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Hey! So for this one you are using the chain rule. The result you want can be found as:
The first of those derivatives, \(\frac{dM}{dv}\), you can calculate by differentiating the expression you were given. The second derivative, \(\frac{dv}{dt}\), you are given! Then you can just substitute in \(0.6c\) for \(v\) into the result once you've multiplied them together!
Another question (Not sure how to do it):
Brine containing 2g of salt/L flows into a container initially filled with 50L of water containing 10g of salt. The brine enters the tank at 5 L/min, the concentration is kept uniform by stirring and the mixture flows out at the same rate. P(t) is the amount of salt (in grams) in the tank after t minutes. Find the amount of salt in the tank after 10 minutes.
Answer: 66.9g (1 d.p.)
This one you can also approach with the change rule, but I prefer to think of it intuitively. So we know that 5L of water flows into the tank per minute, and that each litre contains 2g of salt. So, the salt going into the tank is 10g/min. At this stage:
However, we also have some amount going out of the tank. Let's call this \(\frac{dA}{dt}\), just for reference. At any given time, we know that the amount of water leaving the tank is 5L per minute. The concentration of salt in the tank is always \(\frac{p(t)}{50}\) litres, however much salt there is divided by 50. So the amount of salt leaving the tank is:
Put that together with what we have at the top, and we have the
actual expression we want:
If we're clever, we notice that this matches the expression for exponential growth and/or decay. We can form this equation using our knowledge of that topic (let me know if you need this explained):
Now that we have it, just a matter of substituting \(t=10\):