Maths | Two Point Line formula for derivative Solution

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MATHS
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1MATHS
A fish population grew according to the following quadratic model: the number of fish on
day t is given by P(t)=800t-t2.
(i) Sketch this model for 0<t<800.
Solution: find the two point line formula for the derivative of the P(t) then plot the graph.
P‘(t) = 800−¿2t
For 0<t<800
Let, P’(t) = y and t = x
Now,
=> y = 800-2x
For x=0, y will be 800
Similar for x = 800, y will be -800
Hence the coordinates, (x1,y1) = (0,800) and (x2,y2) = (800,-800)
Hence, the Graph of the model will be:

2MATHS
(ii) What is the average growth rate between t=400 and t=600?
Solution: The Average growth rate in population is total change of population divided by total
change in time.
R= ∆ P
∆ t = P ( t2 ) −P ( t1 )
t2−t1
ATQ,
t2=600 and t1=400
Hence, P ( t2 )= 800×600 − (600)2
=> P ( t2 ) = 480000 – 360000 = 120000

3MATHS
Similarly,
P ( t1 )= 800×400 − (400)2
=> P ( t1 ) = 320000 – 160000 = 160000
Now, R = 120000−160000
600−200 = −40000
200 = -200 Ans.
(iii) What is the initial growth rate at t=0?
Solution: The growth rate at t=0 will be the first order derivative of the P(t) at t=600.
ATQ,
P(t) = 800t – t2
Differentiating both sides w.r.t x.
=> P' ( t )=800−2 t
Hence at t= 0,
=> P' ( t )=800−2 ×0=800 Ans .(which represents that population was increasing at 800 per day)
(iv) What is the growth rate at t=600?
Solution: The growth rate at t=600 will be the first order derivative of the P(t) at t=600.
ATQ,
P(t) = 800t – t2
Differentiating both sides w.r.t x.

4MATHS
=> P' ( t )=800−2 t
Hence at t= 600,
=> P' ( t )=800−2 ×600=−400 Ans .(which represents the decreasing in population)
(v) What is the maximum size that the population grows to?
Solution: for maximum growth rate the first order derivative of the P(t) will be 0.
Hence,
P(t) = 800t – t2
ATQ,
=> P’(t) = 800 – 2t = 0.
=> 800 = 2t
=> t = 800/2 = 400 ans.
∴ Hence, for t=400 the population will grow at max size.