At a summer campfire, the radius of a marshmallow on a stick expands at the rate of
\ {r ' (t)} = \frac{2.4 }{3 t + 6} mm/s
where t is the time of heating in seconds. Initially the radius was 3.8 mm. Find the radius after 16 seconds using the following steps:
When substituting,
du =______ dt
The resulting integral looks like \int \frac{A}{u} \, du where A is a numerical value.
A = _______
The constant of integration, C is
The radius after 16 seconds is
_______mm
r' = 2.4/(3t+6)
r = 0.8 log(3t+6)+c
solve for c at t=0 in
0.8 log6 + c = 3.8
now just plug in t=16 for the final answer.
To find the radius after 16 seconds using the given information, we need to solve the differential equation and apply the initial condition.
Step 1: Substitution
The given rate of change of radius can be represented as:
r'(t) = 2.4 / (3t + 6) mm/s
To solve this, we need to perform a substitution. Let's substitute u = 3t + 6, which means du/dt = 3.
Now, we can rewrite r'(t) as:
r'(t) = 2.4 / u * (du/dt)
Step 2: Integral Form
Using the substitution, the equation can be written as:
r'(t) = 2.4 / u * 3
Integrating both sides with respect to u:
∫ r'(t) du = ∫ (2.4 / u) * 3 du
Step 3: Integration
The integral of 2.4 du is 2.4u, and the integral of (1/u) is ln|u|.
Thus, the resulting integral is:
2.4u * 3 + C = 7.2u + C
Step 4: Determining A and C
Now we need to determine the numerical value A and the constant of integration, C.
Comparing the resulting integral with the form ∫ A/u du, we can see that A = 7.2.
As for the constant of integration, C, we cannot determine its exact value without additional information about the initial conditions. So we leave it as C for now.
Step 5: Evaluating the radius after 16 seconds
To find the radius after 16 seconds, we substitute u = 3t + 6 = 3(16) + 6 = 54 into the integral.
Using A = 7.2, the integral becomes:
7.2u + C = 7.2(54) + C = 388.8 + C
Although we cannot determine the exact value of C, we can say that the radius after 16 seconds is approximately 388.8 + C mm, where C is the constant of integration.