2. Q
• Solution-
As given in the problem ‘m’ be the mass of salt at time t
We have the balance equation i.e.
𝑑𝑚
𝑑𝑡
= 𝐼𝑛𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 − 𝑂𝑢𝑡𝑓𝑙𝑜𝑤 rate
𝑑𝑚
𝑑𝑡
= 𝐶 𝑠 𝑖𝑛 𝑖𝑛𝑓𝑙𝑜𝑤 𝑤𝑎𝑡𝑒𝑟 ∗ 𝑅𝑎𝑡𝑒 𝑜𝑓 𝐼𝑛𝑓𝑙𝑜𝑤 − 𝐶 𝑠 𝑖𝑛 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝑤𝑎𝑡𝑒𝑟 ∗ 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑜𝑢𝑡𝑓𝑙𝑜𝑤
Where 𝐶 𝑠 is concentration of salt.
𝑑𝑚
𝑑𝑡
= 0 ∗ 10 −
𝑚
1000
∗ 10
𝑑𝑚
𝑑𝑡
= −
𝑚
100
i.e. option (b)
Outflow
Inflow
3. Solution-As given in the problem ‘m’ be the mass of salt at time t
We have the balance equation i.e.
𝑑𝑚
𝑑𝑡
= 𝐼𝑛𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 − 𝑂𝑢𝑡𝑓𝑙𝑜𝑤 rate
𝑑𝑚
𝑑𝑡
= 𝐶 𝑠 𝑖𝑛 𝑖𝑛𝑓𝑙𝑜𝑤 𝑤𝑎𝑡𝑒𝑟 ∗ 𝑅𝑎𝑡𝑒 𝑜𝑓 𝐼𝑛𝑓𝑙𝑜𝑤 − 𝐶 𝑠 𝑖𝑛 𝑜𝑢𝑡𝑓𝑙𝑜𝑤 𝑤𝑎𝑡𝑒𝑟 ∗ 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑜𝑢𝑡𝑓𝑙𝑜𝑤
Where 𝐶 𝑠 is concentration of salt.
𝑑𝑚
𝑑𝑡
= 0.05 ∗ 5 −
𝑚
1000
∗ 5
𝑑𝑚
𝑑𝑡
=
50−𝑚
200
i.e. option (c)
4. • Mixing Problems-
1. The tank contains 1000 gal of water in which initially 100 lb of salt is dissolved. Brine runs in at a rate of 10 gal/
min, and each gallon contains 5 lb of dissolved salt. The mixture in the tank is kept uniform by stirring. Brine runs
out at 10 gal /min. Find the amount of salt in the tank at any time t.
2. A tank contains 400 gal of brine in which 100 lb of salt are dissolved. Fresh water runs into the tank at a rate of
2gal/min, The mixture, kept practically uniform by stirring, runs out at the same rate. How much salt will there be
in the tank at the end of 1 hour?
• Exponential growth.
If the growth rate of the number of bacteria at any time t is proportional to the number present at t and doubles in 1
week, how many bacteria can be expected after 2 weeks? After 4 weeks?
• Exponential growth.
If the growth rate of the number of bacteria at any time t is proportional to the number
present at t and doubles in 1 week, how many bacteria can be expected after 2 weeks? After 4 weeks?
5.
6. Solution –
1. Yes
2. Yes
3. No
4. Sketch
5. For long term it always tend to the carrying capacity M=200
6. Q6
7. Point of inflection can be obtain by
𝑑2 𝑦
𝑑𝑥2 = 0 and differentiating the given problem we will get
P= 100
8. The point of inflection is half of the carrying capacity
