A pond initially contains 1,000,000 gal of water and an unknown amount of an undesirable chemical. Water containing 0.01 g of this chemical per gallon flows into the pond at a rate of 300 gal/h. The mixture flows out at the same rate, so the amount of the water in the pond remains constant. Assume that the chemical is uniformly distributed throughout the pond. (a) Write a differential equation for the amount of chemical in the pond at any time.
(b) How much of the chemical will be in the pond after a very long time? Does this limiting amount depend on the amount that was present initially?

(A) Let A(t) denote the amount (in grams, g) of the chemical present in the pond at time t (in hours, h). The starting amount is unknown; call it a, measured in g, so that A(0) = a.

Water containing 0.01 g/gal of the chemical flows in at a rate of 300 gal/h, which increases the amount of the chemical in the pond at a rate of

(0.01 g/gal) • (300 gal/h) = 3 g/h

and flows out at the same rate, so the amount decreases by

(A(t)/1,000,000 g/gal) • (300 gal/h) = 3A(t)/10,000 g/h

Then the net rate of change of the amount of chemical in the pond is given by the ODE,

dA(t)/dt = 3 - 3/10,000 A(t)

(B) Solve the ODE for A(t). There are several ways to do that. For instance, itt's separable, so we have

dA(t) / (3 - 3/10,000 A(t)) = dt

Integrate both sides to get

-10,000/3 ln|3 - 3/10,000 A(t)| = t + C

Solve for A(t) :

ln|3 - 3/10,000 A(t)| = -3/10,000 t + C

3 - 3/10,000 A(t) = exp(-3/10,000 t + C )

3 - 3/10,000 A(t) = exp(-3/10,000 t ) • exp(C )

3 - 3/10,000 A(t) = C exp(-3/10,000 t )

3/10,000 A(t) = 3 - C exp(-3/10,000 t )

A(t) = 10,000 - C exp(-3/10,000 t )

Since A(0) = a, we have

a = 10,000 - C exp(-3/10,000•0)   →   C = 10,000 - a

→   A(t) = 10,000 - (10,000 - a) exp(-3/10,000 t )

As t grows to infinity, the exponential term will vanish, leaving a limiting amount of 10,000 g of the undesirable chemical in the pond, which does not depend on the original amount.

d 25 pi/6

step-by-step explanation:

the diameter is 10 inches

the radius is 1/2 of the diameter or 5 inches

the area of the circle is

a = pi r^2

  = pi 5^2

  = 25 pi

the slice is 60 degrees of the entire circle.

60 degrees out of 360   60/360 = 1/6

the area of the slice is 1/6 of the circle

a slice = 1/6 a circle

a slice = 1/6 * 25 pi

for this case we have that by definition, the surface area of a cylinder is given by:

where:

r: it is the radius of the cylinder

h: it is the height of the cylinder

we can also rewrite it as:

where:

la: it's the lateral area

ba: area of the base

option a