Maths A level Calculus problem?
Would really appreciate it if someone can help me with this question:
The volume of liquid in a tank is given by V=Sqrtroot(2t+t^2) also written (2t+t^2)^1/2 where t is measured in seconds and V is measured in m^3.
I) Write down the volume at time t=0
ii) Find the rate at which the volume is increasing when t=1 second
iii) Given that the capacity of the tank is 10m^3m find how long it takes to fill the tank.
Would really appreciate any help on this.
Thanks
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Answers & Comments
We recommend learning methods as answers only work once.
i) When you substitute t = 0, V = 0
ii) Rate means dV/dt so you must differentiate V.
[Note:Here is a reminder of the the chain rule.
http://www.mathcentre.ac.uk/resources/uploaded/mc-...
In simple terms, if y = Outer function f of [inner function g(t)] then
dy/dt = derivative of outer function * derivative of inner function.
For example to differentiate y = (t^3 + 5t^2)^7
dy/dt = 7(t^3 + 5t^2)^6 * (3t^2 + 10t)
It is a common mistake to miss out multipling by the inner function.]
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Volume V = √(2t + t^2) = (t^2 + 2t)^1/2 where t is secs and V is m^3.
Volume rate = dV/dt = 1/2(t^2 + 2t)^(-1/2) * (2t + 2) = (t + 1)/√(t^2 + 2t)
To find the rate at which the volume is increasing when t = 1 second, substitute t = 1
Rate = (1 + 1)/√(1^2 + 2*1) = 2/√3 ~ 1.1547 m^3/sec
iii) The rate is not constant; it is a function of time, so time to fill is NOT 10 m^3/1.1547 m^3/sec
Instead we have to solve volume √(2t + t^2) = 10
t^2 + 2t - 100 = 0 use quadratic formula and select positive value
t = √(101) - 1 ~ 9.0498 secs
Regards - Ian H