I also need a help with this question (from 2019 NHT exam 1)
7. given that 
find 
at point (1, 1).
Is finding the double derivative by implicit differentiation even part of the course? I've never seen it before...
What I did was implicitly take the first derivative, rearrange for dy/dx in terms of x and y, then take the derivative of both sides again, using chain-rule to simplify all the y terms. I ended up with a dy/dx again, but since I'd already found an expression for it in the previous step, it simplified nicely when I substituted in (1,1). I don't think double derivative by implicit differentiation is explicitly mentioned on the study design, but since it's just applying the chain rule I guess they can do it? Then again, that's true of implicit differentiation itself, but obviously it would be ridiculous to expect you to figure that out based on the chain rule in an exam, so idk. They never put much effort into NHT.
Your first question isn't related rates. You need to find the rate at which salt is flowing in - the rate at which salt is flowing out. The rate at which it's flowing in is 0, because only pure water is flowing in. The rate at which it's flowing out is (rate of all water flowing out)*(concentration of salt in the mix at any point). The question gives you the rate at which water flows out, it's 3 litres per minute. The concentration of salt in the mix at any point is given by (amount of salt)/(amount of water total). The amount of salt is Q, and the amount of water is 16 + 5t - 3t = 16 - 2t, because you start with 16 litres, it flows in at 5l/m and out at 3l/m. So rate at which salt is flowing in - the rate at which salt is flowing out = 0 - 3*(Q/(16 - 2t)) = -3/(16-2t). Pretty sure this would get full marks:
Rate in: 0
Rate out: 3Q/(16+5t-3t) = 3Q/(15+2t)
Therefore dQ/dt = 0 - 3Q/(15+2t) = -3Q/(15+2t)
I know we've already done the exam now, but since it didn't come up on this exam it might be a "show that" question on exam 2, so I'd give it a review.
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