Maths A level Calculus problem?

May 2021 1 102 Report

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

Science & Mathematics / Mathematics

Ian H

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.

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.]

~~~~~~~~~~~~~~~~~~~~~~~~~~

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

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