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Larson and Edwards - Calculus 9/e (Homework)

James Finch

Math - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 32 / 34

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving...  ()

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Question
Points
1 2 3 4 5 6 7 8 9 10
1/2 1/2 8/8 10/10 1/1 1/1 1/1 1/1 7/7 1/1
Total
32/34 (94.1%)

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    Here are some textbook questions from Calculus 9/e by Ron Larson and Bruce H. Edwards published by Brooks/Cole. Click here for a list of all of the questions coded in WebAssign. This demo assignment allows many submissions and allows you to try another version of the same question for practice.

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1. 1/2 points  |  Previous Answers LarCalc9 2.1.057.MI. My Notes
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1/2
 
Identify a function f that has the given characteristics. Then sketch the function.
f(1) = 7
f '(x) = 4, −∞ < x < ∞
f(x) =
4x+3
Correct: Your answer is correct.
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2. 1/2 points  |  Previous Answers LarCalc9 2.1.058. My Notes
Question Part
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1 2
1/1 0/1
1/50 3/50
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1/2
 
Identify a function f that has the given characteristics.
f(–5) = f(4) = 0;  f '(-0.5) = 0,  f '(x) < 0 for x < -0.5;  f '(x) > 0 for x > -0.5
f(x) =
x2+x−20
Correct: Your answer is correct.

Sketch the function.

Incorrect: Your answer is incorrect.
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3. 8/8 points  |  Previous Answers LarCalc9 2.2.062.MI.SA. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8
1/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1
1/50 3/50 1/50 1/50 1/50 1/50 1/50 1/50
Total
8/8
 
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Tutorial Exercise
Determine the point at which the graph of the function below has a horizontal tangent line.
y = 3x2 + 6
Part 1 of 7
We are asked to find any points on the function that have a horizontal tangent line. In other words, the slope of the tangent line is zero Correct: Your answer is correct. seenKey

zero

 .
Part 2 of 7
Since we want to know where the tangent line is zero, we need to find all points where the derivative is zero. Start by finding the derivative of the equation. To do this, we use a combination of the basic differentiation rules. First take the derivative of both sides of the equation.
y = 3x2 + 6
 
dy
dx
 = 
d
dx
 
$$3x2+6
Correct: Your answer is correct. 3 x^2 + 6
Part 3 of 7
Next, apply the Sum Rule to separate the right side into the sum of two derivatives.
dy
dx
 = 
d
dx
[3x2 + 6]
 
dy
dx
 = 
d
dx
$$3x2
Correct: Your answer is correct. 3 x^2
 + 
d
dx
[6]
Part 4 of 7
Now use the Power Rule and the Constant Multiple Rule to determine the derivatives.
dy
dx
 = 
d
dx
[3x2] + 
d
dx
[6]
 
dy
dx
 = 
$$6x
Correct: Your answer is correct. 6x
Part 5 of 7
Now that we have the derivative, we need to set it equal to zero and solve for x.
 
dy
dx
 = 6x = 0 right double arrow implies x = 0 Correct: Your answer is correct. seenKey

0
Part 6 of 7
We now have the x-coordinate of the point where the derivative is zero. To find the y-coordinate, substitute zero for x in the original function.
y = 3x2 + 6
 = 3 · 0 Correct: Your answer is correct. seenKey

0

 2 + 6
 = 6 Correct: Your answer is correct. seenKey

6
Part 7 of 7
We now have both the x- and y-coordinate for the point where the tangent line is horizontal to our function.
The point is
(x, y) = (0,6 Correct: Your answer is correct. seenKey

0, 6

 ).
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4. 10/10 points  |  Previous Answers LarCalc9 2.2.068.MI.SA. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8 9 10
1/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1
1/50 1/50 1/50 1/50 1/50 1/50 1/50 1/50 1/50 1/50
Total
10/10
 
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Tutorial Exercise
Find k such that the line is tangent to the graph of the function.
Function Line
f(x) = k√x y = 4x + 16
Part 1 of 6
We want a value of k that will make the line y tangent to the function f(x). In order for y to be tangent to f(x), then y will touch f(x) at a point. At that point, the two equations are equal.
f(x) = y
k
x
 = 4 Correct: Your answer is correct. seenKey

4

 x + 16 Correct: Your answer is correct. seenKey

16
Part 2 of 6
If y is tangent to f(x) at the point x, the slope of y must equal the derivative of f(x). To determine this we simply take the derivative of each side of the equation.
k
x
 = 4x + 16
 
d
dx
$$k√x
Correct: Your answer is correct. ksqrt(x)
 = 
d
dx
[4x + 16]
Part 3 of 6
To find the derivative of the left side of the equation, rewrite the square root using exponents and use the Power Rule. Use the Sum Rule and the Constant Multiple Rule to solve the right side.
 
d
dx
[k
x
]
=
 
d
dx
[4x + 16]
 
d
dx
kx1/2 Correct: Your answer is correct. seenKey

1/2
=
 
d
dx
 4x + 
d
dx
 16
$$k2√x​
Correct: Your answer is correct. k/2x^(-1/2) 		= 4
Part 4 of 6
Now we solve for k.
k
2
x−1/2		 = 4
 
k
2
x
 = 4
k = 
$$8√x
Correct: Your answer is correct. 8sqrt(x)
Part 5 of 6
We now have a value for k, but it is in terms of x. To determine x we can substitute this value of k into our original equality, and solve the equation for x.
k
x
 
 		=
4x + 16
8
x
(
x
)
=
4x + 16
$$8x
Correct: Your answer is correct. 8x 		=
4x + 16
$$4x
Correct: Your answer is correct. 4x 		= 16
x = 4 Correct: Your answer is correct. seenKey

4
Part 6 of 6
Now that we know x, we can determine k.
k = 8
x
 
k = 8
4
 
k = 16 Correct: Your answer is correct. seenKey

16
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5. 1/1 points  |  Previous Answers LarCalc9 2.2.112. My Notes
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1
1/1
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1/1
 
Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between the object's temperature T and the temperature Ta of the surrounding medium. Write an equation for this law. (Use k if you need to use a constant.)
dT  =
k(T−Ta​)
Correct: Your answer is correct.
dt
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6. 1/1 points  |  Previous Answers LarCalc9 2.3.005.MI. My Notes
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1/1
3/50
Total
1/1
 
Use the Product Rule to differentiate the function.
f(x) = x7 cos(x)
f '(x) =
7x6cos(x)−x7sin(x)
Correct: Your answer is correct.

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7. 1/1 points  |  Previous Answers LarCalc9 2.3.088. My Notes
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1/1
3/50
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1/1
 
Newton's Law of Universal Gravitation states that the force F between two masses, m1 and m2, is given below, where G is a constant and d is the distance between the masses. Find an equation that gives an instantaneous rate of change of F with respect to d. (Assume m1 and m2 represent moving points.)
F =  Gm1m2
d2
F' =
−2Gm1​m2​d3​
Correct: Your answer is correct.
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8. 1/1 points  |  Previous Answers LarCalc9 2.5.004.MI. My Notes
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1/1
 
Find dy/dx by implicit differentiation.
x9 + y6 = 77
dy/dx =
−9x86y5​
Correct: Your answer is correct.

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9. 7/7 points  |  Previous Answers LarCalc9 2.6.019.MI.SA. My Notes
Question Part
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1 2 3 4 5 6 7
1/1 1/1 1/1 1/1 1/1 1/1 1/1
1/50 1/50 1/50 1/50 5/50 1/50 1/50
Total
7/7
 
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Tutorial Exercise
All edges of a cube are expanding at a rate of 3 centimeters per second.
(a) How fast is the surface area changing when each edge is 3 centimeters?

(b) How fast is the surface area changing when each edge is 12 centimeters?
Part 1 of 5
The first piece we need to know in order to solve this problem is how to calculate the surface area of a cube in terms of the length of an edge. A cube is composed of six identical square sides. The surface area of a square whose side has length x is
x2 Correct: Your answer is correct. seenKey

2

 .
Part 2 of 5
If x2 is the area of one side, then the surface area of all six sides combined and thus the cube is
S = 6 Correct: Your answer is correct. seenKey

6

 x2.
Part 3 of 5
The problem tells us that each edge is changing with respect to time at a rate of 3 cm/s. We can express this rate of change as the derivative of x (the length of the edge) with respect to t (time). Thus,
 
dx
dt
 = 3 Correct: Your answer is correct. seenKey

3

 cm/s.
Part 4 of 5
Using the two pieces of information we now have, we can determine how the surface area is changing with respect to time. Take the derivative of both sides of the area equation with respect to t.
S =
6x2
 
dS
dt
 
 		=
 
d
dt
[6x2]
= 12x Correct: Your answer is correct. seenKey

12x

dx
dt
 
= 36x Correct: Your answer is correct. seenKey

36x

Part 5 of 5
For parts (a) and (b), we simply need to substitute the lengths of the sides for x.
(a)    
dS
dt
 = 36x =		36(3)= 108 Correct: Your answer is correct. seenKey

108

 cm2/sec
(b)    
dS
dt
 = 36x =		36(12)= 432 Correct: Your answer is correct. seenKey

432

 cm2/sec
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10. 1/1 points  |  Previous Answers LarCalc9 2.6.030. My Notes
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1/1
 
An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10), the radar detects that the distance s is changing at a rate of 280 miles per hour. What is the speed of the airplane?
Correct: Your answer is correct. mph
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