Querry 05

#$&*

course Mth 272

7/18 around 7 pm

005. `query 5

********************************************* Question: `q 5.1.5 (previously 5.1.12) integrate 3 t^4 dt and check by differentiation YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: antiD of T^4 is T^5 then divide by the power of the antiD 3T^5/5 3/5 t^5 + c Y = 3/5T^5 Y’ = 3/5*5T^4 = 3 T^4 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: `a An antiderivative of the power function t^4 is one power higher so it will be a multiple of t^5. Since the derivative of t^5 is 5 t^4 an antiderivative of t^4 is be t^5 / 5. By the constant rule the antiderivative of 3 t^4 is therefore 3 * t^5 / 5. Adding the arbitrary integration constant we end up with general antiderivative3 t^5 / 5 + c. The derivative of 3/5 t^5 is 3/5 * 5 t^4 = 3 t^4), verifying our antiderivative. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3

********************************************* Question: `q 5.1.7 (previously 5.1.20 (was 5.1.18)) integrate v^-.5 dv and check by differentiation YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: AntiD of V^-.5 V^.5 then divide by the power. V^(.5)/.5 = 2v^.5 +c 2v^.5 Y’ = 2 * .5 v^-.5 Y’ = v^-.5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: `a An antiderivative of this power function is a constant multiple of the power function which is one power higher. The power of the present function is -.5 or -1/2; one power higher is +.5 or 1/2. So you will have a multiple of v^.5. Since the derivative of v^.5 is .5 v^-.5 an antiderivative will be v^.5 / .5 = v^(1/2) / (1/2) = 2 v^(1/2). Adding the arbitrary integration constant we end up with general antiderivative 2 v^(1/2) + c. The derivative of 2 v^(1/2) is 2 * (1/2) v^(-1/2) = v^(-1/2), verifying our antiderivative. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3

********************************************* Question: `q Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:

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Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

"

Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!

`gr31

Querry 05

#$&*

course Mth 272

7/18 around 7 pm

005. `query 5

*********************************************

Question: `q 5.1.5 (previously 5.1.12) integrate 3 t^4 dt and check by differentiation

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

antiD of T^4 is T^5 then divide by the power of the antiD

3T^5/5

3/5 t^5 + c

Y = 3/5T^5

Y’ = 3/5*5T^4 = 3 T^4

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a An antiderivative of the power function t^4 is one power higher so it will be a multiple of t^5. Since the derivative of t^5 is 5 t^4 an antiderivative of t^4 is be t^5 / 5. By the constant rule the antiderivative of 3 t^4 is therefore 3 * t^5 / 5. Adding the arbitrary integration constant we end up with general antiderivative3 t^5 / 5 + c.

The derivative of 3/5 t^5 is 3/5 * 5 t^4 = 3 t^4), verifying our antiderivative. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

ok

------------------------------------------------

Self-critique Rating: 3

*********************************************

Question: `q 5.1.7 (previously 5.1.20 (was 5.1.18)) integrate v^-.5 dv and check by differentiation

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

AntiD of V^-.5

V^.5 then divide by the power.

V^(.5)/.5 = 2v^.5 +c

2v^.5

Y’ = 2 * .5 v^-.5

Y’ = v^-.5

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a An antiderivative of this power function is a constant multiple of the power function which is one power higher. The power of the present function is -.5 or -1/2; one power higher is +.5 or 1/2. So you will have a multiple of v^.5. Since the derivative of v^.5 is .5 v^-.5 an antiderivative will be v^.5 / .5 = v^(1/2) / (1/2) = 2 v^(1/2). Adding the arbitrary integration constant we end up with general antiderivative 2 v^(1/2) + c.

The derivative of 2 v^(1/2) is 2 * (1/2) v^(-1/2) = v^(-1/2), verifying our antiderivative. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

ok

------------------------------------------------

Self-critique Rating:3

*********************************************

Question: `q Add comments on any surprises or insights you experienced as a result of this assignment.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

&#Good work. Let me know if you have questions. &#