MATH 251 Portland Community College Calculating the Gradient of The Tangent Ques

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Quiz 5 MTH251.001 Spring 2021
Printed Name: ___________________________
You do not need to print out this quiz; you may provide your work and answers on your own separate paper.
Upload your solutions as a single PDF file to the “Quiz 5” submission folder in our D2L course.
Your submission should be titled “(your name) MTH251 Quiz 5.”
Due Date: Your quiz results must be submitted by Friday May 21, by 11:59 p.m.
Instructions: – Show all your work; supporting work always needs to be included.
– Clearly indicate (circle or underline) your answer.
– For multiple choice questions, choice the single best answer.
__________________________________________________________________________________________
1.
Match the functions in graphs (A)-(D) with their derivatives (I)-(III) in figure 13. Explain why two of the
(2pt)
functions have the same derivative.
__________________________________________________________________________________________
2.
True or False.
a.
(2pt)
b.
c.
d.
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑥 √21 = √21𝑥 √21−1
𝑑
𝑑
((𝑥 25 ) + (𝑥 −15 )) = 𝑑𝑥 (𝑥 25 ) + 𝑑𝑥 (𝑥 −15 )
𝑒 √21 = √21𝑒 √21−1
𝑑
𝑑
((𝑥 25 )(𝑥 −15 )) = 𝑑𝑥 (𝑥 25 ) 𝑑𝑥 (𝑥 −15 )
__________________________________________________________________________________________
3.
For each of the following find the equation of the indicated derivative.
(6pt)
2
a.
Given 𝑦 = 𝑒 10−𝑥 , find 𝑦 ′ .
b.
Given 𝑓(𝜃) = sin(2𝜃), find 𝑓 (4) (𝜃).
__________________________________________________________________________________________
4.
(5pt)
A suburban community receives its electrical power supply from a local power plant. An
engineer, after studying data on energy usage, determines that a good model for daily energy
usage between 6:00 a.m. and 8:00 p.m. is given by the function
𝑡−13 4
)
7
𝐸(t) = 4 − 2 (
for 6 ≤ 𝑡 ≤ 20
where t = time in hours after midnight and E(t) is in megawatts (so t = 6 is 6:00 am, t = 12 is 12:00 pm
(noon) and so forth).
a. Determine 𝐸′(𝑡).
b. Evaluate 𝐸′(10).
Write a sentence explaining the meaning of the value you found for 𝐸′(10). What is happening to
energy usage at 10 am?
__________________________________________________________________________________________
5.
Consider the relation:
(5pt)
𝑦2 − 𝑦 = 𝑥
The graph of which looks like:
a.
Use Implicit differentiation to find 𝑦′.
Since this graph is not a function, your final answer will likely contain both x’s and y’s.
b.
The point (2,-1) is on this graph (since (-1)2 – (-1) = 2).
Use your result in a) to find the equation of the tangent line at this point.
Quiz 5 MTH251.001 Spring 2021
Printed Name: ___________________________
You do not need to print out this quiz; you may provide your work and answers on your own separate paper.
Upload your solutions as a single PDF file to the “Quiz 5” submission folder in our D2L course.
Your submission should be titled “(your name) MTH251 Quiz 5.”
Due Date: Your quiz results must be submitted by Friday May 21, by 11:59 p.m.
Instructions: – Show all your work; supporting work always needs to be included.
– Clearly indicate (circle or underline) your answer.
– For multiple choice questions, choice the single best answer.
__________________________________________________________________________________________
1.
Match the functions in graphs (A)-(D) with their derivatives (I)-(III) in figure 13. Explain why two of the
(2pt)
functions have the same derivative.
A:We can see on the graph A that the function represented is first increasing and then decreasing. This means
that her derivative has to be first positive and then negative.
This corresponds to the derivative III.
B:We can see on the graph B that the function represented is an increasing function. This so implies that her
derivative is only positive.
This so leads to the derivative I.
C:We can see on the graph C that the function represented is increasing then decreasing and again increasing.
This means that her derivative has to be positive then negative and positive again.
This so leads us to the derivative II.
D:We can see on the graph D that the function represented is first increasing and then decreasing. This is so the
same observation than for the graph A. This so means that the derivative of the graph D has to be positive and
then negative, just as the graph A.
We so have the derivative III which corresponds both for graph A and D
for the previous reasons.
__________________________________________________________________________________________
2.
(2pt)
True or False.
a.
𝑑
𝑑𝑥
𝑥
21
= 21𝑥
21−1
TRUE because as the derivative of a function of the form is and that here we have
u = then is really
b.
𝑑
𝑑𝑥
((𝑥25) + (𝑥−15)) = 𝑑𝑥𝑑 (𝑥25) + 𝑑𝑥𝑑 (𝑥−15)
TRUE as the derivative of a sum is equal to the sum of the derivatives.
c.
𝑑
𝑑𝑥
𝑒
21
= 21𝑒
21−1
FALSE because as the derivative of a function of the form is and that here we have
u = then should be 0 as u’ = 0 because .
d.
𝑑
𝑑𝑥
((𝑥25)(𝑥−15)) = 𝑑𝑥𝑑 (𝑥25) 𝑑𝑥𝑑 (𝑥−15)
FALSE as the derivative of a multiplication is not the multiplication of the
derivatives.
__________________________________________________________________________________________
3.
For each of the following find the equation of the indicated derivative.
(6pt)
2
10−𝑥
‘
a.
Given 𝑦 = 𝑒
, find 𝑦 .
We have y(x) = and we are going to calculate y’. As we have a function of the form then it means that its
derivative is going to be as . Here we have u = so u’ = .
So we can conclude that.
(4)
b.
Given 𝑓(θ) = sin 𝑠𝑖𝑛 (2θ) , find 𝑓 (θ).
𝑓(𝜃)=sin(2𝜃) and we are going to calculate 𝑓(4)(𝜃) :
𝑓’(𝜃)= 2cos(2𝜃) as the derivative of sin is cos
𝑓(2)(𝜃)= 2(-2sin(2𝜃)) = -4sin(2𝜃)
𝑓(3)(𝜃) = -4(2cos(2𝜃)) = -8cos(2𝜃)
𝑓(4)(𝜃) = -8(-2sin(2𝜃)) = 16sin(2𝜃)
so 𝑓(4)(𝜃) = 16sin(2𝜃)
__________________________________________________________________________________________
4.
A suburban community receives its electrical power supply from a local power plant. An
(5pt)
engineer, after studying data on energy usage, determines that a good model for daily energy
usage between 6:00 a.m. and 8:00 p.m. is given by the function
(
𝐸(𝑡) = 4 − 2
𝑡−13 4
7
)
for 6≤𝑡≤20
where t = time in hours after midnight and E(t) is in megawatts (so t = 6 is 6:00 am, t = 12 is 12:00 pm
(noon) and so forth).
a. Determine 𝐸'(𝑡).
We have and we are going to calculate E(t)’ :
So the derivative is
b. Evaluate 𝐸'(10).
Write a sentence explaining the meaning of the value you found for 𝐸'(10). What is happening to
energy usage at 10 am?
First, we calculate E’(10) :
MW/h
The derivative at a precise point is actually the gradient of the tangent from that point.
This means that here, we have calculated the gradient of the tangent at 10am. As this gradient is positive, this
means that at 10am the amount of energy is increasing by 0.09 MW/h.
__________________________________________________________________________________________
5.
Consider the relation:
(5pt)
2
𝑦 −𝑦=𝑥
The graph of which looks like:
a.
Use Implicit differentiation to find 𝑦’.
Since this graph is not a function, your final answer will likely contain both x’s and y’s.
We are going to find y’ while using the implicit differentiation
The derivative of y is so :
b.
The point (2,-1) is on this graph (since (-1)2 – (-1) = 2).
Use your result in a) to find the equation of the tangent line at this point.
In a) we have found that . Here we have find the equation of the tangent line at the point (2 ,-1). We are so
going to start by calculating the gradient of the tangent :
So from here we can deduce the equation of the tangent line, that we are going to call h, at the point (2, -1) :
So finally we have
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Exercice 3
We have y(x) = and we are going to calculate y’. As we have a function of the form then it means
that its derivative is going to be as . Here we have u = so u’ = .
So we can conclude that:
We have f(O)=sin(20) and we are going to calculate /)(O):
(O)=2cos(20) as the derivative of sin is cos
120)=2(-2sin(20)) = -4sin(20)
13)) = -4(2cos(20)) = -8cos(20)
(4)(O) = -86-2sin(20)) = 16 sin(20)
We finally so have /(“@) = 16 sin(20)
Exercice 4:
We have and we are going to calculate E(t)’:
So the derivative is
b) First, we calculate E'(10)
MW/h
The derivative at a precise point is actually the gradient of the tangent from that point.

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