[Technology Exercise]

Draining a tank It take...

Question

[Technology Exercise]

Draining a tank It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth y of fluid in the tank t hours after the valve is opened is given by the formula

y = 6(1 - t/12)² m.

b. When is the fluid level in the tank falling fastest? Slowest? What are the values of dy/dt at these times?

Accepted Answer

Welcome back, everyone. In this problem, a spherical balloon is being deflated for 45 seconds. The volume V in cubic inches of the balloon, 2 seconds after deflation starts is given by a V equals 4/3 of pi multiplied by 3 minus 15th of T cubed for a T greater than or equal to 0 but less than or equal to 45. Determine the time at which the volume is decreasing fastest. What is the value of DVDT at this time? Now, to find when the volume is decreasing fastest, we basically need to find the minimum value of DVDT or the maximum magnitude of the rate since it will be negative. So first, let's differentiate V or volume with respect to time T, and then we'll try to figure out what that or at what point it will be decreasing fastest. Now notice that V is a composite function, so we can differentiate it by using the chain rule. So given the Equals 4/3 of pi multiplied by 3 minus 15th of T or cubed, OK. Then by the chain rule. To find its derivative. First, we're going to differentiate the alter function, so we write back our constant 4/3 of pi, multiplied by 3, multiplied by 3 minus 15th of T raises to the power of 3 minus 1. OK. So we've differentiated our other function. Multiplied by the derivative of our inner function that is 3 minus 15th of t. Now if we look here, notice that uh 3 can 3, OK, so we can have this as 4 pi. Multiplied by 3 minus 15th of t squared multiplied by the derivative of 3 minus 15th of t which would be negative 1/15. And now when we simplify that. Our rate of change of volume with respect to time will be -4 pi divided by 15 multiplied by 3 minus 15th of T squared. Now where do we go from here? Now that we found the derivative. Well, no, we need to find when our derivative is at its minimum value to find when the volume is decreasing fastest. And now, since our derivative or rate of change is negative, which means the volume is decreasing, the minimum value, that is the most negative will occur when our squared term is at its maximum. OK, so. And that's because the expression inside the parenthesis decreases from T equals 0 to T equals 45. So during that window, when will it be maximum? Well, of course, because it's decreasing, that value is going to be maximum at T equals 0, OK? Because the larger T gets, then 3 is going to be subtracting a larger value and thus the result inside the parenthesis is going to be smaller, OK? So. We can say then that 3. Mine, let me put that in red actually. Our expression inside a parenthesis 3 minus 15th of t. Squared will be maximum. And the volume is decreasing fastest, OK. Fastest. At T equals 0, OK? Smallest possible value of our quotient here. So now what is going to be our rate of change at this time? Well, at T equals 0, then our rate of change of volume with respect to time, OK. It's gonna be equal to -4/15. Multiplied by 3 minus 0 divided by 15 square, which would just be 3 squared. OK. That's gonna be negative 4 by 15th multiplied by 9. OK. And now, if we do some transpos if we do some canceling here, 3 goes into 93 times, 3 into 15, 5 times, -4 pi times 3 is -12 pi, and all of that will be divided by 5. I know if we remember our inches because it's our our units because it's changing, the volume is changing with respect to time. It's gonna be cubic inches per second, OK? Thus, the volume is decreasing fastest at T equals 0. At a rate of negative 12/5 of cubic inches per second. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Derivative

Critical Points

Second Derivative Test

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