University of Florida/Egm6341/s10.team3.aks/HW7

Ref Lecture p.40.2

Problem Statement
Integrate the logistic equation and use the code of question 3 for HS and run the code for $$ h = 2^k h $$ and also develop Forward Eular and Back Eular methods for the same step sizes.

Solution
Matlab code with change in step size is given below.

MATLAB CODE:

For k = 1



For k = 2



For k=3



We can see that as we increase the value of k due to decrease in step size our curve shifted to the left side.

Below is the code for Forward Eular method

MATLAB CODE:

Forward Eular breaks down at higher values of step size h .This shows that it is unstable in nature. Below is the code for Backword Eular Method

MATLAB CODE:

For k = 1



For k=2



(12)Convert circumference of ellipse formula from t to α variable
Ref Lecture 42.2

Problem Statement
Prove that $$ a \int\limits_{t=0}^{2\pi}[1-e^2 cos^2 t]^{1/2}dt = 4a \int\limits_{\alpha=0}^{\pi/2} [1-e^2 sin^2 \alpha]^{1/2}d\alpha   $$

Solution
$$ a \int\limits_{t=0}^{2\pi}[1-e^2 cos^2 t]^{1/2}dt $$

$$ = 4a \int\limits_{t=0}^{\pi/2}[1-e^2 cos^2 t]^{1/2}dt $$

Let $$ cos t = Sin \alpha $$

$$ - sin t dt = cos \alpha d\alpha $$

Using above values we obtain

$$ = 4a (-) \int\limits_{\alpha=0}^{\pi/2} [1-e^2 sin^2 \alpha]^{1/2} (-) \frac {cos \alpha}{sin t} d\alpha  $$

$$ = 4a (-) \int\limits_{\alpha=0}^{\pi/2} [1-e^2 sin^2 \alpha]^{1/2} (-) \frac {cos \alpha}{sin t} d\alpha   $$

$$ = 4a \int\limits_{\alpha=0}^{\pi/2} [1-e^2 sin^2 \alpha]^{1/2} \frac {cos \alpha}{(1- sin^2 \alpha)^(1/2)} d\alpha         $$

$$ = 4a \int\limits_{\alpha=0}^{\pi/2} [1-e^2 sin^2 \alpha]^{1/2} \frac {cos \alpha}{cos \alpha} d\alpha        $$

$$ = 4a \int\limits_{\alpha=0}^{\pi/2} [1-e^2 sin^2 \alpha]^{1/2}  d\alpha        $$

Hence Proved