MCV4U0 (Monday, March 7th, 2016) vA
20
𝐴 C:
(Level)
Name:___________________________
TEST: Unit 2 – Derivatives (Application)
For Application Questions:
All work must be shown for full marks. You are being assessed on how you apply the
procedures taught in this unit to real-life situations and to make connections.
For Communication:
Make sure you use good mathematical form in ALL of your work. Show all steps using
correct mathematical techniques and conventions.
1. Choose the best answer for this question from the choices given and justify your answer.
The displacement of a particular amoeba that can be seen through a microscope can be modeled using the
function 𝑓(𝑡) = 𝑡 3 − 2𝑡 2 + 1, where f is the distance (in micrometres) of the amoeba to the right of the centre of
the microscope slide after t seconds have passed. At exactly 1 second, is the amoeba:
YES SLOWING DOWN
Justify your answer below. [2]
____SPEEDING UP
____NEITHER SLOWING DOWN NOR SPEEDING UP
𝒇′ (𝒕) = 𝟑𝒕𝟐 − 𝟒𝒕
𝒇′ (𝟏) = −𝟏
𝒇′′(𝒕) = 𝟔𝒕 − 𝟒
𝒇′′(𝟏) = 𝟐
Since the velocity and acceleration of the object are in opposite directions, the amoeba is slowing down.
2. Determine a quadratic function q(x) that has a tangent line with a slope of 4 at x = 1, a tangent line with a slope
of -8 at x = -1, and passes through the point (2, 15). [4]
𝒒(𝒙) = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄
𝒒′ (𝒙) = 𝟐𝒂𝒙 + 𝒃
𝒒′ (𝟏) = 𝟒
𝟒 = 𝟐𝒂(𝟏) + 𝒃
𝟒 = 𝟐𝒂 + 𝒃
𝒒′ (−𝟏) = −𝟖
−𝟖 = 𝟐𝒂(−𝟏) + 𝒃
−𝟖 = −𝟐𝒂 + 𝒃 [𝟏]
𝟒 = 𝟐𝒂 + 𝒃
−𝟖 = −𝟐𝒂 + 𝒃
−𝟒 = 𝟐𝒃
−𝟐 = 𝒃
[1]
𝟒 = 𝟐𝒂 + (−𝟐)
𝟑=𝒂
[1]
𝒒(𝒙) = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄
𝒒(𝒙) = 𝟑𝒙𝟐 − 𝟐𝒙 + 𝒄
𝒒(𝟐) = 𝟏𝟓
𝟏𝟓 = 𝟑(𝟐)𝟐 − 𝟐(𝟐) + 𝒄
𝟏𝟓 = 𝟏𝟐 − 𝟒 + 𝒄
𝟕=𝒄
𝒒(𝒙) = 𝟑𝒙𝟐 − 𝟐𝒙 + 𝟕 [1]
3. If f and g are the functions whose graphs are shown, let
𝑢(𝑥) = 𝑓(𝑥) ∙ 𝑔(𝑥) and 𝑣(𝑥) =
𝑓(𝑥)
.
𝑔(𝑥)
Find 𝑢′(1) and 𝑣′(5) and show how you determined your answer. (4)
Since 𝒇(𝒙) and 𝒈(𝒙) are piecewise linear function, the derivative at
𝒙 = 𝒂 is equation to the slope of the function at this point.
𝒖′ (𝟏) = 𝒇′ (𝟏)𝒈(𝟏) + 𝒇(𝟏)𝒈′(𝟏)
𝒖′ (𝟏) = (𝟐)(𝟏) + (𝟐)(−𝟏)
𝒖′ (𝟏) = 𝟎
𝒇′ (𝟓)𝒈(𝟓) − 𝒈(𝟓)𝒈′(𝟓)
(𝒈(𝟓))𝟐
𝟏
𝟐
(− ) (𝟑) − 𝟑 ( )
𝟑
𝟑
′ (𝟓)
𝒗
=
𝟐𝟐
𝟑
𝒗′ (𝟓) = −
𝟒
𝒗′ (𝟓) =
More on the next page! Watch your time!
4. The position function of a particle is given by 𝑠(𝑡) = 𝑡 3 − 4.5𝑡 2 − 7𝑡, 𝑡 ≥ 0, where s(t) is the distance from the
starting point (in metres) after t seconds have passed.
A. At what time does the particle reach a velocity of 5 m/s? [3]
𝒗(𝒕) = 𝟑𝒕𝟐 − 𝟗𝒕 − 𝟕
𝟓 = 𝟑𝒕𝟐 − 𝟗𝒕 − 𝟕
𝟎 = 𝟑𝒕𝟐 − 𝟗𝒕 − 𝟏𝟐
𝟎 = 𝟑(𝒕𝟐 − 𝟑𝒕 − 𝟒)
𝟎 = 𝟑(𝒕 − 𝟒)(𝒕 + 𝟏)
[1]
[1]
The particle reaches a velocity of 5 m/s at 4 s. (𝒕 = −𝟏 is not in the domain of 𝒔(𝒕) or 𝒗(𝒕).)
[1]
B. When is the particle neither slowing down nor speeding up? [2]
𝒂(𝒕) = 𝟔𝒕 − 𝟗
𝟎 = 𝟑(𝟐𝒕 − 𝟑)
The particle is neither slowing down or speeding up when acceleration is zero at 1.5 s.
5. Determine an algebraic model that you could use to solve this problem. DO NOT SOLVE. (2)
Suppose that the cost (in dollars) of manufacturing x items is approximated by 𝐶(𝑥) = 625 + 15𝑥 + 0.01𝑥 2 , for
1 ≤ 𝑥 ≤ 500. The unit cost (the cost of manufacturing one item) would then be 𝑈(𝑥) =
should be manufactured in order to ensure that the unit cost is minimized?
𝒙𝑪′ (𝒙) − 𝑪(𝒙)
𝒙𝟐
𝒙(𝟏𝟓 + 𝟎. 𝟎𝟐𝒙) − (𝟔𝟐𝟓 + 𝟏𝟓𝒙 + 𝟎. 𝟎𝟏𝒙𝟐 )
𝑼′(𝒙) =
𝒙𝟐
𝑼′(𝒙) =
𝟎. 𝟎𝟏𝒙𝟐 − 𝟔𝟐𝟓)
𝑼′(𝒙) =
𝒙𝟐
6. A vertical cylindrical tank with a radius of 5 m is being filled with water at a rate of
3 m3/min. How fast is the height of the water increasing (to two decimal places)? [3]
𝑽 = 𝝅𝒓𝟐 𝒉
𝒅𝑽
𝒅𝒕
𝒅𝑽 𝒅𝒉
= 𝒅𝒉 𝒅𝒕
[1]
𝒅𝑽
𝒅𝒉
= 𝝅𝒓𝟐
𝒅𝒕
𝒅𝒕
𝒅𝒉
𝟑 = 𝝅(𝟓)𝟐 𝒅𝒕
[1]
𝒅𝒉
𝒅𝒕
𝟑
𝒅𝒉
=
𝟐𝟓𝝅 𝒅𝒕
𝒅𝒉
≗ 𝟎. 𝟎𝟑𝟖 𝟏𝟗𝟕 𝟏𝟖𝟔 𝟑
𝒅𝒕
𝟑 = 𝟐𝟓𝝅
Therefore, the water level is increasing at an approximate rate of 0.04 m/min. [1]
𝐶(𝑥)
.
𝑥
How many items