Would you please check my answer to make sure I did it right?
Bus travel, a bus company finds that monthly costs for one particular year were given by C(t) = 100 + t^2 dollars after t months. After t months the company had P(t) =1000 +t^2 passengers per month. How fast is its cost per passenger changing after 6 months?
I did c and p prime
C'(t) = 2t
P'(t) = 2t
then i put it into the quotient form
(2t) (1000+ t^2) - (100 + t^2) (2t)/(1000 + t^2)^2
the answer i got was 0.010786021
Please tell me if this is not right. Thank you!
Since both C and P are monthly amounts, I will assume we are looking for the monthly cost per passenger. That would, of course, be
C/P = (100+t^2)/(1000+t^2)
As you say, (C/P)' = 1800t/(1000+t^2)^2
So, at t=6, (C/P)' = 0.01006246
Close, but I wonder what happened down in the details.
To find how fast the cost per passenger is changing after 6 months, we can use the derivative of the cost function with respect to time (t) divided by the derivative of the passenger function with respect to time (t).
Let's go through the solution step by step to check if your answer is correct:
1. The given functions are:
C(t) = 100 + t^2 (cost per month in dollars)
P(t) = 1000 + t^2 (passengers per month)
2. Take the derivatives of the functions:
C'(t) = 2t
P'(t) = 2t
It seems like you have correctly calculated the derivatives.
3. Set up the quotient:
C'(t) / P'(t) = (2t) / (1000 + t^2)
You have correctly set up the quotient.
4. Substitute t = 6 months (since we want to find the rate after 6 months):
C'(6) / P'(6) = (2 * 6) / (1000 + 6^2)
= 12 / (1000 + 36)
= 12 / 1036
≈ 0.01158 (rounded to five decimal places)
It seems like your answer is slightly different from the correct answer. The correct answer should be approximately 0.01158, not 0.01078.
So, the cost per passenger is changing at a rate of approximately 0.01158 dollars per passenger per month after 6 months.
Please note that the rate is given in a unit of dollars per passenger per month.