IB Mathematics Analysis and Approaches Topic AHL 5.14—implicit Functions, Related Rates, Optimisation Questionbank (HL)

Question 1

HLPaper 2

Consider the curve $C$ defined by the equation $x^{2} y + cos (y) = 2$ , where $y > 0$ .

1.

Using implicit differentiation, show that $\frac{d y}{d x} = - \frac{2 x y}{x ^{2} - s i n ( y )}$ .

[5]

2.

Find the coordinates of the point $P$ on $C$ where $y = \frac{π}{2}$ , in the interval $0 < x < 2 π$ .

[6]

Question 2

HLPaper 1

Consider the curve $C$ defined by the equation $x^{2} + y cos (y) = 4 - π$ , where $y > 0$ .

1.

Using implicit differentiation, show that $\frac{d y}{d x} = - \frac{2 x}{c o s ( y ) - y s i n ( y )}$ .

[5]

2.

Find the coordinates of the point $P$ on $C$ where $y = π$ , in the interval $0 < x < 4$ .

[6]

Question 3

HLPaper 1

A cylindrical vase has a height $h$ and a radius $r$ (in arbitrary units), with the constraint $π r^{2} + 4 h = 20$ , where $r, h > 0$ .

1.

Express the volume $V$ of the vase in terms of $r$ .

[3]

2.

Show that $\frac{d V}{d r} = π r (10 - π r^{2})$ .

[4]

3.

Find $r$ and $h$ that maximize $V$ , and find the maximum value of $V$ .

[6]

Question 4

HLPaper 1

Consider the curve $C$ defined by the equation $x^{2}+y^{2}+ \ln (y)=5$ , where $y > 0$ .

1.

Using implicit differentiation, show that $\frac{d y}{d x} = - \frac{2 x y}{2 y ^{2} + 1}$ .

[5]

2.

Find the coordinates of the point $P$ on $C$ where $y = 1$ , in the interval $0 < x < 2 π$ .

[6]

Question 5

HLPaper 3

In the study of kinematics, the position of a moving particle can be described by a position vector that depends on time. If a particle moves through three-dimensional space with constant velocity, its position at time $t$ seconds is given by a vector equation of the form

$r (t) = r_{0} + t v$

where $r_{0}$ is the position of the particle at $t = 0$ and $v$ is its constant velocity vector. Note that while each particle traces out a straight-line path in space (which you are familiar with from the study of vector lines), the key idea here is that the parameter $t$ represents time: at each instant $t$ , the particle occupies a specific point on the line.

This investigation explores what happens when two particles move simultaneously through 3D space, and in particular, how calculus can be used to determine when they are closest together.

Two particles $P$ and $Q$ move through three-dimensional space. At time $t$ seconds ( $t \geq 0$ ), their position vectors are given by

$r_{P} (t) = 314 + t 12 - 1$

$r_{Q} (t) = 152 + t 411$

where distances are measured in metres.

1.

Find the position of each particle at $t = 1$ and at $t = 2$ .

[3]

2.

Find the distance between $P$ and $Q$ at $t = 1$ and at $t = 2$ .

[3]

3.

Show that the displacement vector from $P$ to $Q$ at time $t$ is given by

$d (t) = r_{Q} (t) - r_{P} (t) = - 2 + 3 t 4 - t - 2 + 2 t$

[2]

4.

The distance between $P$ and $Q$ at time $t$ is denoted $D (t) = ∣ d (t) ∣$ .

Show that $[D (t)]^{2} = 14 t^{2} - 28 t + 24$ .

[3]

5.

Since $D (t) \geq 0$ for all $t$ , the value of $t$ that minimises $D (t)$ is the same as the value of $t$ that minimises $[D (t)]^{2}$ . This is a standard technique in optimization: working with the squared distance avoids the complication of differentiating a square root.

Find $\frac{d}{d t} [D (t)]^{2}$ .

[1]

6.

Hence find the value of $t$ at which $P$ and $Q$ are closest together, and determine this minimum distance.

[3]

7.

The velocity vectors of $P$ and $Q$ are $v_{P} = (12 - 1)$ and $v Q = (411)$ respectively. The relative velocity of $Q$ with respect to $P$ is defined as $v r e l = v_{Q} - v_{P}$ .

Find $v_{r e l}$ .

[1]

8.

Find the displacement vector $d (1)$ at the moment of closest approach.

[1]

9.

Hence show that $v_{r e l}$ is perpendicular to $d (1)$ .

[2]

10.

Consider now two particles with general position vectors

$r_{1} (t) = a + t u, r_{2} (t) = b + t w$

where $a$ , $b$ , $u$ , $w$ are constant vectors and $u \neq = w$ .

Define $p = b - a$ and $q = w - u$ .

Show that $[D (t)]^{2} = ∣ p ∣^{2} + 2 t (p \cdot q) + t^{2} ∣ q ∣^{2}$ .

[3]

11.

Hence show that at the time $t^{*}$ when the particles are closest, the relative velocity $q$ is perpendicular to the displacement vector $d (t^{*})$ .

[5]

12.

Explain geometrically why the displacement vector between two particles must be perpendicular to their relative velocity at the instant of closest approach.

[2]

IB Mathematics Analysis and Approaches AHL 5.14—implicit Functions, Related Rates, Optimisation Questionbank

Unlock all 26 questions with a free account

IB Mathematics Analysis and Approaches AHL 5.14—implicit Functions, Related Rates, Optimisation Questionbank

IB Mathematics Analysis and Approaches (AA) Topic AHL 5.14—implicit Functions, Related Rates, Optimisation Questionbank

Unlock all 26 questions with a free account