Applied Mathematics ms

  Applied math is the part of math. That is mostly of the college level students are using the applied mathematics problems. Applied mathematics is very helpful for students for learning about math. The following topics are covered in applied mathematics; they are functions of a complex variable, calculus, ordinary differential equations, partial differential equations and the calculus of variations. Here we will discuss about the applied mathematics ms.

Example problems for applied mathematics ms:

 

Applied mathematics ms – Example: 1
 
   Solve: `g(s) = f(s) + \lambda \int_0^{2\pi} \sin(s) \cos(t) g(t) dt`
 
Solution:
 
`g(s) = f(s) + \lambda \sin(s) \int_0^{2\pi} \cos(t) g(t) dt`
 
Let `c = \int_0^{2\pi} \cos(t) g(t) dt.`
 
Write a new equation:
 
`c = \int_0^{2\pi} \cos(t) \[ f(t) + \lambda \sin(t) c \] dt`
 
`c = \int_0^{2\pi} \cos(t) f(t) dt + \lambda c \int_0^{2\pi} \cos(t) \sin(t) dt`
 
The last integrand is odd and the limits of integration are a multiple of its period, so the integral equals 0.
 
`c = \int_0^{2\pi} \cos(t) f(t) dt`
 
From
 
`g(s) = f(s) + \lambda \sin(s) c`
 
the answer is
 
`g(s) = f(s) + \lambda \sin(s) \int_0^{2\pi} \cos(t) f(t) dt`
 
Applied mathematics ms – Example: 2
 
   Evaluate `\int_{C} F\cdotdr`   where `F=zi+xj+yk`   and C is the arc of the curve `r=\cos t i+\sin t j+tk`   from `t=0`   to `t=\pi`
 
Solution:
 
The given curve is `r=\cos t i+\sin t j+tk`
 
Hence its parametric equations are `x=\cos t,y=\sin t,z=t --(1)`
 
`\int_{C}F\cdot,dr=\int_{C}(zi+xj+yk)\cdot(dxi+dyj+dzk)=\int_{C}(zdx+xdy+ydz)`
 
`=\int_{0}^{2\pi}[z\frac{dx}{dt}+x\frac{dy}{dt}+y\frac{dz}{dt}]dt`
 
`=\int_{0}^{2\pi}[t(-\sin t)+\cos t (\cos t)+\sin t]dt`
 
`=-\int_{0}^{2\pi}t\sin tdt+\frac{1}{2}\int_{0}^{2\pi}(1+\cos 2t)dt+\int_{0}^{2\pi}\sin tdt`
 
`=-[t(-\cos t)_{0}^{2\pi}+(\sin t)_{0}^{2\pi}]+\frac{1}{2}(2\pi)+[-\cos 2\pi+\cos 0]=2\pi+\pi=3\pi`
 
Applied mathematics ms – Example: 3
 
   Prove that `\frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta}=2\sec\theta`
 
Proof:
 
`LHS = \ frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta}`
 
Simplifying,we get
 
`\frac((1+\sin\theta)^2+cos^2 \theta) (\cos\theta(1+\sin\theta))`
 
`\frac{1+\sin^2 \theta+2\sin\theta+\cos^2 \theta}{\cos\theta(1+\sin\theta)}`
 
`\frac{1+1+2\sin\theta}{\cos\theta(1+\sin\theta)}`
 
`\frac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)}`
 
`\frac{2}{\cos\theta}=2\sec\theta=RHS`

 

Practice problems for applied mathematics ms:

 

1. Solve: `u(x) = \int_0^x e^{x-y} u(y) dy, u(0)=0`
 
`Answer: u(x) = 0`
 
2. Evaluate `\int_{C} F\cdotdr`   where `F=yzi+zxj+xyk`  and C is the portion of the curve `r=a\cos t i+b\sin t j+ctk`   from t=0  to `t=\frac{\pi}{2}`
 
`Answer: 0`
 
3. Prove that `\frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=2\csc A`