foundation of the building, EFGH , will be located such that $[\mathrm{EH}]$ is on the southern boundary and point F is on the northern boundary of the property. A possible location of the foundation of the building is shown in the following diagram.

Figure for Question (e) — IB Maths AI HL

The area of the square foundation will be largest when $[\mathrm{GH}]$ lies on $[\mathrm{CD}]$ .

[ 3 ]

(f)

Find the x-coordinate of point E for the largest area of the square foundation of building EFGH.

[ 5 ]

width of the square is $70-x \quad \mathbf{OR}$ the length of the square is $\frac{-x^{2}}{50}+2 x+30$
(M1)
Note: Award (M1) for 70-x seen anywhere. Accept $\frac{-x^{2}}{50}+2 x+30$ but only if this expression is explicitly identified as a dimension of the square.
in term of x, equating the length to the width ED
$\frac{-x^{2}}{50}+2 x+30=70-x$
( $x=14.7920 \ldots$ or 135.21)
$(x=) 14.8 \mathrm{~m}(14.7920 \ldots)$

Note: Award MOMOAO for an unsupported answer of 15. Award at most M1M0A0 for an approach which leads to $A^{\prime}(x)=0$ . This will lead to a square base which extends beyond the east boundary of the property. Similar for any solution where F is not on the northern boundary, or GH is not on the east boundary.

(g)

Find the largest area of the foundation.

[ 5 ]

EITHER
$(70-14.7920 \ldots)^{2}$

OR
$(55.2079 \ldots)^{2}$

OR
$\left(\frac{-(14.7920 \ldots)^{2}}{50}+2(14.7920 \ldots)+30\right)^{2}$

THEN
(Area = ) $3050 \mathrm{~m}^{2}$ (3047.92…)

Note: Follow through from part (d)(i), provided x is between 0 and 70 . Award at most M1A0 if their answer is outside the range of their [0, 4713.33...] from part (b).

IB Mathematics AI HL 5.1 Calculus Question Bank

Exam points

Covered syllabus points

Covered syllabus points

5.1 Calculus - SL content question 1

5.1 Calculus - SL content question 2

5.1 Calculus - SL content question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question (f)

Question (g)