Exam 2 Review Guide for MTH 141 Note: As SI leaders we do not

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Exam 2 Review Guide for MTH 141

Note: As SI leaders we do not see the exam nor know what is on the exam. This is just a guide for practice. Other sources of practice include the course guide, course website, problems from the book, math walk in tutoring, AEC tutoring, Khan Academy,

Mathematica and professor office hours. Good Luck!

Problem 1. Using Implicit Differentiation, find the following:

(π‘Ž) 𝑓𝑖𝑛𝑑 π’…π’š 𝒅𝒙

𝑔𝑖𝑣𝑒𝑛 3

π‘₯

2 𝑦

3

− π‘¦π‘π‘œπ‘ (π‘₯) = 5

(𝑏) 𝑓𝑖𝑛𝑑 𝒅𝒙 π’…π’š

𝑔𝑖𝑣𝑒𝑛 π‘π‘œπ‘ (π‘₯𝑦) = 3π‘₯

3

(𝑐) 𝑓𝑖𝑛𝑑 π’…π’ˆ 𝒅𝒛

𝑔𝑖𝑣𝑒𝑛 𝑧

4

− π‘π‘œπ‘ 

2 (𝑔) = 7𝑧 2 𝑔

Problem 2.

(Hint: need to use Implicit differentiation)

𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘‘β„Žπ‘’ 𝑙𝑖𝑛𝑒 π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘ π‘‘π‘œ π‘‘β„Žπ‘’ π‘π‘’π‘Ÿπ‘£π‘’ 3π‘₯ − 8𝑦

2

+ 11π‘₯

3 𝑦 = 3 π‘Žπ‘‘ π‘‘β„Žπ‘’ π‘π‘œπ‘–π‘›π‘‘ (1,0)

Problem 3.

A jug of water is dropped on the floor, creating a growing circular puddle. When the radius of the puddle is 150 cm, the radius is increasing at a rate of 0.1 cm per minute. At that moment, how fast is the area of the puddle expanding?

Problem 4.

A ladder 20 feet long leans against a house. If the bottom of the ladder slides away from the house horizontally at a rate of 4 ft/sec, how fast is the ladder sliding down the house when the top of the ladder is 8 feet from the ground?

Problem 5.

A tank of water in the shape of a cone pointing downwards is leaking water at a constant rate of 2𝑓𝑑

3

/β„Žπ‘œπ‘’π‘Ÿ . The base radius of the tank is 5 ft and the height of the tank is 14 ft. (Note: Volume of a

1 cone is equal to 𝑉 =

3 πœ‹π‘Ÿ 2 β„Ž )

(a) At what rate is the depth of the water in the tank changing when the depth of the water is 6 ft?

(b) At what rate is the radius of the top of the water in the tank changing when the depth of the water is 6 ft?

Problem 6.

𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘™π‘œπ‘π‘Žπ‘™ π‘™π‘–π‘›π‘’π‘Žπ‘Ÿ π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯π‘–π‘šπ‘Žπ‘‘π‘–π‘œπ‘› π‘œπ‘“ 𝑓(π‘₯) = π‘₯

3

π‘Žπ‘‘ π‘₯ = 1.1 𝑏𝑦 𝑒𝑠𝑖𝑛𝑔 π‘₯

0

= 1

Problem 7.

π‘ˆπ‘ π‘’ π‘™π‘œπ‘π‘Žπ‘™ π‘™π‘–π‘›π‘’π‘Ÿπ‘–π‘§π‘Žπ‘‘π‘–π‘œπ‘› π‘‘π‘œ π‘Žπ‘π‘π‘Ÿπ‘œπ‘₯π‘–π‘šπ‘Žπ‘‘π‘’ 𝑓(π‘₯) = (1 + π‘₯) 15 π‘Žπ‘‘ π‘₯ = 0.965. π‘ˆπ‘ π‘’ π‘₯

0

= 0

Problem 8.

πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ π‘“π‘œπ‘™π‘™π‘œπ‘€π‘–π‘›π‘” π‘™π‘–π‘šπ‘–π‘‘π‘ . π‘ˆπ‘ π‘’ 𝐿

𝐻ôπ‘π‘–π‘‘π‘Žπ‘™

′ 𝑠 𝑅𝑒𝑙𝑒 π‘€β„Žπ‘’π‘› π‘Žπ‘π‘π‘™π‘–π‘π‘Žπ‘π‘™π‘’.

(Hint: Look on page 66 of your course guide for all indeterminate forms)

(π‘Ž) lim

π‘₯→πœ‹

(𝑐) lim

π‘₯→∞

𝑠𝑖𝑛

2

π‘₯ π‘₯ − πœ‹

(𝑏) lim

π‘₯→0

π‘₯

2

− 3

2π‘₯ + 1 𝑒

3π‘₯

π‘₯

2

Problem 9.

πΈπ‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ π‘“π‘œπ‘™π‘™π‘œπ‘€π‘–π‘›π‘” π‘™π‘–π‘šπ‘–π‘‘π‘ . π‘†π‘‘π‘Žπ‘‘π‘’ π‘€β„Žπ‘’π‘‘β„Žπ‘’π‘Ÿ π‘œπ‘Ÿ π‘›π‘œπ‘‘ 𝐿

𝐻ôπ‘π‘–π‘‘π‘Žπ‘™

′ 𝑠 𝑅𝑒𝑙𝑒 π‘€π‘Žπ‘  π‘Žπ‘π‘π‘™π‘–π‘π‘Žπ‘π‘™π‘’.

(π‘Ž) lim

π‘₯→𝑒

𝑙𝑛π‘₯

2

− 2 π‘₯ − 𝑒

(𝑏) lim

π‘₯→0

𝑠𝑖𝑛π‘₯ − π‘₯

7π‘₯

3

(𝑐) lim

π‘₯→∞

4π‘₯

3

− 3π‘₯

5π‘₯

3

2

+ 7π‘₯ − 10

− 2π‘₯

2

+ 8

Problem 10.

In each part, use the graph y=f(x) in the accompanying figure to find the requested information.

(a) Find the intervals on which f is increasing

(b) Find the intervals on which f is decreasing

(c) Find the open intervals on which f is concave up

(d) Find the open intervals on which f is concave down

(e) Find all values of x at which f has an inflection point

Source: Calculus Early Transcendentals Single Variable 10 th

Edition

Anton Bivens Davis

Problem 11.

Find all critical points if 𝑓(π‘₯) = 4π‘₯

4

− 16

π‘₯

2 + 17

Problem 12.

Find all critical points if 𝑓(π‘₯) = π‘₯+1 π‘₯ 2 +3

Problem 13.

Find the absolute maximum and minimum values of

(π‘Ž) 𝑓(π‘₯) = 4

π‘₯

2

− 12π‘₯ + 10 π‘œπ‘› π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘£π‘Žπ‘™ [1,2]

(𝑏) 𝑓(π‘₯) = 8π‘₯ −

π‘₯

2 π‘œπ‘› π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘£π‘Žπ‘™ [0,6]

(𝑐) 𝑓(π‘₯) = (π‘₯ − 2) 3

π‘œπ‘› π‘‘β„Žπ‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘£π‘Žπ‘™ [1,4]

Problem 14.

Find the maximum volume of an open-topped square-based box that can be made out of

1200cm 2 of sheet metal.

Problem 15.

Find the minimum amount of sheet metal needed to create a closed-topped rectangularbased box (one side of the base is twice the length of the other) if the volume of the box must be 9000 in 3 .

(Remember: 𝑉 = 𝑙 βˆ™ 𝑀 βˆ™ β„Ž π‘Žπ‘›π‘‘ 𝑆𝐴 = 2(π‘€β„Ž + 𝑙𝑀 + π‘™β„Ž)

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