Asked by ricky
At a summer campfire, the radius of a marshmallow on a stick expands at the rate of
\ {r ' (t)} = \frac{2.1 }{1 t + 5} mm/s
where t is the time of heating in seconds. Initially the radius was 3.8 mm. Find the radius after 27 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 27 seconds is
mm
\ {r ' (t)} = \frac{2.1 }{1 t + 5} mm/s
where t is the time of heating in seconds. Initially the radius was 3.8 mm. Find the radius after 27 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 27 seconds is
mm
Answers
Answered by
Steve
r' = 2.1/(t+5)
r = 2.1 log(t+5) + C
at t=0,
3.8 = 2.1 log5 + C
C = 3.8-2.1log5 = 0.42
so,
r = 2.1 log(t+5)+0.42
now just plug in t=27
r = 2.1 log(t+5) + C
at t=0,
3.8 = 2.1 log5 + C
C = 3.8-2.1log5 = 0.42
so,
r = 2.1 log(t+5)+0.42
now just plug in t=27
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.