MCV4U: Calculus: Culminating Assignment Draft Name: ____________________________ In the 1920’s, a scientist (J.P. Maxfield) at Bell Laboratory did research to determine the most optimal design for...

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Answer To: MCV4U: Calculus: Culminating Assignment Draft Name: ____________________________ In the 1920’s, a...

Asif answered on Jun 28 2021
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APPLICATION OF CALCULUS
Mathematics deals with several operations one of which is optimizing ‘continuous change’. Geometry deals with shapes, algebra deals with generalization of arithmetic operation, whereas calculus deals with ‘continuous change’. Calculus can be segregated into several categories :
Namely:
· RATE OF CHANGE:
Differentiation and Integration can be used to estimate the rate of change in various aspects namely rate
of change of population which might help us t determine the population of an entire city, rate of change of loss or profit in a certain company ,etc.
For example - The population of town is modelled by P(t) = 6t2+110t+3000P(t) = 6t2 + 110t + 3000 where ‘P’ is population and ‘t’ is number of years since 1990.
So if we have to calculate the population of the city in 2021 we have to take t=31 (2021-1990)
So by replacing t with 31 we get the answer as 12,176, which is the estimated population for the town in 2021.
Now we can also calculate the rate of population change in 2005,
For that we need to take the derivative of the equation:
dp/dt = 12t + 110
So in 2005 , t = 15
So, dp/dt = 12*15 + 110 = 290
To compute the average change of population at  t = 15 , we simply choose another point t and so the Average Rate of Change will be :
change of population/time passed = P(t) − P(15)/(t−15) = 6t^2 + 110t + 3000 − 6000/(t − 15) = 6t^2 + 110t−3000/(t−15)
Now we have the average rate of change at t = 15.
To have the instantaneous rate of change at t = 15, we will choose some values of ‘t’ that will get closer and closer to 15 on both sides → t = 14.99, t = 15.01 etc. [1]
· LIMITS:
By using limits we can estimate the instantaneous velocity ie, the velocity of an object at a given moment at a given time.
For example –
If we throw a ball from the roof , we can calculate the velocity of the ball exactly when it reaches the ground which cannot be calculated by using the simple formula
Velocity = Distance / time
So here we can apply calculus as ;
V final = lim time→0 distance/time
In other words, as we measure the time it takes as we get closer and closer to the ground, the time approaches 0 (and so does the distance).
In calculus, this would be asking for the derivative of the distance, s, travelled with respect to the time, t, it takes. Using derivative notation, the final velocity can be expressed as:
V final = ds/dt
The right hand side represents the instantaneous velocity at a given moment in time. And the derivative is defined using limits. [2]
· CONTINUITY:

Precise definition of can be provided by calculus using limits.
A function is only said to be continuous in the interval [a, b] if it is continuous at every point inside the given interval. This definition is also made by assuming that both the functions, f(a) and lim x → af(x) , lim x → a , exist. If the above conditions are not met, then the given function will not be continuous at x = a .

In the above picture it is evident that the function is continuous for the given interval.
· DERIVATIVES:
If there exists a variable ‘y’ which varies with another quantity ‘x’ which can be given as y = f(x) , ie dy/dx = f ‘(x). This represents the rate of change of ‘y’ with respect to ‘x’.
Again, dy/dx x=x0 = f ‘(x0) which represents the rate of change of y with respect to x when x = x 0.
To apply it in mathematics we can look at a simple example which demonstrates the rate of change of circumference of the circle when the radius is increasing at the rate of 0.7cm/s.
Here the rate of change in radius is given as 0.7cm/s
At time ‘t’ , the radius is given as ‘r’
Circumference of the circle = 2пr
Differentiating both sides with respect to ‘t’ , we get ,
dC/dt = d/dt (2пr) = 2п*dr/dt .
now, dr/dt = 0.7
So by substituting the value in above equation we get the rate of change of circumference with respect to time as 2*п*0.7 = 4.396cm/s.
· HIGHER ORDER DERIVATIVES:
Higher order derivatives can also be used in several aspects for example for calculating the instantaneous acceleration or instantaneous force. If velocity = ds/dt then acceleration = d2s/dt2.
For example if s(t) = 10t2 , then v(t) = s ‘(t) =20t and a(t) = s’’(t) or v’(t) = 20.
[3]
· OPTIMIZATION:
Optimization usually estimated the largest value or the smallest value...
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