What is integreation and differentiation and how to apply it inn follwing problem.If x=6t2 +5t then find velocity
kvnmurty:
which class are you in ? please inform. do you need explanation for integration? do you understand the concept explanation i gave for derivative and velocity.
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let us take a small time interval Δt like 0.001 sec..
At time t, distance x is x1 = 6 t² + 5 t
at t + Δt ie., t+0.001 sec, x2 = 6 (t +Δt)² + 5 t = 6 t² + 12 t Δt + (Δt)² + 5 t + 5 Δt
Distance travelled in Δt duration is : x2 - x1 = 12 t Δt + (Δt)² + 5 Δt
This is denoted by Δx. So Δx = x2 - x1
Now find x2 - x1 / Δt = Δx / Δt = 12 t + Δt + 5
Velocity is defined as Limiting value of Δx / Δt as Δt approaches 0
so velocity = limit Δt -> 0, (12 t + Δt + 5 ) = 12 t + 5
That is you take Δt = 0.00001 sec, find Δx. take Δt = 0.000000001 sec, find Δx. Then Δx / Δt value approaches a fixed value. That value is 12 t + 5.
This is also called derivative of x with respect to t. Limit as Δt -->0 the value of Δx / Δt. it is denoted by d x / d t.
the reverse process of derivative is called integration.
At time t, distance x is x1 = 6 t² + 5 t
at t + Δt ie., t+0.001 sec, x2 = 6 (t +Δt)² + 5 t = 6 t² + 12 t Δt + (Δt)² + 5 t + 5 Δt
Distance travelled in Δt duration is : x2 - x1 = 12 t Δt + (Δt)² + 5 Δt
This is denoted by Δx. So Δx = x2 - x1
Now find x2 - x1 / Δt = Δx / Δt = 12 t + Δt + 5
Velocity is defined as Limiting value of Δx / Δt as Δt approaches 0
so velocity = limit Δt -> 0, (12 t + Δt + 5 ) = 12 t + 5
That is you take Δt = 0.00001 sec, find Δx. take Δt = 0.000000001 sec, find Δx. Then Δx / Δt value approaches a fixed value. That value is 12 t + 5.
This is also called derivative of x with respect to t. Limit as Δt -->0 the value of Δx / Δt. it is denoted by d x / d t.
the reverse process of derivative is called integration.
do you understand the derivative concept now clearly? which class are you? do you want a simple explanation of integration? is it in your syllabus?
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