MCV4U · Unit 4 test review

Applications of Derivatives: Test Review

Everything for the Unit 4 test on one page: the key facts from all 4 lessons, the mistakes that cost the most marks, and a practice test that points you back to the lessons you need.

Key facts

Motion: Displacement, Velocity and Acceleration

$v (t) = s^{'} (t)$ and $a (t) = v^{'} (t) = s^{''} (t)$ ; speed is $∣ v (t) ∣$ .
At rest: $v (t) = 0$ . Changes direction: $v$ changes sign.
Speeding up when $v$ and $a$ have the same sign; slowing down when they have opposite signs.
Displacement $= s (end) - s (start)$ . Total distance adds $∣Δ s ∣$ over each piece between turning points.
For $h (t) = - 4.9 t^{2} + v_{0} t + h_{0}$ : maximum height where $h^{'} (t) = 0$ , landing where $h (t) = 0$ .

Rates of Change in Context

The units of $Q^{'} (t)$ are (units of $Q$ ) per (unit of $t$ ); the sign says increasing or decreasing.
$\frac{d}{d t} b^{t} = b^{t} ln b$ and $\frac{d}{d t} e^{k t} = k e^{k t}$ : exponential growth and decay rates are proportional to the amount present.
For a sinusoidal model $A sin (k (t - c)) + d$ , the rate is $A k cos (k (t - c))$ (radians); it is fastest at the midline and zero at a maximum or minimum.
Inflation: $P^{'} (t) > 0$ ; deflation: $P^{'} (t) < 0$ ; inflation is fastest where $P^{''} (t) = 0$ and $P^{''}$ changes from positive to negative.
Relative (percent) rate of change: $\frac{Q ^{'} ( t )}{Q ( t )} \times 100%$ .

Optimization Problems

On a closed interval, the absolute maximum and minimum occur at critical numbers or endpoints: evaluate them all and compare.
Use the constraint to write the quantity to optimize as a function of ONE variable, then state its domain.
Reject critical numbers outside the domain; confirm a max or min with the first or second derivative test.
Open box from an $L \times W$ sheet: $V (x) = x (L - 2 x) (W - 2 x)$ with $0 < x < \frac{min ( L , W )}{2}$ .
To minimize a distance, minimize $d^{2}$ instead; the same $x$ gives the minimum.

Optimization in Business and Science

Revenue $R = (price) (quantity sold)$ ; profit $P (x) = R (x) - C (x)$ .
Marginal revenue is $R^{'} (x)$ and marginal cost is $C^{'} (x)$ ; a maximum profit inside the domain occurs where $R^{'} (x) = C^{'} (x)$ .
Average cost is $\frac{C ( x )}{x}$ ; differentiate it (it is a rational function) to find the lowest cost per item.
For $R (x) = k x b^{- x}$ with $k > 0$ and $b > 1$ , the maximum is at $x = \frac{1}{ln b}$ ; for the bus fare, $\frac{1}{ln 1.15} \approx 7.16$ .
Compare numeric (table), graphical and algebraic (derivative) strategies: the algebraic one gives the exact value.
Always answer in context with units, a domain check and sensible rounding.

Where marks get lost

Deciding "slowing down" from $a < 0$ alone. If $v < 0$ too, the object is speeding up in the negative direction. Always compare the signs of $v$ and $a$ . (Motion: Displacement, Velocity and Acceleration)
Answering a total-distance question with the displacement $s (end) - s (start)$ , which cancels the trip back. Split at every turning point first. (Motion: Displacement, Velocity and Acceleration)
Assuming every zero of $v$ is a change of direction. Check the sign of $v$ on both sides; $v (t) = (t - 2)^{2}$ touches zero without changing sign. (Motion: Displacement, Velocity and Acceleration)
Substituting the time of maximum height into $v (t)$ instead of $h (t)$ : $v$ is $0$ there by definition, so that tells you nothing about the height. (Motion: Displacement, Velocity and Acceleration)
Giving a negative speed. Speed is $∣ v ∣$ ; the sign belongs to velocity and shows direction. (Motion: Displacement, Velocity and Acceleration)
Treating $0. 6^{t}$ like a power function and writing its derivative as $t (0.6)^{t - 1}$ . The variable is in the exponent, so $\frac{d}{d t} (0.6)^{t} = (0.6)^{t} ln 0.6$ . (Rates of Change in Context)
Forgetting the inner derivative in a sinusoidal model: $\frac{d}{d t} sin (\frac{2 π t}{12.4})$ is $\frac{2 π}{12.4} cos (\frac{2 π t}{12.4})$ , not just $cos (\frac{2 π t}{12.4})$ . (Rates of Change in Context)
Evaluating trig models in degree mode. The derivative rules for sine and cosine hold only in radians. (Rates of Change in Context)
Reporting a rate without units or interpretation, or mixing units: a daylight rate in hours per day must be multiplied by 60 to give minutes per day. (Rates of Change in Context)
Thinking the fastest change happens at the highest point. At a maximum of a smooth model the rate is $0$ ; for tides and daylight the fastest change is at the midline. (Rates of Change in Context)
Leaving the function in two variables and differentiating anyway. Use the constraint to eliminate one variable first. (Optimization Problems)
Keeping a critical number outside the domain, such as $x = 15$ for the $30$ cm box, which leaves no base at all. (Optimization Problems)
Forgetting the endpoints of a closed interval: the absolute extreme can be at an endpoint, not at a critical number. (Optimization Problems)
Stopping at $x = 5$ when the question asks for the maximum volume or the dimensions. Reread the question and answer it in a sentence with units. (Optimization Problems)
Assuming the best rectangle is always a square. With a river on one side, the best field is twice as long as it is wide. (Optimization Problems)
Maximizing the wrong quantity: the number of riders or the energy $E (t)$ instead of revenue or the AVERAGE rate $\frac{E ( t )}{t + 2}$ . Reread what the question wants as large as possible. (Optimization in Business and Science)
Differentiating $(1.15)^{- x}$ as $- x (1.15)^{- x - 1}$ (the power rule). The variable is in the exponent, so the derivative is $- (1.15)^{- x} ln 1.15$ . (Optimization in Business and Science)
Setting the whole product $1200 (1.15)^{- x} (1 - x ln 1.15)$ equal to zero and trying to solve $1200 (1.15)^{- x} = 0$ . An exponential is never zero; only the bracket can be. (Optimization in Business and Science)
Leaving the answer as $x \approx 7.155$ without saying it is a fare in dollars, or rounding a number of items to a decimal. Interpret in context and round sensibly. (Optimization in Business and Science)
Ignoring the domain: prices, quantities and times cannot be negative, and on a closed domain the endpoints must be compared. (Optimization in Business and Science)

Practice test

24 questions from across the unit. Each one shows the lesson it comes from.

Motion: Displacement, Velocity and Acceleration
1
At some instant an object has $v = - 3$ m/s and $a = - 2 m/s^{2}$ . Which statement is true?
Motion: Displacement, Velocity and Acceleration
2
An object has position $s (t) = 2 t^{3} - 21 t^{2} + 60 t$ metres. It is at rest at two times. What is the LATER one, in seconds?
s
Motion: Displacement, Velocity and Acceleration
3
A ball's height is $h (t) = - 4.9 t^{2} + 24.5 t + 2$ metres. Find its maximum height, to one decimal place.
m
Motion: Displacement, Velocity and Acceleration
4
An object has velocity $v (t) = (t - 2)^{2}$ m/s. What happens at $t = 2$ s?
Motion: Displacement, Velocity and Acceleration
5
Put the steps in order to find the total distance travelled between two times.
1. 1Solve $v (t) = 0$ to find the turning times inside the interval
2. 2Add the absolute differences between consecutive positions
3. 3Evaluate $s (t)$ at the start, each turning time and the end
4. 4Differentiate to find $v (t)$
Motion: Displacement, Velocity and Acceleration
6
The derivative of velocity with respect to time is called what?
Rates of Change in Context
7
$P (t)$ is a town's population and $t$ is in years. What are the units of $P^{'} (t)$ ?
Rates of Change in Context
8
For the butterfly model $P (t) = \frac{6000}{1 + 49 ( 0.6 ) ^{t}}$ , find $P^{'} (10)$ to the nearest whole number.
butterflies/day
Rates of Change in Context
9
A drink cools as $T (t) = 20 + 70 e^{- 0.05 t}$ . At what time, in minutes, is it cooling at $2° C$ per minute? Give one decimal place.
min
Rates of Change in Context
10
A tank holds $V (t) = 1000 (1 - \frac{t}{40})^{2}$ litres of water as it drains, $0 \leq t \leq 40$ minutes. Find $V^{'} (10)$ in litres per minute.
L/min
Rates of Change in Context
11
The graph of a price index rises for 10 years in an S shape: slowly at first, then more and more steeply, then more gently again. At which point on the graph was inflation fastest?
Rates of Change in Context
12
On an S-shaped (logistic) population graph, the point where the population is growing fastest is called the point of ______.
Optimization Problems
13
Equal squares of side $x$ cm are cut from the corners of a $24$ cm by $24$ cm sheet to make an open box. What value of $x$ gives the maximum volume?
cm
Optimization Problems
14
A farmer has $1000$ m of fence to enclose a rectangular field along a straight river (no fence on the river side). What is the maximum area, in square metres?
m²
Optimization Problems
15
Two positive numbers $x$ and $y$ add to $30$ . Find the maximum value of $x y^{2}$ .
Optimization Problems
16
For the $30$ cm by $30$ cm open box, $V^{'} (x) = (30 - 2 x) (30 - 6 x) = 0$ gives $x = 5$ or $x = 15$ . Why is $x = 15$ rejected?
Optimization Problems
17
Match each problem to the function to optimize.
1. A. $A (x) = x (600 - 2 x)$
2. B. $f (x) = (x - 3)^{2} + x$
3. C. $S (r) = 2 π r^{2} + \frac{1000}{r}$
4. D. $V (x) = x (30 - 2 x)^{2}$
Open box from a $30$ cm square sheet
Field beside a river with $600$ m of fence
Closed can holding $500 cm^{3}$
Point on $y = x$ closest to $(3, 0)$
Optimization Problems
18
On a closed interval, the absolute maximum can occur at a critical number or at one of the ______ of the interval.
Optimization in Business and Science
19
At a price of $p$ dollars, $q = 1200 - 40 p$ tickets sell. What price, in dollars, maximizes revenue?
dollars
Optimization in Business and Science
20
Bus riders per day are $1200 (1.15)^{- x}$ at a fare of $x$ dollars. Find the fare that maximizes revenue, to the nearest cent (in dollars, e.g. 1.23).
dollars
Optimization in Business and Science
21
The cost of making $x$ items is $C (x) = 4000 + 10 x + 0.1 x^{2}$ dollars. How many items minimize the average cost per item?
items
Optimization in Business and Science
22
For the foraging bird, $A (t) = \frac{3000 t}{( t + 4 ) ( t + 2 )}$ . Find the time in the patch, in minutes, that maximizes $A (t)$ , to 2 decimal places.
min
Optimization in Business and Science
23
A company's profit is $P (x) = R (x) - C (x)$ . At a production level that maximizes profit (not at an endpoint of the domain), which is true?
Optimization in Business and Science
24
What is the derivative of a revenue function, $R^{'} (x)$ , called? (Two words.)

Endless drill

Fresh numbers every time, for the calculation questions in this unit.

Endless drillnew numbers every time

Generating a problem…