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# 2015 AP Calculus AB/BC 1c

Minimum water in the pipe.

## Want to join the conversation?

• around how come the time interval where the amount of water is at a minimum happens to be where the two lines intersect? ,is it because at that time the area between the curves (representing the amount of water inside) is at a minimum?
• The points of minimum and maximum are the points where the derivative of the function is equal to 0 (0 raise over any run, or just a flat horizontal line). For more detailed description, I would advise to review "Optimization" in "Differential calculus" section on Khan Academy.
• When you are finding where there is a local minimum (about ), how do you know that the value t=3.272 is a local minimum and not a local maximum? When the derivative of a function is 0 the function can have either a max or min. Wouldn't you have to check to make sure it is indeed a min and not a max?
• I feel like that is a point of interest, because I found that a little bit strange as well. One thing to point out is that one item to look at would be in part b, where we found that at t = 3, the graph was decreasing. While this isn't to say that this is always 100 % accurate, you can look at the trend and see that when t = 3.272, the derivative = 0 and the graph was decreasing moments before it became 0, you can reason that the point at t will most likely be equal to the minimum point on the graph.
• Wouldn't it be easier to just create a new function:
N(t) = R(t) - D(t)
Graph N(t), find where N(t) equals zero (it equals 0 at t=0 and t=3.272). Then find which zero it changes from negative to positive at (t=3.272), then call that the answer? And reason that the endpoints t=0 and t=8 can be disregarded because the graph of N(t), which is the rate of change of the water, goes down is negative before hitting t=3.272 and then continues on positive till t=8?
• 1.How did you get 3.272?
2. How can you find w(3.272) with a TI-84 Plus calculator? The one used in this video has different features.
• At ish Khan says it could be either a maximum or a minimum but because its lower than both the endpoints its clearly a minimum, what if the limits of integration on the BC exam are like 0 to 45 then should we find all the points where the f'(x) graph reaches 0 and try all of them or is there a different method anyone knows about?
• Would it be possible to evaluate R(t) - D(t) = 0, as well as the integral of (sin(t^2/35)dt) without a calculator?
• I'm pretty sure it would be possible. But I don't know how.
• At , and during the entire rest of the problem, wouldn't it be easier simply to find the intercept between the two functions R(t), D(t)? Is it only a coincidence that it worked for this problem.
• I think your question needs a bit more elaboration. Wouldn't it be easier to find what exactly? There are 3 parts to this question and all are slightly different in what they're asking.