Complete solutions to Tyler’s Quiz questions

8/12/2019 Complete solutions to Tylers Quiz questions

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Complete solutions to Tylers

Quiz questions


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Q6

According to one theory of learning, thenumber of items, (), that a person canlearn after hours of instruction is given

by = 15 for 0 64. Find therate of learning at the end of eight hoursof instruction.

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Q6

As soon as you see the word rate inany word problem, you should bethinking slope.

Then the question becomes, Are welooking for an average rate (algebra) oran instantaneous rate (calculus)?

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Q6

If this were an algebra question, wewould usually see the word average inthe wording of the problem.

Since they are requesting the rate atone specific time (that is, we are onlygiven one point), this is an instantaneous

rate of change problem.

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Q6

We need calculus.

We are looking for the instantaneous

rate of change of()

when = 8

.

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Q6

= 15

= 15

= 15 23

= 151

23

= 10

=10

=

10

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Q6

= 8 =10

8 =

10

2 = 5

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Q8

Find the average rate of change for thefunction over the given interval.

= 4 2 + 2between

= 9and = 6.

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Q8

Average rate of change (ARC) is analgebra question.

Instantaneous rate of change (IRC) is acalculus question.

Rates of change are slopes of graphs.

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Q8

=

We need the -coordinates of the pointsat the beginning and end of the giveninterval.= = 9 = 4 9

2 9 + 2

= 4 729 2 81 + 2 = 3076 9, 3076

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Q8

= = 6 = 4 6 2 6 + 2

= 4 216 2 36 + 2 = 934 6, 934

=934 (3076)

6 (9) =

934 + 3076

6 + 9

=21423

= 714

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Q9

Find the slope of the secant line joining(2, (2))and (3, (3)) for () = 3 8.

We need calculus to find the slopes oftangent lines because the tangent lineonly touches the graph of , rather thanpassing through.

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Q9

In this problem, we are looking for theslope of a secant line.

Secant lines pass through curves atmore than one point, so this is analgebra question, not a calculus question.

=

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Q9

We need the -coordinates of the twopoints. 2 = 3 2 8 = 3 4 8 = 12 8

= 20 2, 20 3 = 3 3 8 = 3 9 8 = 27 8

= 35 3, 35

=35 (20)3 2 =35 + 20

1 = 15

1= 15

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Q 10

Find the equation of the tangent line at = 2for = 4 + 2 3. Writeanswer in the form = + .

Solution:

We need two things to find the equationof any line (1) the slope and (2) a point.

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Q 10

(1)To find the slope of a tangent line to agraph, we use calculus by finding thederivative of the function. = = 0 + 1 2 2 3 () = 1 4 9

To find at = 2, we plug in 2for

= 2 = 1 4 2 9 2 = 1 8 9 4 = 1 8 36 = 43

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Q 10

(2)To find the point, we plug = 2into()to get the coordinate. coordinate = 2 = 4 + 2 2 2

3 2 = 4 + 2 8 24 = 26 The point is (2, 26)

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Q 10

(2)To find the equation of the line, usethe formula from algebra

= ( )

where (, )is the point we found above.

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Q 10

26 = 43 2 + 26 = 43 + 86

+ 26 26 = 43 + 86 26

= 43 + 60

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Q 11

Find the equation of the tangent line tothe curve when has the given value.

() = 3 ; = 4

Solution:

We need two things to find the equationof any line (1) the slope and (2) a point.

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Q 11

(1)To find the slope of a tangent line to agraph, we use calculus by finding thederivative of the function. = = 0 2 = 2 = 2

To find at = 4, we plug in 4for

= (4) = 2(4) = 8

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Q 11

(2)To find the point, we plug = 4into()to get the coordinate. coordinate = (4) = 3 4 = 3 16 =

13 The point is (4, 13)

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Q 11

To find the equation of the line, use theformula from algebra

= ( )

where (, )is the point we found above.

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Q 11

13 = 8 4 + 13 = 8 + 32

+ 13 13 = 8 + 32 13

= 8 + 19

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Complete solutions to Tyler’s Quiz questions

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