Derivative and water flow

1. Jan 31, 2008

lgen0290

1. The problem statement, all variables and given/known data

Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 6.00t 0.8 - 3.35t + 23.00, with t 0, m in grams, and t in seconds.
(a) At what time is the water mass greatest?

(b) What is that greatest mass?

(c) In kilograms per minute, what is the rate of mass change at t = 2.00 s?

(d) In kilograms per minute, what is the rate of mass change at t = 5.00 s?
2. Relevant equations
The first derivative would be 4.8t^-.2-335

3. The attempt at a solution

a)I set the derivative equal to zero and figured t to be 2.3375, but it says thats wrong.
b)I assume I'd plug a into the original and try tht, but I can't get a.
c and d)I tried to put 2 and 5 into t as the origional, but they are not right.

2. Jan 31, 2008

blindside

a) is, as you said, the solution for t when

$$\frac{dm}{dt}=0.$$

However, the solution to

$$t^{0.2}=\left[\frac{4.8}{3.35}\right]$$

is not 2.3375, check your algebra.

b), as you said is $$m(t)$$ for the solution above
c) and d) can both be found by substituting the times into the expression for $$\frac{dm}{dt}.$$

3. Jan 31, 2008

lgen0290

Thanks. Where did the 3.35 come from? How would I solve t^-.2?

4. Feb 1, 2008

blindside

The 3.35 is from the original expression for m, unless you've mistyped it. I used logarithms and the change of base rule to solve for t.