Math Examples of Application of Integration

First we will calculate velocity vector.
To calculate the velocity vector we have to differentiate the Position vector x(t) and y(t), the process of calculation of velocity vector is shown below:
Therefore velocity vector v(t)= d(t⁵)/dt, d(t³)/dt
=> v(t) = d(t⁵)/dt, d(t³)/dt = (5t⁴), (3t²),
Therefore the velocity vector is equals to v(t) = (5t⁴),(3t²),  
Now we will calculate the acceleration vector.
To calculate the acceleration vector we have to differentiate the velocity vector, therefore we get- 
a(t) = d(5t⁴)/dt, d(3t²)/dt = 20t³, 6t
Therefore the acceleration vector is equals to a (t) = 20t³, 6t.

First we will calculate velocity vector.
 
To calculate the velocity vector we have to differentiate the Position vector x(t) and y(t), the process of calculation of velocity vector is shown below-
 
Therefore velocity vector v(t) = d(t³ + 1) / dt, d(t²)/dt
 
=> v(t) = d(t³ + 1)/dt, d(t²)/dt = (3t²), (2t)
 
Therefore the velocity vector is equals to v(t) = (3t²), (2t)  
 
Now we will calculate the acceleration vector.
 
To calculate the acceleration vector we have to differentiate the velocity vector, therefore we get-
 
a(t) =  d(3t²)/dt, d(2t)/dt = 6t, 2
 
Therefore the acceleration vector is equal to a(t) = 6t, 2.

First we will calculate velocity vector.
 
To calculate the velocity vector we have to differentiate the Position vector x(t) and y(t), the process of calculation of velocity vector is shown below-
 
Therefore velocity vector v(t) = d(t⁴ + 3 t)/dt, d(t³ – t²)/dt
 
=> v(t)= d(t⁴ + 3t)/dt, d(t³ – t²)/dt = (4t³ + 3) , (3t² + t)
 
When t = 2 then velocity vector will be equals to-
 
=> v (2) = (4(2) 3 + 3),(3(2)²+ 2) = 35, 14
 
Therefore the velocity vector is equals to 35, 14 when t = 2.
 
Now we will calculate the acceleration vector.
 
To calculate the acceleration vector we have to differentiate the velocity vector, therefore we get-
 
a(t) =  d(4t³ + 3)/dt, d(3t² + t)/dt = 12t², 6t + 1
 
When t = 3 then acceleration vector a(t) will be equals to:
=> a(3) = 108, 19
 
Therefore the acceleration vector is equals to 108, 19 when t = 3.

First we will calculate the velocity vector.
 
To calculate the velocity vector we will need to differentiate the Position vectors, the process of calculation of velocity vector is shown below-
 
velocity vector v(t) = r'(t) = [d (4 sin (t)/dt, d(9 cos t)/dt]
 
=> [d (4sin(t)/dt, d(9cost)/dt] = [4 cos (t), - 9sin(t)]
 
Therefore velocity vector for our problem will be equals to v(t) = [ 4 cos ( t ) , - 9 sin (t)]
 
Now we will calculate the acceleration vector.
 
Acceleration vector can be calculated by differentiating the velocity vector v(t), the process is shown below-
 
a(t) = r'' (t) = [d (4 cos(t)) /dt, d(- 9 sin(t))/dt]
 
=>  a(t) = r'' (t) = [d (4 cos(t))/dt, d(- 9sin(t))/dt] = [- 4 sin(t), - 9cos(t)].
 
Therefore acceleration vector  a(t) is equals to  [- 4 sin (t), -9 cos(t)].

Velocity vector can be calculated by differentiating the Position vector.
 
 
Therefore velocity vector will be equals to v(t) = [d(sin (3t))/dt,
 d(cos (5t))/dt].
 
=> v(t) = [3 cos(3t), -5sin(5t)],
 
Therefore velocity vector is equals to v(t) = [3cos(3t), -5sin(5t)].
 
Now we will calculate acceleration vector.
 
Acceleration vector can be calculated by differentiating the velocity vector  or double differentiating the position vector, above process is shown below-
 
Acceleration, a(t) = [d(3 cos(3t))/dt, d(-5 sin(5t))/dt = [-9 sin (3t), -25cos(5t)].
 
Therefore acceleration vector of the moving body is equals to a(t) = [- 9sin(3t), -5cos(5t)].

To calculate the velocity vector we will differentiate the given coordinates, therefore we get-
 
[d(sin t)/dt, d(t²/2)/dt], therefore we will get (cos t, t).
 
Therefore velocity vector v(t) = (cos t, t)
 
Now we will find the acceleration vector.
 
To find acceleration vector we will need to differentiate the velocity vector, therefore we get-
 
[d(cos t)/dt, d(t)/dt] = (–sin t, 1)
 
Therefore acceleration vector will be equals to a(t) = (-sint , 1).

Velocity vector can be calculated as-
 
=> v(t) = (dx/dt, dy/dt) = [d(t³ + 4t²)/dt, d(t⁴ – t³)] = [3t² + 8t, 4t³ – 3t²].
 
Now velocity vector at t = 1 is given by-
 
v(1) = (11, 1)
 
Now finding acceleration vector-
 
a(t) = (d²x/dt, d²y/dt) = [d(3t² + 8t)/dt, d(4t³ – 3t²)] = (6t + 8, 12t² – 6t)
 
When t = 2 then acceleration vector will be-
 
a(2) = (20, 36).

Here our Parametric Equations will be x = t and y = t³
 ,
Now, velocity vector can be calculated as v(t) = r ' ( t ) = i + 3t² j + 3k,
 
and Acceleration vector can be calculated as a(t) = r'' ( t ) = 6t j.
 
Now when t = 1, then velocity vector is given as-
 
v(t) = r' (1) = i + 3j + 3k,
 
Acceleration vector curves can be given as-
 
a(1) = r''(1) = 6 j.
 
Hence velocity vector is equals to (i + 3j + 3k) and acceleration vector is equals to 6 j.

Here our Parametric Equations are x = t² – 4 and y = t, so we can find that the curve is a Parabola represented by x = y² – 4 and velocity vector can be given as-
 
v(t) = r' (t) = (2t)i + j,
 
And acceleration vector will be-
 
a(t) = r'' (t) = 2i.
 
Now when t = 0, the velocity and acceleration vectors will be-
 
Velocity vector = v (0) = 2 (0)i + j = j,
 
Acceleration vector = a (0) = 2 I,
 
Now when t = 2, the velocity and acceleration vectors will be-
 
Velocity vector = v (2) = 2 (2) i + j = 4i + j,
 
Acceleration vector = a (2) = 2 I.

Velocity vector can be represented by:
 
v(t) = r '(t) = (cos t/2)i – (sin t/2)j
 
And the acceleration vector can be represented by:
 
a(t) = r'' (t) = (- 1/2 sin t/2) i – (1/2 cos t/2)j
 
Parametric equations for the curve are-
 
x = 2sin t/2 and y = 2cos t/2.
 
Now eliminate the parameter 't', we get the rectangular equation given below-
 
x² + y² = 4.
 
Therefore curve is a Circle having radius 2 and center at origin.
 
As we know velocity vector is v(t) = (cos t/2)i – sin (t/2)j
 
Here the magnitude is constant but the direction changes as 't' increases, so particle revolves around the circle at constant speed.

Here acceleration a = t² + 1,
We know that a = dv / dt,
Therefore dv = a . dt,
Or v = ∫₂³ a . dt,
So we get-
v = ∫₂³ (t² + 1). dt,
v = 2t |₂³,
v = 2 [3] - 2 [2],
v = 2 m / s.
Therefore velocity of the body is 2 m / s.

Here 2 ≤ t ≤ 4.
Now dividing [2, 4] into subintervals of length Δ t and suppose t _k be the time in k th subinterval. The amount of consumption of potato during this interval is given by-
R (t _k) Δ t million bushels.
The amount consumed for 2 ≤ t ≤ 4 is given by -
∑ R (t _k) Δ t million bushels.
Now our Definite Integral for amount of consumption from t = 2 to t = 4 is given by-
∫₂⁴ R (t) dt = ∫₂⁴ (2.2 + 1.1 ^t) dt million bushels
After evaluation we get-
7.066 million bushels.
Therefore the consumption is equals to 7.066 million bushels.

Let f (x) be the force required to stretch the spring by 'x' meters from its natural length. Using the Hooke's Law, that is f (x) = k x where ' k ' is constant.
Therefore f (2) = 10 = k. 2,
So k = 5 and f (x) = 5 x for this spring.
Now writing the integral for the work done to apply force 'f' on interval from x = 0 to x = 4.
Now dividing or partitioning the intervals into small subintervals on each of which 'f' is constant so we can apply constant - force formula for work. If ‘x_k’ is any Point in k th subinterval, For whole interval 'f' is approximately f (x _k) = 5 x_k. The amount of work done by force 'f' on interval is 5 x _k Δ x, here Δ x denotes the length of the interval. Therefore sum is
∑ f (x_k) Δ x = ∑  5 x _k  Δ x is used for approximation of the work done by ' f ' from  x = 0 to x = 4.
Now integrating we get-
∫₀⁴ f (x) dx = ∫₀⁴ 5 x dx = 5 x² / 2 | ⁴₀ = 40.
Hence amount of work done to stretch the spring is equals to 40.

Acceleration for any length of time 't' adds to the velocity of car we get, v (t) = 5 + ∫ 2.4 u du = 5 + ∫₀^t 2.4 u . du = 5 + 1.2 t² mph.
Therefore the distance covered from t = 0 to t = 8 sec is,
∫₀⁸| v (t) |dt = ∫₀⁸ (5 + 1.2 t²) dt,
=> ∫₀⁸ (5 + 1.2 t²) dt = [5 t + 0.4 t³] ₀⁸ = 244 mph * seconds = 0.068 mile.
Hence the distance traveled during those 8 seconds is 0.068 mile.

Calculate the net change in velocity, when acceleration is constant velocity change = acceleration * time applied,
Partition time [0, 8] into small subintervals of length Δ t. The acceleration is constant at each sub interval; therefore ‘t _k’ is any Point in the k th subinterval, the change in velocity due to acceleration in subinterval is,
a (t_k)  Δ t mph.
And net change in velocity for 0 ≤ t ≤ 8 is equals to-
∑ a ( t_k )  Δ t mph.
Now write the Definite Integral we get
∫ a ( t ) dt,
Net change in velocity = ∫₀⁸ 2.4 t dt = 1.2 t² |₀⁸ = 76.8 mph.
Hence the velocity of the car when 8 seconds are up is 76.8 mph.