More Applications of Definite Integrals 333Answer: Total number of gallons leaked=∫ 1005 e−^0.^1 tdt.- How much money should Mary invest at 7.5% interest a year compounded
continuously so that she will have $100,000 after 20 years.
Answer: y(t)=y 0 ekt,k= 0 .075, andt= 20 .y(20)=100, 000=y 0 e(0.075)(20). Thus,
using a calculator, you obtainy 0 ≈22313, or $22,313. - Given
dy
dx
=
x
y
andy(1)=0, solve the differential equation.Answer: y d y=xdx⇒∫
ydy=∫
xdx⇒
y^2
2=
x^2
2
+c⇒ 0 =1
2
+c⇒c=−1
2
Thus,
y^2
2=
x^2
2−
1
2
ory^2 =x^2 −1.- Identify the differential equation for the
slope field shown.
Answer: The slope field suggests a hyperbola
of the formy^2 −x^2 =k,so2y
dy
dx
− 2 x= 0and
dy
dx=
x
y.
- Find the solution of the initial value problem
dP
dt
=. 75 P
(
1 −P
2500
)
withP 0 =10.Answer:
dP
dt=. 75 P
(
1 −P
2500
)
⇒2500
P(2500−P)
dP=. 75 dt⇒
∫
2500
P(2500−P)
dP=∫. 75 dt⇒
∫ (
1
P+
1
2500 −P
)
dP=∫. 75 dt
⇒ln|P|−ln∣∣
2500 −P∣∣
=. 75 t+C 1 ⇒ln∣
∣∣
∣P
2500 −P
∣
∣∣
∣=.^75 t+C^1⇒P
2500 −P
=C 2 e.^75 t⇒P(t)=2500 C 2 e.^75 t
1 +C 2 e.^75 t⇒P(0)=
2500 C 2
1 +C 2
= 10
⇒ 2500 C 2 = 10 + 10 C 2 ⇒ 2490 C 2 = 10 ⇒C 2 =
1
249
.
Therefore,P(t)=
2500 (1/249)e.^75 t
1 +( 1 / 249 )e.^75 t=
2500 e.^75 t
249 +e.^75 t
soP(t)=2500
249 e−.^75 t+ 1