## MATH1111 Quizzes

Measuring Speed Quiz
Web resources available Questions

This quiz tests the work covered in Lecture 7 and corresponds to Section 2.1 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There is a web quiz at Wiley. It is the same quiz for each section of Chapter 2 and you should probably wait until the end of lecture 9 before you attempt it.

Be aware that it doesn’t seem to accept the written answers so you will have to check whether your answers are correct when they print the correct answer. Questions 11 and 12 were illegible on 14/11/05.

The Learning Hub (Mathematics) has a booklet on differentiation Introduction to Differential Calculus which covers all of the topics for the next few lectures. In particular, Chapter 1 of the booklet covers this topic.

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

The distance, $s\phantom{\rule{0.3em}{0ex}},$ a car has travelled from its starting point on a trip is shown in the table below as a function of time, $t\phantom{\rule{0.3em}{0ex}},$ since the trip started.
 $t$ (hours) 0 1 2 3 4 5 $s$ (km) 0 45 135 220 300 400
Which of the following is the average velocity between $t=1$ and $t=3\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) 80 km/hr b) 88.3 km/hr c) 87.5 km/hr d) 85 km/hr

Choice (a) is incorrect
Try again, this is the average velocity over the whole trip.
Choice (b) is incorrect
Try again, this is the average velocity between $t=2$ and $t=5\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is correct!
Average velocity = $\frac{220-45}{2}=\frac{175}{2}=87.5$ km/hr.
Choice (d) is incorrect
Try again, this is the average velocity between $t=2$ and $t=3\phantom{\rule{0.3em}{0ex}}?$
In a time of $t$ seconds, a particle moves a distance of $s$ metres from its starting point, where $s=3{t}^{2}+4$.
Which one of the following sets of statements is correct? Exactly one option must be correct)
 a) The average velocity between $t=1$ and $t=1+h$ at $h=0.1$ is $6.3\phantom{\rule{0.3em}{0ex}},$at $h=0.01$ is $6.03\phantom{\rule{0.3em}{0ex}},$ at $h=0.001$ is $6.003$and we estimate the instantaneous velocity at of the particle at $t=1$ to be 6 m/sec. b) The average velocity between $t=1$ and $t=1+h$ at $h=0.1$ is $6.1\phantom{\rule{0.3em}{0ex}},$at $h=0.01$ is $6.01\phantom{\rule{0.3em}{0ex}},$ at $h=0.001$ is $6.001$and we estimate the instantaneous velocity at of the particle at $t=1$ to be 6 m/sec. c) The average velocity between $t=1$ and $t=1+h$at $h=0.1$ is $3.1\phantom{\rule{0.3em}{0ex}},$at $h=0.01$ is $3.01\phantom{\rule{0.3em}{0ex}},$ at $h=0.001$ is $3.001$and we estimate the instantaneous velocity at of the particle at $t=1$ to be 3 m/sec. d) The average velocity between $t=1$ and $t=1+h$ at $h=0.1$ is $76.1\phantom{\rule{0.3em}{0ex}},$at $h=0.01$ is $706.01\phantom{\rule{0.3em}{0ex}},$ at $h=0.001$ is $7006.001$and we cannot estimate the instantaneous velocity at of the particle at $t=1\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is correct!
The average velocity =$\frac{3{\left(1+h\right)}^{2}+4-7}{h}=\frac{3+6h+3{h}^{2}+4-7}{h}=3h+6$
Choice (b) is incorrect
Try again, you may not have expanded $3{\left(1+h\right)}^{2}$ correctly.
Choice (c) is incorrect
Try again, you may not have expanded $3{\left(1+h\right)}^{2}$ correctly.
Choice (d) is incorrect
Try again, you seem to have forgotten to substitute $t=1$ into the formula.
Consider the graph below

At which labelled points does the graph have negative slope? Exactly one option must be correct)
 a) A, B, D. b) B, C, E, F. c) A, D. d) C, E, F.

Choice (a) is incorrect
Try again, B has zero slope which we do not consider to be negative.
Choice (b) is incorrect
Try again, all of these slopes are greater than, or equal to zero.
Choice (c) is correct!
The graph is decreasing as $x$ increases at these points so the slope is negative.
Choice (d) is incorrect
Try again, these slopes are all positive.
Using algebra which of the following evaluates the limit  $\underset{h\to 0}{lim}\frac{{\left(2+h\right)}^{3}-8}{h}\phantom{\rule{1em}{0ex}}$ correctly?
Exactly one option must be correct)
 a) $\underset{h\to 0}{lim}\frac{{\left(2+h\right)}^{3}-8}{h}=\underset{h\to 0}{lim}\left(h+4\right)=4\phantom{\rule{0.3em}{0ex}}.$ b) $\underset{h\to 0}{lim}\frac{{\left(2+h\right)}^{3}-8}{h}=\underset{h\to 0}{lim}\left({h}^{2}+6h+12\right)=12\phantom{\rule{0.3em}{0ex}}.$ c) $\underset{h\to 0}{lim}\frac{{\left(2+h\right)}^{3}-8}{h}=\underset{h\to 0}{lim}\left({h}^{2}+8h+10\right)=10\phantom{\rule{0.3em}{0ex}}.$ d) $\underset{h\to 0}{lim}\frac{{\left(2+h\right)}^{3}-8}{h}=\underset{h\to 0}{lim}\left({h}^{2}+6h+12\right)=17\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again, you seem to have expanded ${\left(2+h\right)}^{3}$ incorrectly.
Choice (b) is correct!
$\underset{h\to 0}{lim}\frac{{\left(2+h\right)}^{3}-8}{h}=\underset{h\to 0}{lim}\frac{8+12h+6{h}^{2}+{h}^{3}-8}{h}=\underset{h\to 0}{lim}\left({h}^{2}+6h+12\right)=12\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, you seem to have expanded ${\left(2+h\right)}^{3}$ incorrectly.
Choice (d) is incorrect
Try again, you seem to have substituted $h=1$ into the formula instead of $h=0\phantom{\rule{0.3em}{0ex}}.$