The Dope Knight

HW4/README.md

@@ -0,0 +1,37 @@
+# Homework #3
+## due 3/1/17 by 11:59pm
+
+
+1. Use your repository 'roots_and_optimization'. Document all the HW4 work under the
+heading `# Homework #4` in your `README.md` file
+
+    a. Create a function called 'collar_potential_energy' that computes the total
+    potential energy of a collar connected to a spring and sliding on a rod. As shown in
+    the figure given a position, xc, and angle, theta:
+
+    ![Collar-mass on an inclined rod](collar_mass.png)
+
+    The spring is unstretched when x_C=0.5 m. The potential energy due to gravity is:
+
+    PE_g=m x_C\*g\*sin(theta)
+
+    where m=0.5 kg, and g is the acceleration due to gravity, 
+
+    and the potential energy due to the spring is:
+
+    PE_s=1/2\*K \*(DL)^2
+
+    where DL = 0.5 - sqrt(0.5^2+(0.5-x_C)^2) and K=30 N/m.
+
+    b. Use the `goldmin.m` function to solve for the minimum potential energy at xc when
+    theta=0. *create an anonymous function with `@(x) collar_potential_energy(x,theta)` in
+    the input for goldmin. Be sure to include the script that solves for xc*
+
+    c. Create a for-loop that solves for the minimum potential energy position, xc, at a
+    given angle, theta, for theta = 0..90 degrees. 
+
+    d. Include a plot of xc vs theta. `plot(theta,xc)` with
+
+    `![Steady-state position of collar on rod at angle theta](plot.png)`
+
+3. Commit your changes to your repository. Sync your local repository with github.

README.md

@@ -74,15 +74,15 @@ general, I will not post homework solutions.
 |3|1/31||Consistent Coding habits|
 |   |2/2|5|Root Finding| 
 |4|2/7|6|Root Finding con’d|      
-|   |2/9|7|Optimization|      
-|5|2/14||Intro to Linear Algebra|
+|   |2/9|7| **Snow Day**|      
+|5|2/14|| Optimization |
 |   |2/16|8|Linear Algebra|
 |6|2/21|9|Linear systems: Gauss elimination|
 |   |2/23|10|Linear Systems: LU factorization|
-|7|2/28||Midterm Review|
-|   |3/2||Midterm|
-|8|3/7|11|Linear Systems: Error analysis|
-|   |3/9|13|Eigenvalues|
+|7|2/28|11|Linear Systems: Error analysis|
+|   |3/2|12|Eigenvalues|
+|8|3/7|1-10 |Midterm Review|
+|   |3/9|1-10|Midterm|
 |9|3/14| N/A |Spring Break!|
 |   |3/16| N/A |Spring Break!|
 |10|3/21|12|Linear Systems: Iterative methods|

lecture_08/goldmin.m

@@ -33,4 +33,4 @@
   if xopt~=0, ea = (2 - phi) * abs((xu - xl) / xopt) * 100;end
   if ea <= es | iter >= maxit,break,end
 end
-x=xopt;fx=f(xopt,varargin{:});
+x=xopt;fx=f(xopt,varargin{:});