open qa 21

#$&*

course Mth 271

Question: `q **** problem 1 7th edition Query 2.8.4 dy/dt for (3,4) with x'=8; dx/dt for (4,3) with y'=-2 ****

What are your solutions?

Your solution:

d/dy (x^2) + d/dt (y^2) = d/dt (25)

2x dx/dt + 2y dy/dt = 0

2 * 3 * 8 + 2 * 4 * dy/dt = 0

48 + 8 dy/dt = 0

8 dy/dt = -48

Dy/dt = -6

2 * 4 * dx/dt + 2 * 3 * -2 = 0

8 dx/dt - 12 = 0

8 dx/dt = 12

Dx/dt = 12/8 = 3/2

Confidence Rating: 3

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

Given Solution:

`a At (3,4) you are given dx/dt as x ' = 8.

Since 2x dx/dt + 2y dy/dt = 0 we have

2(3) * 8 + 2 * 4 dy/dt = 0 so

dy/dt = -48/8 = -6.

At (4,3) you are given dy/dt as y' = -2. So you get

2 * 4 dx/dt + 2 * 3 * -2 = 0 so

8 dx/dt - 12 = 0 and therefore

8 dx/dt = 12. Solving for dx/dt we get

dx/dt = 12/8 = 3/2. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `q **** problem 2 7th edition Query 2.8.6 roc of volume if r increases at rate 2 in/min, if r= 6 in and if r = 24 in **** What is the rate of volume change if r is increasing at 2 inches / minute?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Dr/dt = 2

1. r = 6

2. r = 24

V = 4/3 Pi r^3

d/dt V = d/dt (4/3 Pi r^3)

dV/dt = 4/3 Pi * 3r^2 * dr/dt

dV/dt = 4/3 Pi * 3 * 36 * 2

dV/dt = 288 Pi

r = 24

dV/dt = 4608 Pi

confidence rating #$&*:

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

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

Given Solution:

`a The shape is a sphere. The volume of a sphere, in terms of its radius, is

V = 4/3 `pi r^3.

Taking the derivative with respect to t, noting that r is the only variable, we obtain

dV/dt = ( 4 `pi r^2) dr/dt

You know that r increases at a rate of 2 in / min, which means that dr/dt = 2.

Plugging in dr/dt = 2 and r = 6 gives 4 pi (6^2) * 2 = 288 pi = 904 approx.

Plugging in dr/dt = 2 and r = 24 gives 4 pi (24^2) * 2 = 4 pi (576)(2) = 4608 pi = 14,476 approx.. **

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

Self-critique (if necessary):

"

Self-critique (if necessary):

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

Self-critique rating:

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

Question: `q **** problem 2 7th edition Query 2.8.6 roc of volume if r increases at rate 2 in/min, if r= 6 in and if r = 24 in **** What is the rate of volume change if r is increasing at 2 inches / minute?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Dr/dt = 2

1. r = 6

2. r = 24

V = 4/3 Pi r^3

d/dt V = d/dt (4/3 Pi r^3)

dV/dt = 4/3 Pi * 3r^2 * dr/dt

dV/dt = 4/3 Pi * 3 * 36 * 2

dV/dt = 288 Pi

r = 24

dV/dt = 4608 Pi

confidence rating #$&*:

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

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

Given Solution:

`a The shape is a sphere. The volume of a sphere, in terms of its radius, is

V = 4/3 `pi r^3.

Taking the derivative with respect to t, noting that r is the only variable, we obtain

dV/dt = ( 4 `pi r^2) dr/dt

You know that r increases at a rate of 2 in / min, which means that dr/dt = 2.

Plugging in dr/dt = 2 and r = 6 gives 4 pi (6^2) * 2 = 288 pi = 904 approx.

Plugging in dr/dt = 2 and r = 24 gives 4 pi (24^2) * 2 = 4 pi (576)(2) = 4608 pi = 14,476 approx.. **

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

Self-critique (if necessary):

"

Self-critique (if necessary):

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

Self-critique rating:

#*&!

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

Question: `q **** problem 2 7th edition Query 2.8.6 roc of volume if r increases at rate 2 in/min, if r= 6 in and if r = 24 in **** What is the rate of volume change if r is increasing at 2 inches / minute?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Dr/dt = 2

1. r = 6

2. r = 24

V = 4/3 Pi r^3

d/dt V = d/dt (4/3 Pi r^3)

dV/dt = 4/3 Pi * 3r^2 * dr/dt

dV/dt = 4/3 Pi * 3 * 36 * 2

dV/dt = 288 Pi

r = 24

dV/dt = 4608 Pi

confidence rating #$&*:

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

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

Given Solution:

`a The shape is a sphere. The volume of a sphere, in terms of its radius, is

V = 4/3 `pi r^3.

Taking the derivative with respect to t, noting that r is the only variable, we obtain

dV/dt = ( 4 `pi r^2) dr/dt

You know that r increases at a rate of 2 in / min, which means that dr/dt = 2.

Plugging in dr/dt = 2 and r = 6 gives 4 pi (6^2) * 2 = 288 pi = 904 approx.

Plugging in dr/dt = 2 and r = 24 gives 4 pi (24^2) * 2 = 4 pi (576)(2) = 4608 pi = 14,476 approx.. **

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

Self-critique (if necessary):

"

Self-critique (if necessary):

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

Self-critique rating:

#*&!#*&!

&#This looks very good. Let me know if you have any questions. &#