Sec2Sess44

Unformatted notes below. Notes on my shorthand method
Session 44: Example
Rate of change of area with respect to angle
area of triangle = 1/2 * side_a * side_b * sin(@) where side_a is the hypotenuse and @ is the angle between side_a and side_b. 
Assume it is a right triangle and therefore a = b * cos(@). The constraint is a = b * cos(@).
Area is a function of a, b, and @. 
3 general approaches could be applied to solve the problem
1. Treat a,b,@ as independent variables
Pd(area)/pd(@) keeping a,b constant
1/2*ab*cos(@)
2. Keep a constant and allow b to vary
pd(area)/pd(@)_a
b = a / cos (@) = a * sec (@)
1/2*ab*cos(@) = 1/2*a^2*sin(@)/cos(@) = 1/2*a^2*tan(@)
d(tan()) = cos^2
pd(area)/pd(@) = 1/2*a*cos^2(@)
3. Keep b constant and allow a to vary
2 systematic ways to solve for area (total differentials or implicit differentials):
1st Method: Total differentials. Keep a constant. a = b cos(@)
find derivative:
d(a) = d(b)*cos(@)-b*sin(@)*d(@)
since a is constant than a = 0 = d(b)*cos(@)-b*sin(@)*d(@)
d(b) can be found now
cos(@)*d(b) = b*sin(@)*d(@)
d(b) = b*(sin(@)/cos(@))*d(@)
d(b) = b*tan(@)*d(@)
function A=1/2*a*b*sin(@)
substituting variables produces:
d(area)=1/2*b*sin(@)*d(a)+1/2*a*sin(@)*d(b)+1/2*a*b*cos(@)*d(@)
d(a) cancels out because it is held constant and is therefore 0
d(b) can be substituted for the equation earlier described and the only variable to solve for is @ (angle, theta) at that point
d(area) = 1/2*a*sin(@)*(b*tan(@)*d(@)) + 1/2*a*b*cos(@)*d(@) = 1/2*a*b*(sin(@)*tan(@)+cos(@))*d(@)
sin(@)*tan(@)+cos(@) = sec(@)
therefore pd(area)/pd(@)_a=1/2*a*b*sec(@)
Overall: by differentiating the constraint equation and holding a constant d(b) can be written in terms of @. d(a) cancels out. pd(area) can then be found only using @.
Summary:
1) write d(area) in terms of d(a), d(b) and d(@)
2) a=constraint=>set d(a)=0
3) differentiate constraint => solve for d(b) in terms of d(@). Apply substitutions to solve for d(area)
General Purpose Summary:
1) write f() in terms of x, y, and z
2) x=constraint=>set d(x)=0
3) differentiate constraint => solve for d(y) in terms of d(z). Apply substitutions to solve for d(f)
If more variables are used than more constraints can be used to solve for variables able to substitute in for other variables. Eventually the method enables the solution to be found using few or one variable because the variables are codependent, and therefore solving for one allows for solving for the others. It is known how others will change with respect to each variable due to the dependency.
2cnd Method: Implicit differentiation and the chain rule: pd(area)/pd(@)_a in formula for A (area)
pd(area)/pd(@)_a = A_@*(pd(@)/pd(@))_a + A_a*(pd(a)/pd(@))_a + A_b*(pd(b)/pd(@))_a where A_<variable> is a description of a function that produces the portion of the answer that corresponds to the <variable>.
(pd(@)/pd(@))_a in words is the rate of change of theta(@) with respect to @ with a constant. The result is 1 because @ depends on @ at a rate of 1.
(pd(a)/pd(@))_a = 1 because a is constant
Therefore 
pd(area)/pd(@)_a = (pd(b)/pd(@))_a which is the same formula seen in the 1st method for finding a solution.
Some valuable things to remember when solving dependent variable problems are:
1. If variables are codependent and functions f() and g() are given that contain the same variables than the functions are interchangeable. I.e., the functions are equal. One function may have more input variables f(x,y,z) compared to g(x,y) but g() can find the same answer with less variables in that case. A function, e.g., g(), may have a constraint which needs to be adhered to for it to equal the other equation e.g., f(). Importantly once a solution to one is found that solution can be applied to the other function and can assist with finding the answer to the overall question a given problem is asking.
2. A reminder that a gradient can be used to quickly find partial derivatives using its co-factors. Vice versa is also true and therefore when trying to find a gradient for a problem or if a gradient can be found in a problem that method can be used to help find answers to the overall problem asked.