If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# 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.
• 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.
• Around , Sal uses R(x) - D(x) in the integral rather than R(t)-D(t). Can someone please explain?
(1 vote)
• As he explains, he as used the variable t for the upper bondary of the integral (W is expressed as a function of time so t is a common choice here), so another variable name has to be chosen for the integration. He's using x, for it's the most conventional choice, but any other letter could do the job.