on a day when the air temperature is 34 degrees celsius, a can of soft drink is taken from the fridge at a temperature of 4 degrees celsius at 2pm. Thereafter the temperature of the can satisfies the differential equation; dT/dt=0.1(34-T), where t is the number of minutes after 2pm. What is the temperature of the can at 2:15pm? At what time does the temperature reach 30 degrees?

dT/dt=0.1(34-T)

dT/(34-T) = 0.1dt
Integrate both sides
-ln(34-T)=0.1t+c
34-T=Ce^(-0.1t)
T=34-Ce^(-0.1t)
Initial conditions:
t=0, T=4 => 4=34-C => C=30
so
T=34-30e^(0.1t)
A) Substitute t=15 to solve for T.
B) Substitute T=30 to solve for t.

wow thank you so much

You're welcome!

To solve the differential equation dT/dt = 0.1(34 - T), we can use separation of variables method.

Step 1: Separate the variables.
dT/(34 - T) = 0.1 dt

Step 2: Integrate both sides.
∫(1/(34 - T)) dT = ∫0.1 dt

Step 3: Simplify and evaluate the integrals.
ln|34 - T| = 0.1t + C1
ln|34 - T| = 0.1t + C2 (Combining constants)

Step 4: Raise both sides as exponents.
|34 - T| = e^(0.1t + C2)
|34 - T| = e^0.1t * e^C2
|34 - T| = Ce^0.1t (Where C = e^C2)

Step 5: Perform case analysis.
Case 1: (34 - T) > 0
34 - T = Ce^0.1t

Case 2: (34 - T) < 0
T - 34 = Ce^0.1t

Step 6: Solve for T.

Now, let's find the temperature of the can at 2:15 pm (15 minutes after 2 pm).

Using t = 15, we will evaluate the temperature T at that time.

Case 1: (34 - T) > 0
34 - T = Ce^0.1(15)

Solving for C:
4 (initial temperature) - C * e^(0.1 * 0) = C
4 - C = C
2C = 4
C = 2

Substituting C = 2 back into the equation:
34 - T = 2e^0.1(15)

34 - T = 2e^1.5

Solving for T:
T = 34 - 2e^1.5

Thus, the temperature of the can at 2:15 pm is 34 - 2e^1.5 degrees Celsius.

To find the time it takes for the temperature to reach 30 degrees, we need to find t when T = 30.

Case 1: (34 - T) > 0
34 - T = Ce^0.1t

Substituting T = 30 and solving for t:
34 - 30 = Ce^0.1t

4 = Ce^0.1t

Using C = 2 (from before):

4 = 2e^0.1t

2 = e^0.1t

Taking the natural logarithm of both sides:

ln(2) = 0.1t

Solving for t:

t = ln(2)/0.1

Thus, the temperature reaches 30 degrees approximately ln(2)/0.1 minutes after 2 pm.