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Project: Calculus

Path: 2022-07-22-215148.ipynb

Views: ²⁵
Image: ubuntu2004

Kernel: SageMath 9.2

Shri Ramdeobaba College of Engineering and Management,Nagpur

$\hspace{85pt}$ Department of Mathematics

$\hspace{105pt}$ Computational Mathematics Lab

Name:Sharvari Gohane

Roll Number:67

Section: C

$\hspace{50pt}$ Experiment 3: Differential Calculus and its Applications

Aim: To Learn Calculus with SageMath

1. Limit

To Evaluate $\lim_{x \to a} f(x)$ syntax is:

$\hspace{115 pt}$ Syntax: limit(f(x),x=a, dir='+'or '-')

"dir" - (default: None); dir may have the value 'plus' (or '+' or 'right' or 'above') for a limit from above, 'minus' (or '-' or'left' or 'below') for a limit from below, or may be omitted (implying a two-sided limit is to be computed).

Problem: Find the limit of $lim_{x \to 0} \frac{x}{|x|}$

In [1]:

f(x)=x/abs(x)
plot(f(x),(x,-2,2))

Out[1]:

In [2]:

limit(f(x),x=0)

Out[2]:

und

In [3]:

limit(f(x),x=0,dir = "+")

Out[3]:

1

In [4]:

limit(f(x),x=0,dir = "-")

Out[4]:

-1

Problem:

Evaluate $lim_{x \to 2} \frac{x^2-4}{x-2}$

In [5]:

plot((x^2-4)/(x-2),0,4,figsize=3)

Out[5]:

In [6]:

limit((x^2-4)/(x-2),x=2)

Out[6]:

4

Problem:

Evaluate $lim_{x \to a} x^a$ for various values of $a$ .

In [7]:

var('a')

Out[7]:

a

In [8]:

assume(a>0)

In [9]:

limit(x^a,x=infinity)

Out[9]:

+Infinity

In [10]:

forget()

In [11]:

assume(a<0)

In [12]:

limit(x^a,x=oo)

Out[12]:

0

Exercise 3.1

Find the following limits

$lim_{x \to 0} (x^2-\frac{2^x}{1000})$

Solution-1

In [28]:

f(x)=(x^2-(2^x)/(1000))

In [29]:

limit(f(x),x=0)

Out[29]:

-1/1000

In [30]:

limit(f(x),x=0,dir= "+")

Out[30]:

-1/1000

In [31]:

limit(f(x),x=0,dir= "-")

Out[31]:

-1/1000

$lim_{x \to 0.5^-} (\frac{2x-1}{|2x^3-x^2|})$

Solution 2

In [11]:

limit((2*x-1)/abs(2*x^3-x^2),x=0.5,dir='-')

Out[11]:

-4.0

2.Derivatives

The syntax for derivative of f(x) is

$\hspace{80pt}$ diff(f, args)

Repeated differentiation is supported by the syntax given in the examples below.

In [1]:

f(x)=x*sin(1/x)

In [2]:

f.diff(x) # first order derivative of f(x)

Out[2]:

x |--> -cos(1/x)/x + sin(1/x)

In [3]:

f.diff() # first order derivative of f(x)

Out[3]:

x |--> -cos(1/x)/x + sin(1/x)

In [4]:

diff(f(x),x ) # first order derivative of f(x)

Out[4]:

-cos(1/x)/x + sin(1/x)

In [5]:

f.diff(4) # or diff(f(x),x,4) gives fourth order derivative of f(x)

Out[5]:

x |--> -12*sin(1/x)/x^5 - 8*cos(1/x)/x^6 + sin(1/x)/x^7

Problem

Consider a function $f(x)=sin(cos(5x))-e^{(x-a)^{2}}$ . Plot the graph of $f (x)$ along with first two derivative in $[0.5, 2]$ .

In [2]:

f(x) = sin(cos(5*x))-exp((x - 1)^2)

In [9]:

f1(x) = f.diff()

In [10]:

f2(x) = f.diff(2)

In [11]:

p = plot(f(x),0.5,2,color='red')
p1 = plot(f1(x),0.5,2,color='green')
p2 = plot(f2(x),0.5,2,color='black')
show(p+p1+p2, ymax=20,ymin=-30,figsize=5)

Out[11]:

Exercise 3.2

Find the first four derivatives of $f(t)=ln(1+t^2)$ and plot them along with the graph of $f(t)$ .

Solution-1

In [17]:

f(t)=ln(1+t^2)

In [24]:

f1=f.diff(t)
show(f1(t))

Out[24]:

\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{2 \, t}{t^{2} + 1}

In [25]:

f2=f.diff(2)
show(f2(t))

Out[25]:

\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{4 \, t^{2}}{{\left(t^{2} + 1\right)}^{2}} + \frac{2}{t^{2} + 1}

In [26]:

f3=f.diff(3)
show(f3(t))

Out[26]:

\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{16 \, t^{3}}{{\left(t^{2} + 1\right)}^{3}} - \frac{12 \, t}{{\left(t^{2} + 1\right)}^{2}}

In [27]:

f4=f.diff(4)
show(f4(t))

Out[27]:

\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{96 \, t^{4}}{{\left(t^{2} + 1\right)}^{4}} + \frac{96 \, t^{2}}{{\left(t^{2} + 1\right)}^{3}} - \frac{12}{{\left(t^{2} + 1\right)}^{2}}

In [30]:

p1=plot(f(t),-2,2,color='red')
p2=plot(f1(t),-2,2,color='blue')
p3=plot(f2(t),-2,2,color='green')
p4=plot(f3(t),-2,2,color='black')
p5=plot(f4(t),-2,2,color='orange')
show(p1+p2+p3+p4+p5,ymin=-25,ymax=25,figsize=5)

Out[30]:

Find the first four derivative of $f(x)=sin(3x)+e^{-2x}+ln(7x)$ .

Solution-2

In [2]:

f(x)=sin(3*x)+e^(-2*x)+ln(7*x)

In [3]:

f.diff()

Out[3]:

x |--> 1/x + 3*cos(3*x) - 2*e^(-2*x)

In [4]:

f.diff(2)

Out[4]:

x |--> -1/x^2 + 4*e^(-2*x) - 9*sin(3*x)

In [5]:

f.diff(3)

Out[5]:

x |--> 2/x^3 - 27*cos(3*x) - 8*e^(-2*x)

In [6]:

f.diff(4)

Out[6]:

x |--> -6/x^4 + 16*e^(-2*x) + 81*sin(3*x)

3. Partial Derivatives

For $f(x)= \frac{xy}{x+y}$ find $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial^2 f}{\partial x^2},\frac{\partial^2 f}{\partial y^2},\frac{\partial^2 f} {\partial x \partial y},\frac{\partial^2 f}{\partial y \partial x}$

In [10]:

var('x,y')
f(x,y) = (x*y)/(x+y)
fx = f.diff(x) # Gives first order derivative of f with respect to x
show(fx(x,y))
fy = f.diff(y) # Gives first order derivative of f with respect to y
show(fy(x,y))
fxx=f.diff(x,2) # Gives Second order derivative of f with respect to x . Instead of fxx we can give any other name also.
show(fxx(x,y))
fyy=f.diff(y,2) # Gives Second order derivative of f with respect to y . Instead of fyy we can give any other name also.
fxy=f.diff(x,y) # Here we are differentiating function first with respect to x and then we differentiate the result with respect to y.
show(fxy(x,y))
fyx=f.diff(y,x) # Here we are differentiating function first with respect to y and then we differentiate the result with respect to x.
show(fxy(x,y))

Out[10]:

\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{y}{x + y} - \frac{x y}{{\left(x + y\right)}^{2}}

\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{x}{x + y} - \frac{x y}{{\left(x + y\right)}^{2}}

\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{2 \, y}{{\left(x + y\right)}^{2}} + \frac{2 \, x y}{{\left(x + y\right)}^{3}}

\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{x + y} - \frac{x}{{\left(x + y\right)}^{2}} - \frac{y}{{\left(x + y\right)}^{2}} + \frac{2 \, x y}{{\left(x + y\right)}^{3}}

\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{x + y} - \frac{x}{{\left(x + y\right)}^{2}} - \frac{y}{{\left(x + y\right)}^{2}} + \frac{2 \, x y}{{\left(x + y\right)}^{3}}

Problem

For $f(x)=f(x,y)=4xye^{(-x^2-y^2)}$ , find all first order and second order derivatives at point (1,2).

In [11]:

var('x,y')
f(x,y)=4*x*y*exp(-x^2-y^2)
show(f(x,y))

Out[11]:

\renewcommand{\Bold}[1]{\mathbf{#1}}4 \, x y e^{\left(-x^{2} - y^{2}\right)}

In [12]:

show(f.diff(x)) # Gives Derviative of f with respect to x

Out[12]:

\renewcommand{\Bold}[1]{\mathbf{#1}}\left( x, y \right) \ {\mapsto} \ -8 \, x^{2} y e^{\left(-x^{2} - y^{2}\right)} + 4 \, y e^{\left(-x^{2} - y^{2}\right)}

In [13]:

show(f.diff(x)(x,y)) # Gives Derviative of f with respect to x

Out[13]:

\renewcommand{\Bold}[1]{\mathbf{#1}}-8 \, x^{2} y e^{\left(-x^{2} - y^{2}\right)} + 4 \, y e^{\left(-x^{2} - y^{2}\right)}

In [14]:

f.diff(x)(x=1.0,y=2.0) # Gives Derviative of f with respect to x at (1,2)

Out[14]:

-0.0539035759926837

In [15]:

f.diff(y)(x=1,y=2) # Gives Derviative of f with respect to y at (1,2)

Out[15]:

-28*e^(-5)

In [16]:

f.diff(x,2)(x=1,y=2) # Gives double Derviative of f with respect to x at (1,2).

Out[16]:

-16*e^(-5)

In [17]:

f.diff(x,y)(x=1,y=2) # Gives Derviative of f first with respect to x and then with respect to y at (1,2).

Out[17]:

28*e^(-5)

In [18]:

f.diff(y,x)(x=1,y=2) # Gives Derviative of f first with respect to y and then with respect to x at (1,2).

Out[18]:

28*e^(-5)

Exercise 3.3

Let $f(x,y)=ln(x^2+y^2)$ . Show that $\frac{\partial^2 f}{\partial x^2 }+\frac{\partial^2 f}{\partial y^2 }=0$ .

Solution-1

In [2]:

var('x,y')
f(x,y)=ln(x^2+y^2)
show(f(x,y))

Out[2]:

\renewcommand{\Bold}[1]{\mathbf{#1}}\log\left(x^{2} + y^{2}\right)

In [3]:

fxx = f.diff(x,2)
show(fxx(x,y))

Out[3]:

\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{4 \, x^{2}}{{\left(x^{2} + y^{2}\right)}^{2}} + \frac{2}{x^{2} + y^{2}}

In [4]:

fyy = f.diff(y,2)
show(fyy(x,y))

Out[4]:

\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{4 \, y^{2}}{{\left(x^{2} + y^{2}\right)}^{2}} + \frac{2}{x^{2} + y^{2}}

In [5]:

show(fxx+fyy)

Out[5]:

\renewcommand{\Bold}[1]{\mathbf{#1}}\left( x, y \right) \ {\mapsto} \ -\frac{4 \, x^{2}}{{\left(x^{2} + y^{2}\right)}^{2}} - \frac{4 \, y^{2}}{{\left(x^{2} + y^{2}\right)}^{2}} + \frac{4}{x^{2} + y^{2}}

Find all the first order partial derivatives of $w=x^2y-10y^2z^3+43x-7tan(4y)$

Solution-2

In [15]:

var('x,y,z')
w=x^2*y-10*y^2*z^3+43*x-7*tan(4*y)
show(w)

Out[15]:

\renewcommand{\Bold}[1]{\mathbf{#1}}-10 \, y^{2} z^{3} + x^{2} y + 43 \, x - 7 \, \tan\left(4 \, y\right)

In [21]:

var('x,y,z')
f(x,y) = x^2*y-10*y^2*z^3+43*x-7*tan(4*y)
fx = f.diff(x) # Gives first order derivative of f with respect to x
show(fx(x,y,z))
fy = f.diff(y) # Gives first order derivative of f with respect to y
show(fy(x,y,z))
fz = f.diff(z) # Gives first order derivative of f with respect to z
show(fz(x,y,z))
fxx=f.diff(x,2) # Gives Second order derivative of f with respect to x . Instead of fxx we can give any other name also.
show(fxx(x,y,z))
fyy=f.diff(y,2) # Gives Second order derivative of f with respect to y . Instead of fyy we can give any other name also.
show(fyy(x,y,z))
fzz=f.diff(z,2) # Gives Second order derivative of f with respect to z . Instead of fzz we can give any other name also.
show(fzz(x,y,z))
fxy=f.diff(x,y) # Here we are differentiating function first with respect to x and then we differentiate the result with respect to y.
show(fxy(x,y,z))
fyz=f.diff(y,x) # Here we are differentiating function first with respect to y and then we differentiate the result with respect to x.
show(fxy(x,y,z))
fzx=f.diff(z,x) # Here we are differentiating function first with respect to z and then we differentiate the result with respect to x.
show(fzx(x,y,z))

Out[21]:

\renewcommand{\Bold}[1]{\mathbf{#1}}2 \, x y + 43

\renewcommand{\Bold}[1]{\mathbf{#1}}-20 \, y z^{3} + x^{2} - 28 \, \tan\left(4 \, y\right)^{2} - 28

\renewcommand{\Bold}[1]{\mathbf{#1}}-30 \, y^{2} z^{2}

\renewcommand{\Bold}[1]{\mathbf{#1}}2 \, y

\renewcommand{\Bold}[1]{\mathbf{#1}}-20 \, z^{3} - 224 \, {\left(\tan\left(4 \, y\right)^{2} + 1\right)} \tan\left(4 \, y\right)

\renewcommand{\Bold}[1]{\mathbf{#1}}-60 \, y^{2} z

\renewcommand{\Bold}[1]{\mathbf{#1}}2 \, x

\renewcommand{\Bold}[1]{\mathbf{#1}}2 \, x

\renewcommand{\Bold}[1]{\mathbf{#1}}0

4. Implicit Derivative

Find the slope formula for the Folium of Descartes implicitly defined by x^3 + y^3 = 6xy.( Find $\frac{dy}{dx}$ )

In [19]:

x = var('x')
y = var('y')
f(x,y) = x^3 + y^3 - 6*x*y
y=function('y')(x)
temp=diff(f(x,y))
solve(temp,diff(y))

Out[19]:

[diff(y(x), x) == -(x^2 - 2*y(x))/(y(x)^2 - 2*x)]

Exercise 3.4

Find the $\frac {dy}{dx}$ for the implicit function $2(x^2+y^2)^2=25(x^2-y^2)$ .

Solution-1

In [6]:

x = var('x')
y = var('y')
f(x,y) = (2*(x^2+y^2)^2) - 25*(x^2-y^2)
y=function('y')(x)
temp=diff(f(x,y))
show(solve(temp,diff(y)))

Out[6]:

\renewcommand{\Bold}[1]{\mathbf{#1}}\left[\frac{\partial}{\partial x}y\left(x\right) = -\frac{4 \, x^{3} + 4 \, x y\left(x\right)^{2} - 25 \, x}{4 \, y\left(x\right)^{3} + {\left(4 \, x^{2} + 25\right)} y\left(x\right)}\right]

Find the $\frac {dy}{dx}$ for the implicit function $y^2=x^3(2-x)$ .

Solution-2

In [7]:

x = var('x')
y = var('y')
f(x,y) = x^3*(2-x)-y^2
y=function('y')(x)
temp=diff(f(x,y))
show(solve(temp,diff(y)))

Out[7]:

\renewcommand{\Bold}[1]{\mathbf{#1}}\left[\frac{\partial}{\partial x}y\left(x\right) = -\frac{2 \, x^{3} - 3 \, x^{2}}{y\left(x\right)}\right]

5. Local Maximum and Minimum

Syntax for finding local maximum value of function $f(x)$ in interval $(a,b)$ is

$\hspace{105 pt}$ f.find_local_maximum(a,b).

Syntax for finding local minimum value of function $f(x)$ in interval $(a,b)$ is

$\hspace{105 pt}$ f.find_local_minimum(a,b).

Problem:

Find the local maximum and local minimum of the function $f(x)=e^{-x/2}+e^{-2x^2}$ .

In [21]:

f(x) = exp(-x/2) + exp(-2*x^2)
f.plot(-2,2,figsize =4)

Out[21]:

In [22]:

f.find_local_maximum(-1,1)

Out[22]:

(2.0340673102820146, -0.13932332835389183)

In [23]:

f.find_local_minimum(-1.5,-0.5)

Out[23]:

(1.7650250464715878, -0.8669457216285031)

In [24]:

f.find_local_minimum(1,2)

Out[24]:

(0.3682149119773104, 1.999999956179319)

In [25]:

f.find_local_maximum(1,2)

Out[25]:

(0.7418659187092465, 1.0000000286997572)

In [26]:

plot(f.diff(),-2,2,figsize=4)

Out[26]:

In [11]:

c1 = f.diff().find_root(-2,-0.5);c1

Out[11]:

-0.8669457228341761

In [9]:

c2 = f.diff().find_root(-0.5,0);c2

Out[9]:

-0.13932333585693482

In [10]:

f.diff(2)(x=c1), f.diff(2)(x=c2)

Out[10]:

(2.17068359111117, -3.280901640208612)

Problem:

Consider a function $f(x)=2x^3-9x^2+12x-3$

(a) Find the intervals on which f is increasing or decreasing.

(b) Find the local maximum and minimum values of f.

In [12]:

f(x) = 2*x^3 - 9*x^2 + 12*x - 3
pf = plot(f,0,3)
show(pf,figsize=4)

Out[12]:

In [19]:

df = f.diff()
d2f = f.diff(2)

In [20]:

cpts = solve(df==0,x,solution_dict=True)

In [21]:

a,b = cpts[0][x],cpts[1][x]

In [22]:

infl = solve(d2f==0,x,solution_dict=True);infl

Out[22]:

[{x: 3/2}]

In [23]:

c = infl[0][x]

In [24]:

pf = plot(f,0,3,thickness=0.5)
pt1 = point((a,f(a)),color='red',size=20) # Gives point where function has maximum value
pt2 = point((b,f(b)),color='green',size=30)# Gives point where function have minimum value
pt3 = point((c,f(c)),color='black',size=20) # gives point of inflection
pdf = plot(df,0,3,thickness=0.5,color='red')
pd2f = plot(d2f,0,3,thickness=0.5,color='black')
show(pf+pt1+pt2+pt3+pdf+pd2f,ymin=-5,ymax=8,figsize=4)

Out[24]:

Problem:

The Hubble Space Telescope was deployed on April 24, 1990, by the space shuttle Discovery. Amodel for the velocity of the shuttle during this mission, from liftoff at $t=0$ until the solid rocket boosters were jettisoned at $t=126$ seconds, is given by

v(t)=0.001302t^3-20.09029t^2+23.61t-3.083

(in feet per second). Using this model, estimate the absolute maximum and minimum values of the acceleration of the shuttle between liftoff and the jettisoning of the booster.

In [26]:

var('t')
v(t) =0.001302*t^3-0.09029*t^2+23.61*t-3.083

In [27]:

plot(v,0,126,figsize=4)

Out[27]:

In [28]:

a = v.diff()
a.plot(0,126,figsize=4)

Out[28]:

In [29]:

a.find_local_minimum(0,126)

Out[29]:

(21.52288169482847, 23.115720101631652)

In [30]:

a(0),a(23.115720101631652), a(126)

Out[30]:

(23.6100000000000, 21.5228816948285, 62.8685760000000)

Exercise 3.5

Find the local maximum and local minimum of $f(x)=x^4e^{-x}$ .

Solution-1

In [20]:

f(x)=x^4*exp(-x)
show(f(x))
f.plot((-2,2),figsize=4)

Out[20]:

\renewcommand{\Bold}[1]{\mathbf{#1}}x^{4} e^{\left(-x\right)}

In [18]:

f.find_local_maximum(-2,2)

Out[18]:

(118.22488359077923, -1.9999999605494494)

In [19]:

f.find_local_minimum(-2,2)

Out[19]:

(2.8279164949881223e-43, 2.3060389863085893e-11)

The power needed to propel an airplane forward at velocity $v$ is

P=Av^3+\frac{BL^2}{v}

where $A$ and $B$ are positive constants specific to the particular aircraft and L is the lift, the upward force supporting the weight of the plane. Find the speed that minimizes the required power.

Solution-2

In [13]:

var('p,v')
A=3
B=4
L=5

In [14]:

p(v) = A*v^3 + (B*L^2)/v
show(p(v))

Out[14]:

\renewcommand{\Bold}[1]{\mathbf{#1}}3 \, v^{3} + \frac{100}{v}

In [15]:

p.plot(0,100,figsize=5)

Out[15]:

In [16]:

p.find_local_minimum(0,100)

Out[16]:

(73.02967433402215, 1.8257418601157738)

Learning:

In this experiment we learnt to find out the limits,derivatives of a given function,also their partial and implicit derivatives. We also learned to find out the local maximum and local minimum of the functions using sagemath and also plot their graphs.

In [0]:

Real-time collaboration for Jupyter Notebooks, Linux Terminals, LaTeX, VS Code, R IDE, and more, all in one place. Commercial Alternative to JupyterHub.

Shri Ramdeobaba College of Engineering and Management,Nagpur

\hspace{85pt} Department of Mathematics

\hspace{105pt} Computational Mathematics Lab

Name:Sharvari Gohane

Roll Number:67

Section: C

\hspace{50pt}Experiment 3: Differential Calculus and its Applications

Aim: To Learn Calculus with SageMath

1. Limit

\hspace{115 pt} Syntax: limit(f(x),x=a, dir='+'or '-')

Problem: Find the limit of limx→0x∣x∣ lim_{x \to 0} \frac{x}{|x|}limx→0​∣x∣x​

Problem:

Evaluate limx→2x2−4x−2 lim_{x \to 2} \frac{x^2-4}{x-2}limx→2​x−2x2−4​

Problem:

Evaluate limx→axa lim_{x \to a} x^alimx→a​xa for various values of aaa.

Exercise 3.1

Find the following limits

limx→0(x2−2x1000) lim_{x \to 0} (x^2-\frac{2^x}{1000})limx→0​(x2−10002x​)

Solution-1

limx→0.5−(2x−1∣2x3−x2∣) lim_{x \to 0.5^-} (\frac{2x-1}{|2x^3-x^2|})limx→0.5−​(∣2x3−x2∣2x−1​)

Solution 2

2.Derivatives

Problem

Exercise 3.2

Solution-1

Solution-2

3. Partial Derivatives

Problem

Exercise 3.3

Solution-1

Solution-2

4. Implicit Derivative

Exercise 3.4

Solution-1

Solution-2

5. Local Maximum and Minimum

Problem:

Problem:

Problem:

Exercise 3.5

Solution-1

Solution-2

Learning:

Real-time collaboration for Jupyter Notebooks, Linux Terminals, LaTeX, VS Code, R IDE, and more,
all in one place. Commercial Alternative to JupyterHub.

$\hspace{85pt}$ Department of Mathematics

$\hspace{105pt}$ Computational Mathematics Lab

$\hspace{50pt}$ Experiment 3: Differential Calculus and its Applications

$\hspace{115 pt}$ Syntax: limit(f(x),x=a, dir='+'or '-')

Problem: Find the limit of $lim_{x \to 0} \frac{x}{|x|}$

Evaluate $lim_{x \to 2} \frac{x^2-4}{x-2}$

Evaluate $lim_{x \to a} x^a$ for various values of $a$ .

$lim_{x \to 0} (x^2-\frac{2^x}{1000})$

$lim_{x \to 0.5^-} (\frac{2x-1}{|2x^3-x^2|})$