?How does calculus help in understanding the motion of objects under constant acceleration

Calculus is essential in understanding the motion of objects under constant acceleration because it allows us to derive and analyze the kinematic equations that describe such motion. Using calculus, we can integrate the acceleration function to obtain the velocity function, and then integrate the velocity function to find the position function. For example, if an object is accelerating at a constant rate 𝑎 a, the velocity 𝑣 ( 𝑡 ) v(t) can be found by integrating the acceleration: 𝑣 ( 𝑡 ) = ∫ 𝑎 𝑑 𝑡 = 𝑎 𝑡 + 𝑣 0 v(t)=∫adt=at+v 0 where 𝑣 0 v 0 is the initial velocity. Further, integrating the velocity function gives the position 𝑥 ( 𝑡 ) x(t): 𝑥 ( 𝑡 ) = ∫ 𝑣 ( 𝑡 ) 𝑑 𝑡 = ∫ ( 𝑎 𝑡 + 𝑣 0 ) 𝑑 𝑡 = 1 2 𝑎 𝑡 2 + 𝑣 0 𝑡 + 𝑥 0 x(t)=∫v(t)dt=∫(at+v 0 )dt= 2 1 at 2 +v 0 t+x 0 where 𝑥 0 x 0 is the initial position. This process illustrates how calculus provides a systematic way to derive these relationships and analyze motion in a precise manner.

.Explain the concept of electric flux and how Gauss's law uses calculus to relate it to electric fields

Electric flux is a measure of the number of electric field lines passing through a given surface. Mathematically, it is defined as the surface integral of the electric field 𝐸 E over a surface 𝑆 S: Φ 𝐸 = ∫ 𝑆 𝐸 ⋅ 𝑑 𝐴 Φ E =∫ S E⋅dA where 𝑑 𝐴 dA is a vector representing an infinitesimal element of the surface area with a direction normal to the surface. Gauss's law relates electric flux to the charge enclosed by a surface. It states that the total electric flux through a closed surface is equal to the enclosed charge 𝑄 Q divided by the permittivity of free space 𝜖 0 ϵ 0 : Φ 𝐸 = ∮ 𝑆 𝐸 ⋅ 𝑑 𝐴 = 𝑄 enc 𝜖 0 Φ E =∮ S E⋅dA= ϵ 0 Q enc Using calculus, Gauss's law allows us to calculate the electric field for symmetrical charge distributions by evaluating the surface integral. For example, for a point charge 𝑞 q inside a spherical surface of radius 𝑟 r, the electric field is radially symmetric, and Gauss's law simplifies the calculation by using the symmetry to evaluate the integral, leading to: 𝐸 ⋅ 4 𝜋 𝑟 2 = 𝑞 𝜖 0 E⋅4πr 2 = ϵ 0 q 𝐸 = 𝑞 4 𝜋 𝜖 0 𝑟 2 E= 4πϵ 0 r 2 q This shows how calculus is used to connect electric fields and charges in a fundamental way.

Calculus based physics fundamentals Tutorial

What will you learn in this course?

Apply differential calculus to analyze motion, forces, and energy in physics problems using calculus techniques
Utilize integral calculus to calculate electric and magnetic fields, heat transfer, and wave phenomena in physical systems
Solve complex physics problems involving mechanics, electromagnetism, thermodynamics, and waves through advanced mathematical methods
Interpret physical phenomena by modeling with calculus-based equations in real-world physics scenarios

Develop problem-solving skills for physics using calculus to predict and explain natural phenomena
Enhance understanding of physics principles by integrating calculus concepts into the study of fundamental laws and theories
Use calculus to derive and analyze equations governing physical systems in mechanics, electromagnetism, and thermodynamics
Apply mathematical modeling and calculus tools to analyze and solve physics problems in research and practical applications

How to Get The Certificate

You must have an account Register
Watch All Lessons
Watch at least 50% of Lesson Duration
you can follow your course progress From Your Profile
You can Register With Any Course For Free
The Certificate is free !

Lessons | 20