Starting from:
$30
CS3430-Assignment 3 Theory of the Firm, Implicit Differentiation, Related Rates Solved
Problem 1: Product and Quotient Rules
Extend your differentiation engine in deriv.py with the product and quotient rules of differentiation. Specifically, your implementation of prod_deriv should be able to handle not just products of constants and powers but also products of a sum and a sum, a sum and a power, a sum and a product, a sum and a quotient, a power and a sum, a power and a power, a power and a product, a power and a quotient, a product and a sum, a product and a power, a product and a product, a product and a quotient, a quotient and a sum, a quotient and a power, a quotient and a product, and a quotient and a quotient. This explicit enumeration is a rather windy way of what can be stated succinctly: your differentiation engine must be commutative. Use the implementation of the quotient class in quot.py and make appropriate changes in tof.py to convert the outputs of your deriv function to Python functions.
Test 01
Let’s differentiate
def test_01():
print(’\n***** Test 01 ***********’) e1 = make_plus(make_pwr(’x’, 1.0), make_const(1.0)) e2 = make_pwr(’x’, 3.0) e3 = make_prod(make_const(5.0), make_pwr(’x’, 1.0)) e4 = make_plus(e2, e3) e5 = make_plus(e4, make_const(2.0)) e6 = make_prod(e1, e5)
# 1) print the expression we just constructed print(’-- function expression is:\n’) print(e6)
# 2) differentiate and make sure that it not None drv = deriv(e6) assert not drv is None print(’-- derivative is:\n’) print(e6)
# 3) convert drv into a function e6f = tof(drv) assert not e6f is None
# steps 2) and 3) can be combined into tof(deriv(e6)). # 4) construct the ground truth function gt = lambda x: 4.0*(x**3) + 3*(x**2) + 10.0*x + 7.0 # 5) compare the ground gruth with what we got in # step 3) on an appropriate number range. print(’\n Comparison with ground truth:\n’) err = 0.00001 for i in range(10):
print(e6f(i), gt(i)) assert abs(e6f(i) - gt(i)) <= err
print(’Test 01: pass’)
Here is the output of test_01 in the Py shell.
***** Test 01 ***********
-- function expression is:
(((x^1.0)+1.0)*(((x^3.0)+(5.0*(x^1.0)))+2.0)) -- derivative is:
(((x^1.0)+1.0)*(((x^3.0)+(5.0*(x^1.0)))+2.0)) --comparison with ground truth:
(7.0, 7.0)
(24.0, 24.0)
(71.0, 71.0)
(172.0, 172.0)
(351.0, 351.0)
(632.0, 632.0)
(1039.0, 1039.0)
(1596.0, 1596.0)
(2327.0, 2327.0)
(3256.0, 3256.0) Test 01: pass
Test 02
Let’s do dxd (2x4 − x + 1)(−x5 + 1).
def test_02():
print(’\n***** Test 02 ***********’) e1 = make_prod(make_const(2.0), make_pwr(’x’, 4.0)) e2 = make_prod(make_const(-1.0), make_pwr(’x’, 1.0)) e3 = make_plus(e1, e2) e4 = make_plus(e3, make_const(1.0)) e5 = make_prod(make_const(-1.0), make_pwr(’x’, 5.0)) e6 = make_plus(e5, make_const(1.0)) e7 = make_prod(e4, e6) print(’-- function expression is:\n’) print(e7) drv = deriv(e7)
assert not drv is None print(’\n-- derivative is:\n’) print(drv) e7f = tof(drv) assert not e7f is None gt = lambda x: -18.0*(x**8) + 6.0*(x**5) - 5.0*(x**4) + 8.0*(x**3) - 1.0 err = 0.00001 print(’\n--comparison with ground truth:\n’) for i in range(10): print(e7f(i), gt(i)) assert abs(e7f(i) - gt(i)) <= err
print(’Test 02: pass’)
Here is the output of test_02 in the Py shell.
***** Test 02 ***********
-- function expression is:
((((2.0*(x^4.0))+(-1.0*(x^1.0)))+1.0)*((-1.0*(x^5.0))+1.0)) -- derivative is:
(((((2.0*(x^4.0))+(-1.0*(x^1.0)))+1.0)*((-1.0*(5.0*(x^4.0)))+0.0))
+(((-1.0*(x^5.0))+1.0)*(((2.0*(4.0*(x^3.0)))+
(-1.0*(1.0*(x^0.0))))+0.0)))
--comparison with ground truth:
(-1.0, -1.0)
(-10.0, -10.0)
(-4433.0, -4433.0)
(-116830.0, -116830.0)
(-1174273.0, -1174273.0)
(-7014626.0, -7014626.0)
(-30191185.0, -30191185.0)
(-103674838.0, -103674838.0)
(-301809665.0, -301809665.0)
(-774513658.0, -774513658.0) Test 02: pass
Test 03
Let’s do .
def test_03():
print(’\n***** Test 03 ***********’) q = make_quot(make_plus(make_pwr(’x’, 1.0),
make_const(11.0)),
make_plus(make_pwr(’x’, 1.0), make_const(-3.0))) pex = make_pwr_expr(q, 3.0) print(’-- function expression is:\n’) print(pex) pexdrv = deriv(pex) assert not pexdrv is None print(’\n-- derivative is:\n’) print(pexdrv) pexdrvf = tof(pexdrv) assert not pexdrvf is None gt = lambda x: -42.0*(((x + 11.0)**2)/((x - 3.0)**4)) err = 0.00001 print(’\n--comparison with ground truth:\n’) for i in range(10):
if i != 3.0:
print(pexdrvf(i), gt(i)) assert abs(pexdrvf(i) - gt(i)) <= 0.001
print(’Test 03: pass’)
Here is the output of test_03 in the Py shell.
***** Test 03 ***********
-- function expression is: ((((x^1.0)+11.0)/((x^1.0)+-3.0))^3.0) -- derivative is:
((3.0*((((x^1.0)+11.0)/((x^1.0)+-3.0))^2.0))*(((((x^1.0)+-3.0)
*((1.0*(x^0.0))+0.0))+(-1.0*(((x^1.0)+11.0)*((1.0*(x^0.0))+0.0))))/
(((x^1.0)+-3.0)^2.0)))
--comparison with ground truth:
(-62.74074074074073, -62.74074074074074)
(-378.0, -378.0)
(-7098.0, -7098.0)
(-9450.0, -9450.0)
(-672.0, -672.0)
(-149.85185185185188, -149.85185185185185)
(-53.15625, -53.15625)
(-24.259200000000003, -24.2592)
(-12.962962962962964, -12.962962962962962)
Test 03: pass
Let’s put to use our differentiation engine to solve some real world scientific computing problems. In each of the three problems below, you will need to abstract the statement of the problem into a Python function to solve a specific type of problem.
Problem 2: Quantitative Theory of the Firm:
The demand equation for X Logistics, a small trucking company, is
60,
where x is the number of rides per day. The constraint says that the company cannot do more than 60 rides per day (e.g., it does not have enough trucks). Find the value of x and the corresponding price the company must charge to maximize its daily revenue.
Generalize the above problem into a function maximize_revenue that takes an expression of the demand equation and a keyword that specifies the constraint on the values of the variable in the demand equation and returns a tuple of three constants: the number of units that results in maximum revenue, i.e., the value of x, the value of maximum revenue, and the unit price when the revenue is maximum. The constraint should be specified either as a lambda or a one-parameter named function. For example, in the above problem, the constraint is lambda x: 0 <= x <= 60. These constraints are needed to weed out negative values, because in the quantitative theory of the firm, as we do function optimization, we typically ignore negative numbers of units. Who is interested in -2 rides per day anyway? What does it even mean? Save your implementation of maximize_revenue in hw03.py. Here is a test that applies maximize_revenue to the above problem.
from hw03 import maximize_revenue def max_rev_test():
print(’\n***** Max Revenue Test ************’) e1 = make_prod(make_const(1.0/12.0), make_pwr(’x’, 2.0)) e2 = make_prod(make_const(-10.0), make_pwr(’x’, 1.0)) sum1 = make_plus(e1, e2) dmndf_expr = make_plus(sum1, make_const(300.0)) num_units, rev, price = \ maximize_revenue(dmndf_expr, constraint=lambda x: 0 <= x <= 60)
print(’x = ’, num_units.get_val()) print(’rev = ’, rev.get_val()) print(’price = ’, price.get_val()) print(’Max Revenue Test: pass’)
Here is the output of max_rev_test in the Py shell.
***** Max Revenue Test ************
(’x = ’, 20.0)
(’rev = ’, 2666.6666666666665)
(’price = ’, 133.33333333333331)
Max Revenue Test: pass
This output has a straightforward interpretation: to maximize its daily revenue, X Logistics must do 20 rides per day, at a price of ≈ $133.33 per ride for a maximum revenue of ≈ $2,666.67 per day.
Develop a couple, at least two, of your own unit tests for this type of problem from the problems we worked out on the board in class and the week 3 reading handouts.
Problem 3: Science and Geology:
Due to an oil tanker accident, the leaking oil is forming an approximately circular disk on the surface of the ocean. The response team has determined with its measurement equipment that when the radius of the disk r is equal to ≈ 150 meters, the radius is increasing at a rate of ≈ 20 meters per hour. The volume of the disk is V ≈ 0.02πr2. How fast is the volume of the disk changing when the response team measures the disk’s radius?
Recall that we worked out this problem on the board at the end of lecture 6. Generalize the above problem into a function dydt_given_x_dxdt(yt, x, dxdt), where yt is a function expression that relates y to x, and x and dxdt are constant objects. Notationally, dydt denotes the rate of change of y with respect to t, i.e., , and dxdt – the rate of change of x with respect to t, i.e., .
Also, note that x is just a variable name. In the above problem, for example, x is the radius r of the oil disk and dxdt is the same as , i.e., 20 meters per hour, and yt is specified by V ≈ 0.02πr2.
The function dydt_given_x_dxdt(yt, x, dxdt) returns the value of when the value of x is specified by its second parameter and the value of is given by its third parameter, i.e., dxdt. Let’s write a test for the above problem.
from hw03 import dydt_given_x_dxdt def oil_disk_test():
yt = make_prod(make_const(0.02*math.pi),
make_pwr(’r’, 2.0))
print(yt) dydt = dydt_given_x_dxdt(yt, make_const(150.0), make_const(20.0))
assert not dydt is None assert isinstance(dydt, const) print(dydt)
Here is the output of oil_disk_test in the Py shell.
(0.0628318530718*(r^2.0))
376.991118431
Thus, the volume of the oil disk is increasing (because the value is positive) at ≈ 377 m3 per hour.
Save your implementation of dydt_given_x_dxdt(yt, x, dxdt) in hw03.py. A word of caution: your implementation of dydt_given_x_dxdt does not have to solve the generic implicit differentiation problem; it just needs to be powerful enough to handle only this type of problem.
Problem 4: Radiology and Health Care:
A patient is receiving radiation treatment for a spherical tumor on her arm. The doctor has determined that when the radius of the tumor is ≈ 10.3 mm, the radius is decreasing at a rate of ≈ 1.75 mm per week. What is the rate at which the volume of the tumor is changing when the doctor measures its radius if the volume of the tumor is given by V ≈ 0.003πr3?
Use your implementation of dydt_given_x_dxdt from Problem 3 to solve this problem by writing a function arm_tumor_test in hw03.py that, similarly to oil_disk_test, prints out the rate at which the patient’s tumor is changing.
Extend your differentiation engine in deriv.py with the product and quotient rules of differentiation. Specifically, your implementation of prod_deriv should be able to handle not just products of constants and powers but also products of a sum and a sum, a sum and a power, a sum and a product, a sum and a quotient, a power and a sum, a power and a power, a power and a product, a power and a quotient, a product and a sum, a product and a power, a product and a product, a product and a quotient, a quotient and a sum, a quotient and a power, a quotient and a product, and a quotient and a quotient. This explicit enumeration is a rather windy way of what can be stated succinctly: your differentiation engine must be commutative. Use the implementation of the quotient class in quot.py and make appropriate changes in tof.py to convert the outputs of your deriv function to Python functions.
Test 01
Let’s differentiate
def test_01():
print(’\n***** Test 01 ***********’) e1 = make_plus(make_pwr(’x’, 1.0), make_const(1.0)) e2 = make_pwr(’x’, 3.0) e3 = make_prod(make_const(5.0), make_pwr(’x’, 1.0)) e4 = make_plus(e2, e3) e5 = make_plus(e4, make_const(2.0)) e6 = make_prod(e1, e5)
# 1) print the expression we just constructed print(’-- function expression is:\n’) print(e6)
# 2) differentiate and make sure that it not None drv = deriv(e6) assert not drv is None print(’-- derivative is:\n’) print(e6)
# 3) convert drv into a function e6f = tof(drv) assert not e6f is None
# steps 2) and 3) can be combined into tof(deriv(e6)). # 4) construct the ground truth function gt = lambda x: 4.0*(x**3) + 3*(x**2) + 10.0*x + 7.0 # 5) compare the ground gruth with what we got in # step 3) on an appropriate number range. print(’\n Comparison with ground truth:\n’) err = 0.00001 for i in range(10):
print(e6f(i), gt(i)) assert abs(e6f(i) - gt(i)) <= err
print(’Test 01: pass’)
Here is the output of test_01 in the Py shell.
***** Test 01 ***********
-- function expression is:
(((x^1.0)+1.0)*(((x^3.0)+(5.0*(x^1.0)))+2.0)) -- derivative is:
(((x^1.0)+1.0)*(((x^3.0)+(5.0*(x^1.0)))+2.0)) --comparison with ground truth:
(7.0, 7.0)
(24.0, 24.0)
(71.0, 71.0)
(172.0, 172.0)
(351.0, 351.0)
(632.0, 632.0)
(1039.0, 1039.0)
(1596.0, 1596.0)
(2327.0, 2327.0)
(3256.0, 3256.0) Test 01: pass
Test 02
Let’s do dxd (2x4 − x + 1)(−x5 + 1).
def test_02():
print(’\n***** Test 02 ***********’) e1 = make_prod(make_const(2.0), make_pwr(’x’, 4.0)) e2 = make_prod(make_const(-1.0), make_pwr(’x’, 1.0)) e3 = make_plus(e1, e2) e4 = make_plus(e3, make_const(1.0)) e5 = make_prod(make_const(-1.0), make_pwr(’x’, 5.0)) e6 = make_plus(e5, make_const(1.0)) e7 = make_prod(e4, e6) print(’-- function expression is:\n’) print(e7) drv = deriv(e7)
assert not drv is None print(’\n-- derivative is:\n’) print(drv) e7f = tof(drv) assert not e7f is None gt = lambda x: -18.0*(x**8) + 6.0*(x**5) - 5.0*(x**4) + 8.0*(x**3) - 1.0 err = 0.00001 print(’\n--comparison with ground truth:\n’) for i in range(10): print(e7f(i), gt(i)) assert abs(e7f(i) - gt(i)) <= err
print(’Test 02: pass’)
Here is the output of test_02 in the Py shell.
***** Test 02 ***********
-- function expression is:
((((2.0*(x^4.0))+(-1.0*(x^1.0)))+1.0)*((-1.0*(x^5.0))+1.0)) -- derivative is:
(((((2.0*(x^4.0))+(-1.0*(x^1.0)))+1.0)*((-1.0*(5.0*(x^4.0)))+0.0))
+(((-1.0*(x^5.0))+1.0)*(((2.0*(4.0*(x^3.0)))+
(-1.0*(1.0*(x^0.0))))+0.0)))
--comparison with ground truth:
(-1.0, -1.0)
(-10.0, -10.0)
(-4433.0, -4433.0)
(-116830.0, -116830.0)
(-1174273.0, -1174273.0)
(-7014626.0, -7014626.0)
(-30191185.0, -30191185.0)
(-103674838.0, -103674838.0)
(-301809665.0, -301809665.0)
(-774513658.0, -774513658.0) Test 02: pass
Test 03
Let’s do .
def test_03():
print(’\n***** Test 03 ***********’) q = make_quot(make_plus(make_pwr(’x’, 1.0),
make_const(11.0)),
make_plus(make_pwr(’x’, 1.0), make_const(-3.0))) pex = make_pwr_expr(q, 3.0) print(’-- function expression is:\n’) print(pex) pexdrv = deriv(pex) assert not pexdrv is None print(’\n-- derivative is:\n’) print(pexdrv) pexdrvf = tof(pexdrv) assert not pexdrvf is None gt = lambda x: -42.0*(((x + 11.0)**2)/((x - 3.0)**4)) err = 0.00001 print(’\n--comparison with ground truth:\n’) for i in range(10):
if i != 3.0:
print(pexdrvf(i), gt(i)) assert abs(pexdrvf(i) - gt(i)) <= 0.001
print(’Test 03: pass’)
Here is the output of test_03 in the Py shell.
***** Test 03 ***********
-- function expression is: ((((x^1.0)+11.0)/((x^1.0)+-3.0))^3.0) -- derivative is:
((3.0*((((x^1.0)+11.0)/((x^1.0)+-3.0))^2.0))*(((((x^1.0)+-3.0)
*((1.0*(x^0.0))+0.0))+(-1.0*(((x^1.0)+11.0)*((1.0*(x^0.0))+0.0))))/
(((x^1.0)+-3.0)^2.0)))
--comparison with ground truth:
(-62.74074074074073, -62.74074074074074)
(-378.0, -378.0)
(-7098.0, -7098.0)
(-9450.0, -9450.0)
(-672.0, -672.0)
(-149.85185185185188, -149.85185185185185)
(-53.15625, -53.15625)
(-24.259200000000003, -24.2592)
(-12.962962962962964, -12.962962962962962)
Test 03: pass
Let’s put to use our differentiation engine to solve some real world scientific computing problems. In each of the three problems below, you will need to abstract the statement of the problem into a Python function to solve a specific type of problem.
Problem 2: Quantitative Theory of the Firm:
The demand equation for X Logistics, a small trucking company, is
60,
where x is the number of rides per day. The constraint says that the company cannot do more than 60 rides per day (e.g., it does not have enough trucks). Find the value of x and the corresponding price the company must charge to maximize its daily revenue.
Generalize the above problem into a function maximize_revenue that takes an expression of the demand equation and a keyword that specifies the constraint on the values of the variable in the demand equation and returns a tuple of three constants: the number of units that results in maximum revenue, i.e., the value of x, the value of maximum revenue, and the unit price when the revenue is maximum. The constraint should be specified either as a lambda or a one-parameter named function. For example, in the above problem, the constraint is lambda x: 0 <= x <= 60. These constraints are needed to weed out negative values, because in the quantitative theory of the firm, as we do function optimization, we typically ignore negative numbers of units. Who is interested in -2 rides per day anyway? What does it even mean? Save your implementation of maximize_revenue in hw03.py. Here is a test that applies maximize_revenue to the above problem.
from hw03 import maximize_revenue def max_rev_test():
print(’\n***** Max Revenue Test ************’) e1 = make_prod(make_const(1.0/12.0), make_pwr(’x’, 2.0)) e2 = make_prod(make_const(-10.0), make_pwr(’x’, 1.0)) sum1 = make_plus(e1, e2) dmndf_expr = make_plus(sum1, make_const(300.0)) num_units, rev, price = \ maximize_revenue(dmndf_expr, constraint=lambda x: 0 <= x <= 60)
print(’x = ’, num_units.get_val()) print(’rev = ’, rev.get_val()) print(’price = ’, price.get_val()) print(’Max Revenue Test: pass’)
Here is the output of max_rev_test in the Py shell.
***** Max Revenue Test ************
(’x = ’, 20.0)
(’rev = ’, 2666.6666666666665)
(’price = ’, 133.33333333333331)
Max Revenue Test: pass
This output has a straightforward interpretation: to maximize its daily revenue, X Logistics must do 20 rides per day, at a price of ≈ $133.33 per ride for a maximum revenue of ≈ $2,666.67 per day.
Develop a couple, at least two, of your own unit tests for this type of problem from the problems we worked out on the board in class and the week 3 reading handouts.
Problem 3: Science and Geology:
Due to an oil tanker accident, the leaking oil is forming an approximately circular disk on the surface of the ocean. The response team has determined with its measurement equipment that when the radius of the disk r is equal to ≈ 150 meters, the radius is increasing at a rate of ≈ 20 meters per hour. The volume of the disk is V ≈ 0.02πr2. How fast is the volume of the disk changing when the response team measures the disk’s radius?
Recall that we worked out this problem on the board at the end of lecture 6. Generalize the above problem into a function dydt_given_x_dxdt(yt, x, dxdt), where yt is a function expression that relates y to x, and x and dxdt are constant objects. Notationally, dydt denotes the rate of change of y with respect to t, i.e., , and dxdt – the rate of change of x with respect to t, i.e., .
Also, note that x is just a variable name. In the above problem, for example, x is the radius r of the oil disk and dxdt is the same as , i.e., 20 meters per hour, and yt is specified by V ≈ 0.02πr2.
The function dydt_given_x_dxdt(yt, x, dxdt) returns the value of when the value of x is specified by its second parameter and the value of is given by its third parameter, i.e., dxdt. Let’s write a test for the above problem.
from hw03 import dydt_given_x_dxdt def oil_disk_test():
yt = make_prod(make_const(0.02*math.pi),
make_pwr(’r’, 2.0))
print(yt) dydt = dydt_given_x_dxdt(yt, make_const(150.0), make_const(20.0))
assert not dydt is None assert isinstance(dydt, const) print(dydt)
Here is the output of oil_disk_test in the Py shell.
(0.0628318530718*(r^2.0))
376.991118431
Thus, the volume of the oil disk is increasing (because the value is positive) at ≈ 377 m3 per hour.
Save your implementation of dydt_given_x_dxdt(yt, x, dxdt) in hw03.py. A word of caution: your implementation of dydt_given_x_dxdt does not have to solve the generic implicit differentiation problem; it just needs to be powerful enough to handle only this type of problem.
Problem 4: Radiology and Health Care:
A patient is receiving radiation treatment for a spherical tumor on her arm. The doctor has determined that when the radius of the tumor is ≈ 10.3 mm, the radius is decreasing at a rate of ≈ 1.75 mm per week. What is the rate at which the volume of the tumor is changing when the doctor measures its radius if the volume of the tumor is given by V ≈ 0.003πr3?
Use your implementation of dydt_given_x_dxdt from Problem 3 to solve this problem by writing a function arm_tumor_test in hw03.py that, similarly to oil_disk_test, prints out the rate at which the patient’s tumor is changing.
1 file (30.5KB)
