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- A major scientific use of computers is in finding numerical solutions to mathematical problems which have no analytical solutions (i.e. solutions which may be written down in terms of polynomials and standard mathematical functions). In this chapter we look briefly at some areas where
*numerical methods*have been highly developed, e.g. solving nonlinear equations, evaluating integrals, and solving differential equations.

Use the Bisection method to find the square root of 2, taking 1 and 2 as initial values of xL and xR. Continue bisecting until the maximum error is less than 0.05.

Successive bisections are: 1.5, 1.25, 1.375, 1.4375 and 1.40625. The exact answer is 1.414214..., so the last bisection is within the required error

Use the Trapezoidal rule to evaluate ∫_{0}^{4} x^{2}dx , using a step-length of h=1.

22 (exact answer is 21.3333)

The basic equation for modeling radioactive decay is [dx/dt]=−rx, where x is the amount of the radioactive substance at time t, and r is the decay rate.

Some radioactive substances decay into other radioactive substances, which in turn also decay. For example, Strontium 92 (r1 =0.256 per hr) decays into Yttrium 92 (r2 =0.127 per hr), which in turn decays into Zirconium.

Write down a pair of differential equations for Strontium and Yttrium to describe what is happening. Starting at t =0 with 5×1026 atoms of Strontium 92 and none of Yttrium, use the Runge-Kutta method (ode23) to solve the equations up to t =8 hours in steps of 1/3 hr. Also use Euler's method for the same problem, and compare your results.

Some radioactive substances decay into other radioactive substances, which in turn also decay. For example, Strontium 92 (r1 =0.256 per hr) decays into Yttrium 92 (r2 =0.127 per hr), which in turn decays into Zirconium.

Write down a pair of differential equations for Strontium and Yttrium to describe what is happening. Starting at t =0 with 5×1026 atoms of Strontium 92 and none of Yttrium, use the Runge-Kutta method (ode23) to solve the equations up to t =8 hours in steps of 1/3 hr. Also use Euler's method for the same problem, and compare your results.

The differential equations to be solved are

dS/dt = -r1S

dY/dt = r1S × r2Y

The exact solution after 8 hours is S = 6.450*10^{25} and Y = 2.312*10^{26}

dS/dt = -r1S

dY/dt = r1S × r2Y

The exact solution after 8 hours is S = 6.450*10

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Integration 0:14
- Non-Well Behaved Functions
- Example: Integration
- Trapezoidal Rule
- Numerical Method
- Quad Function
- Symbolic Variables
- Numerical Differentiation
- Diff
- First Order Differential Equation 12:40
- Euler’s Method
- Exponential Growth
- Limitations of Euler’s

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