## MATH1111 Quizzes

Interpretations of the Derivative Quiz
Web resources available Questions

This quiz tests the work covered in Lecture 10 and corresponds to Section 2.4 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

I can’t find any sites appropriate for this topic but it would be worth going to the sites from the previous topic to consolidate what you have learned.

You should be able to attempt web quiz at Wiley.

The Mathematics Learning Centre booklet Introduction to Differential Calculus covers all of the topics for these lectures. In particular, Chapters 1 to 3.1 of the booklet cover the topics from Week 3.

The site http://www.math.uncc.edu/$\sim$bjwichno/fall2004-math1242-006/Review˙Calc˙I/lec˙deriv.htm covers some of the material in Section 2.1-2.3

There is an applet that lets you sketch the derivative of a given function at http://www.ltcconline.net/greenl/java/Other/DerivativeGraph/classes/DerivativeGraph.html After you have mastered the topic you might like to try the tests at http://www.univie.ac.at/future.media/moe/tests/diff1/defabl.html and http://www.univie.ac.at/future.media/moe/tests/diff1/poldiff.html and the puzzle at http://www.univie.ac.at/future.media/moe/tests/diff1/ablerkennen.html

Suppose $y=f\left(x\right)\phantom{\rule{0.3em}{0ex}}.$ Which of the following statements is correct? Exactly one option must be correct)
 a) $\frac{dy}{dx}=f\left(x\right)\phantom{\rule{0.3em}{0ex}}.$ b) $\frac{dy}{dt}={f}^{\prime }\left(x\right)\phantom{\rule{0.3em}{0ex}}.$ c) $\frac{dy}{dx}=f\left(t\right)\phantom{\rule{0.3em}{0ex}}.$ d) $\frac{dy}{dx}={f}^{\prime }\left(x\right)\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again,$\frac{dy}{dx}$ is the derivative of $f$ with respect to $x\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, $y$ is a function of $x\phantom{\rule{0.3em}{0ex}},$ not $t\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, $f$ is a function of $x\phantom{\rule{0.3em}{0ex}},$ not $t\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is correct!
Both notations are equivalent.
Suppose $y=f\left(x\right)\phantom{\rule{0.3em}{0ex}}.$ Which of the following correctly interprets ${\frac{dy}{dx}|}_{x=2}=12\phantom{\rule{0.3em}{0ex}}?$
There may be more than one correct answer. (Zero or more options can be correct)
 a) The value of the derivative of $f$ at  $x=12$  is 2. b) The value of the derivative of $f$ at  $x=2$  is 12. c) The function $f$ is increasing at  $x=2\phantom{\rule{0.3em}{0ex}}.$ d) The function $f$ is increasing at  $x=12\phantom{\rule{0.3em}{0ex}}.$

There is at least one mistake.
For example, choice (a) should be False.
Try again,  ${\frac{dy}{dx}|}_{x=2}$ gives the value of the derivative at  $x=2\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (b) should be True.
${\frac{dy}{dx}|}_{x=2}$ gives the value of the derivative at  $x=2\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (c) should be True.
When the derivative is positive at a point the function is increasing.
There is at least one mistake.
For example, choice (d) should be False.
Try again, we have no information about what is happening at  $x=12\phantom{\rule{0.3em}{0ex}}.$
Recall ${\frac{dy}{dx}|}_{x=2}$ gives the value of the derivative at  $x=2\phantom{\rule{0.3em}{0ex}}.$
Correct!
1. False Try again,  ${\frac{dy}{dx}|}_{x=2}$ gives the value of the derivative at  $x=2\phantom{\rule{0.3em}{0ex}}.$
2. True ${\frac{dy}{dx}|}_{x=2}$ gives the value of the derivative at  $x=2\phantom{\rule{0.3em}{0ex}}.$
3. True When the derivative is positive at a point the function is increasing.
4. False Try again, we have no information about what is happening at  $x=12\phantom{\rule{0.3em}{0ex}}.$
Recall ${\frac{dy}{dx}|}_{x=2}$ gives the value of the derivative at  $x=2\phantom{\rule{0.3em}{0ex}}.$
If an object is placed in a medium with constant temperature $A$ then $T\left(t\right)\phantom{\rule{1em}{0ex}}$ describes the temperature of the object, in degrees Celsius, $t$ hours later.
Newton’s law of cooling tells us that if the object is cooling then  $\frac{dT}{dt}=-k\left(T-A\right)$  where $k$ is a constant of proportionality.
Which of the following gives the correct interpretation of  $\frac{dT}{dt}\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) $\frac{dT}{dt}$ describes the number of degrees Celsius loss of temperature per second for the object. b) $\frac{dT}{dt}$ describes the number of hours it takes for the object to lose one degree Celsius of temperature. c) $\frac{dT}{dt}$ describes the number of seconds it takes for the object to lose one degree Celsius of temperature. d) $\frac{dT}{dt}$ describes the number of degrees Celsius loss of temperature per hour for the object.

Choice (a) is incorrect
Try again, you do not have the correct units for the time.
Choice (b) is incorrect
Try again, we are looking at the change in temperature per unit time.
Choice (c) is incorrect
Try again, we are looking at the change in temperature per unit time.
Choice (d) is correct!
Since $T$ is measured in degrees Celsius and $t$ is measured in hours then $\frac{dT}{dt}$ describes the rate of change of temperature, that is the number of degrees Celsius loss of temperature per hour for the object.
If an object is placed in a medium with constant temperature $A$ then $T\left(t\right)\phantom{\rule{1em}{0ex}}$ describes the temperature of the object, in degrees Celsius, $t$ hours later.
Newton’s law of cooling tells us that if the object is cooling then  $\frac{dT}{dt}=-k\left(T-A\right)$  where $k$ is a constant of proportionality.
Which of the following describes the meaning of  ${\frac{dT}{dt}|}_{t=1}=-0.676\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) After 1 hour the object is losing approximately 0.676 degree of temperature per hour. b) The object is losing 1 degree of temperature every 0.676 hours. c) The temperature after 1 hour is 0.676 degrees less than its original temperature. d) Without knowing the values of $A$ and $k$ we cannot draw a conclusion.

Choice (a) is correct!
${\frac{dT}{dt}|}_{t=1}$ tells us the rate of change of temperature at $t=1$ hour.
So after 1 hour the object is losing approximately 0.676 degree of temperature per hour.
Choice (b) is incorrect
Try again, ${\frac{dT}{dt}|}_{t=1}$ tells us the rate of change of temperature at $t=1$ hour.
Choice (c) is incorrect
Try again, ${\frac{dT}{dt}|}_{t=1}$ tells us the rate of change of temperature at $t=1$ hour, not what the temperature loss has been in that hour.
Choice (d) is incorrect
Try again, we have enough information to describe the meaning of  ${\frac{dT}{dt}|}_{t=1}=-0.676\phantom{\rule{0.3em}{0ex}}.$