QUERY 21

course MTH 271

11/10, 1430

assignment #021021. `query 21

Applied Calculus I

11-08-2009

......!!!!!!!!...................................

23:47:58

**** Query 2.8.4 dy/dt for (3,4) with x'=8; dx/dt for (4,3) with y'=-2 **** What is dy/dt if x=3, y=4 and dx/dt = 8?

......!!!!!!!!...................................

RESPONSE -->

Original equation: x^2 + y^2 = 25

d/dt[x^2] + d/dt[y^2] = d/dt[25]

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

x=3, y=4, dx/dt=8

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 = (-48/8) = -6

x=4, y=3, dy/dt= -2

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

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: 2

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

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

......!!!!!!!!...................................

23:48:11

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. **

......!!!!!!!!...................................

RESPONSE -->

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

Self-critique Rating:ent: 3

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

......!!!!!!!!...................................

23:50:58

**** 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?

......!!!!!!!!...................................

RESPONSE -->

Sphere: V= (4/3) pi r^3

dr/dt = 2 in/min

V= 4/3 pi r^3

dV/dt= 4 pi r^2 dr/dt

r=6

dV/dt = 4 pi (6^2) (2)

dV/dt = 288 pi in/min

r=24

dV/dt= 4 pi (24^2) (2)

dV/dt= 4608 pi in/min

confidence rating: 2

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

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

......!!!!!!!!...................................

23:53:46

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.. **

......!!!!!!!!...................................

RESPONSE -->

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

Self-critique Rating:ent: 3

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

"

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