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C3 rates of change question help

c3 q 2.jpg Hi guys, how do you do Q6 and Q8 using the chain rule?
Original post by Jellyxx
Hi guys, how do you do Q6 and Q8 using the chain rule?


Well Q8 is straight forward. You let u=2x+1u=2x+1 and the equation turns into y=u1/2+4u−1/2y=u^{1/2}+4u^{-1/2} then just chain rule it.

Which part of Q6 is confusing you?
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Jellyxx
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Original post by RDKGames
Well Q8 is straight forward. You let u=2x+1u=2x+1 and the equation turns into y=u1/2+4u−1/2y=u^{1/2}+4u^{-1/2} then just chain rule it.

Which part of Q6 is confusing you?


Thanks! I realise what you mean now about q8

For q6. I got part a) but I am confused on part b). Would 0.8 litres/s refer to the rate at which volume is changing? How does part a tie in with part b.
(edited 7 years ago)
Original post by Jellyxx
*Would 0.8 litres/s refer to the rate at which volume is changing?


yes.

the cross section is a trapezium with height h. you need to find an expression for the area of the trapezium and multiply by the length of the prism to get the volume of water.

then you can find dh/dt by saying

dh/dt = dh/dv x dv/dt*
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Jellyxx
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Original post by the bear
yes.

the cross section is a trapezium with height h. you need to find an expression for the area of the trapezium and multiply by the length of the prism to get the volume of water.

then you can find dh/dt by saying

dh/dt = dh/dv x dv/dt*


Thank you very much!

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