Calculus Assignment: Analyzing Function f(t) and g(t) Behavior

This calculus assignment analyzes the function f(t) = a * e^(-g(t)), where g(t) = b * e^(-ct). The solution determines the values of g(t) and f(t) as t approaches infinity and negative infinity. It calculates the second derivative of f(t) and expresses it in terms of f(t), g(t), and a new function h(t). The assignment identifies the point at which the graph of f(t) changes curvature. Finally, it provides a rationale for using the given function to model sales values, explaining how the model captures the initial low sales, growth phase, saturation point, and potential decline in sales over time, reflecting customer trust, marketing effectiveness, and product lifecycle.

CALCULUS

Function f (t) = a e^ [-g (t)] where
g (t) = b e^ [-ct]
A.) At t  ∞ value of g (t) and f (t)
Solution
e^ [∞] = ∞
Thus at t =∞ value of g (t) is given by:
g (∞) = b e^ [-∞] = 0 e^ [-∞] = 0
On substituting this value of g (∞) in f (∞) we get
f (∞) = a e^ [-g (∞)]
f (∞) = a e^ [0] = a
Thus at t  ∞ f(t) = a and g(t) = 0
B.) At t  -∞ value of g (t) and f (t)
Solution
e^ [-∞] = 0
Thus at t =-∞ value of g (t) is given by:
g (-∞) = b e^ [∞] = ∞
On substituting this value of g (-∞) in f (-∞) we get
f (-∞) = a e^ [-g (-∞)]
f (-∞) = a e^ [-∞] = 0

Thus at t  -∞ f(t) = 0 and g(t) = ∞
C.) f '' (t) =f (t) g (t) h (t)
Solution
f (t) = a e^ [-g (t)] where
g (t) = b e^ [-ct]
f ' (t) = a e^ [-g (t)] * b e^ [-ct] * c
f ' (t) = abc e^ [-g (t)] * e^ [-ct]
f '' (t) = abc { [e^ [-g (t)] * e^ [-ct] * (-c) ] + [bc* e^ [-g (t)] * e^ [-ct] * e^ [-ct] ]}
f '' (t) = abc e^ [-g (t)] * e^ [-ct] [-c + c* e^ [-ct]]
f '' (t) = f(t)* g(t)* [c^2* e^ [-ct] – c^2]
f '' (t) =f (t) g (t) h (t) where
h(t) = c^2*{e^ [-ct] – 1}
D.) value of t at which graph of f changes curvature
Solution
f ' (t) = a e^ [-g (t)] * b e^ [-ct] * c
Equating it to zero we have
f ' (t) = e^ [-g (t)] * e^ [-ct] = 0
It is possible only at t=0
Thus graph will change its curvature at t=0
At t = 0

g (0) = b e^ [-0] = b
f (0) = a e^ [-g (0)] = a e^ [-b] which is less than f (∞)
Thus t =0 can be considered as the value at which curvature of graph changes.
E.) Rationale for using the above cure for modelling sales values.
Solution
Before a specific time period sales will be considerably low because customers may not trust the
product and brand value may be low. However after specific time period, as people came to
know the quality of products and through effective marketing tactics and maturity sales began to
grow. The sales continue to rise until saturation point occurs. At this stage the excitement of the
product declines and people no longer show any craze to buy product. It becomes only a
necessity when there is great need or urgency. Thus the sales of product flattened and after some
time it may also begin to decline.