 When air expands diabetically, which is just a fancy way of saying that as it expands it's neither gaining or losing heat Then its pressure and volume are related to each other by the equation PV to the 1.4 power is equal to C where C is some constant Suppose that at a certain instant of time the volume is 400 cubic centimeters and the pressure is 80 KPa and is decreasing at a rate of 10 KPa per minute at what rate is the volume increasing at this time This is a related rates problem because we're given information about the rate of change of one Quantity and we want to know about its relationship to another quantity So what quantity are we trying to figure out here? What rate is the volume increasing at a specific time? So we're trying to figure out what is the change of volume with respect to time at this moment of time Well, what else do we what so that's what we need to know? What do we actually know about this problem? We know that the volume at that moment of time is going to be 400 cubic centimeters So V equals 400 cubic centimeters We know that the pressure at this moment of time is going to be 80 kilo Pascal's so P equals 80 kilo Pascal's and we also know that the rate in which the pressure is changing is It's decreasing 10 kilo Pascal's per minute So we get that Dp over Dt at this moment is negative 10 Kilo Pascal's per minute You notice I have put a negative 10 the negative 10 represents that the change of pressure is decreasing with respect to time So we're only make we've able to observe the given information and what we need to figure out Now we've reached what's typically the hardest part of a related rates problem is actually coming up with the equation that Relates together pressure and volume the two quantities in play here, right? We need to relate together the derivative of volume with the derivative of pressure So you find an equation that relates pressure and volume and then take the derivative implicitly But the good news is and this is what's so beautiful about this problem Is they're giving you the equation it just handed to you on a silver platter Wow, how amazing and that does happen sometimes that because of the scientific applications The equation that we have to relate the quantities is essentially already given to us We don't have to search for it that really removes the hardest part of this problem. Definitely We're now in the position where we need to take the derivative of both sides of this equation So we take the derivative of the left-hand side with respect to T We're gonna take the derivative of the right-hand side with respect to T As we take the derivative of the left-hand side because we have a product to two things in order to expand the derivative We're gonna use the product rule so we're gonna get the derivative pressure times volume to the 1.4 and Then we're gonna get pressure times the derivative of volume to the 1.4 like so on the right-hand side We have to take the derivative of the constant But notice they didn't give us the derivative of the constant and guess what don't know Don't care about that constant because its derivative is gonna equal zero. That's why I wasn't given to us Next we need to finish calculating the derivative of volume to the 1.4 Be aware that we are taking the derivative with respect to time not with respect to volume So the chain rule comes into play here. The derivative will look like 1.4 times V to the 0.4 power That's the outer derivative. We're gonna lower the power by 1 there and then we times that by the inner derivative V prime Now at this moment, we've now finished all the derivative calculations We're gonna plug in the information that we know into this into this equation here So the change of pressure remember with respect to time with a negative 10 kilo Pascal's per minute The volume was 400 cubic centimeters and I'm just gonna leave that right now as 400 to the 1.4 power I'm gonna practice what in computation theory is called lazy computation. It sounds like a bad thing I mean, how could lazy be good, but the idea is we're not gonna compute it until we actually need it It turns out we're gonna wait because later on if we just if we postpone the calculation We might find it easier to do it later than it is now. What's the hurry? There isn't one So then we're gonna plug in the pressure which we know that to be 80 kilo Pascal's then we have 1.4 times V to the 0.4 so again, we're gonna get 400 to the 0.4 power again I'm gonna be lazy in my computation. We're gonna times that by V prime, which we don't know So now when you look at this we got a bunch of numbers We could try to crunch them in our calculator and go from there But essentially we have to solve this linear equation for V prime which we don't know So we're gonna move this quantity to the other side of the equation It's we have a negative 10 times 400 to the 1.4. So we move to the other side of the equation It'll become positive. So we get positive 10 times 400 to the 1.4 power What's left on the left hand side notice that 80 could be factors 8 times 10 The reason that that advantage is is it times 10 by 1.4. You're actually get 14 So you get 8 times 14 That way you can avoid the fractions right 400 to the point for power and then times that by V prime We then need to divide both sides of the equation by this coefficient right here divide both sides by the 8 times 14 times 400 400 to the point 4 8 times 14 times 400 to the point 4 and so now we're gonna see that our lazy computation actually paid off Notice how we have a 400 to the 1.4 on the top. We have a 400 to the point 4 on the bottom This is an exponential expression We have the same base and since we're across the fraction bar We're gonna subtract the exponents there and so we end up with the change of volume to equal 10 times 400 to the first power over 8 times 14 you'll notice that at this moment We don't we just have to take form to the first power. That's pretty easy. That's just 400 So we don't have to compute any x what it actually canceled each other The next ones can't show each other out and so no irrational numbers were ever necessary because in the end the irrational numbers cancel out I would recommend this, you know having an algebraic mindset instead of a numerical mindset Not mindset can help us with these lazy computations And so if we continue on with this thing, we're gonna end up with Well, we simplify the fraction we're getting 250 over 7. Whoops over 7. Don't worry about the arithmetic That's not the point of this video 257 a 250 over 7 what this is gonna be cubic centimeters per minute Although an approximate solution is probably more advantage advantageous for us here So we're gonna get thirty five point seven one four and this will be centimeters cube per minute as Then that's gonna be how quickly the volume of this gas is increasing at this specific moment in time