Tutorial 1

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Limits: Computing limits using limit definition
• Derivatives: First derivatives using differentiation rules
• Implicit Differentiation: Finding derivatives of implicitly defined functions
• Parametric Equations: Derivatives of parametric curves

Introduction

Sympy: SymPy is a Python library for symbolic mathematics. We will use this library to solve all the tutorials.

import sympy as sy
from sympy.abc import x, y, t                     # variable x
from sympy.solvers import solve                   # Solving equations
from sympy import sqrt,tan,sin,cos,sec,pi,root,ln
from sympy import log,exp,atan,asinh,atanh,asin   # Import all required math function
from sympy import diff, idiff                     # Solving differentiation
sy.init_printing(use_latex=True)                  # Show it in natural display

How to use sympy to solve limit questions?

1. Declare the symbol to be used in the function. For example, if you wanted to use the symbol $x$, just use x as usual because we have import the variable x earlier. Alternatively, you can declare $x$ such that x = sy.symbols('x'). This will make the variable x a representation of $x$ in the function to be solved. Note that sy is used because we import sympy as sy.

2. Type the function. For example, if the function is $\frac{1}{x}$, type it as 1/x. Note, for power, python uses ** instead of ^.

3. Solve the limit with sympy library by using sy.limit(function,symbol,limit). sy.limit takes three parameters, which are the function, the symbols used in the function and the limit. The return of this fuction will be the answer.

How to use sympy to solve derivative questions?

Just diff(function,symbol)!

Some cheat sheet!

To make sure if you input the equation correctly, you can use the display(function) function. For example,

expr = sqrt(x)
display(expr)

will return:

$\sqrt{x}$

The representation of mathematics function in Python:

Mathematics Function	Python representation
$x^2$	x**2
$\sqrt{x}$	sqrt(x)
$\sqrt[3]{x}$	root(x,3)
$\tan{x}$	tan(x)
$\sec{x}$	sec(x)
$\pi$	pi

Note that the sqrt, tan, sec, pi used in this notebook are from sympy library, not from math library.

That's it! Let's use this knowledge and solve them.

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Question 1

Use the limit definition to evaluate:

a. $\lim\limits_{x \to 5} \frac{x^2 - 25}{x^2 + x - 30}$

lim = 5
func = (x**2-25)/(x**2+x-30)
sy.limit(func,x,lim)

b. $\lim\limits_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$

lim = 9
func = (sqrt(x)-3)/(x-9)
sy.limit(func,x,lim)

c. $\lim\limits_{x \to 0} \frac{x}{3 - \sqrt{x + 9}}$

lim = 0
func = x/(3-sqrt(x+9))
sy.limit(func,x,lim)

Question 2

Find the derivative of $\log(4+\cos x)$

diff(log(4+cos(x),10))

Question 3

Find $\frac{dy}{dx}$ for $\cos(x^2)=xe^y$

eqn=cos(x**2)-x*exp(y)
idiff(eqn,y,x)

Question 4

Find $\frac{dy}{dx}$ for $x^3y^3-2y=x$

eqn=x**3*y**3-2*y-x
idiff(eqn,y,x)

Question 5

A curve in the plane is defined parametrically by the equations $x=7\ln(t)$ and $y=\sqrt{1-4t}$. Find $\frac{dy}{dx}$

dydt = sqrt(1-4*t).diff(t)
dxdt = 7*ln(t).diff(t)
dydt/dxdt

Question 6

A curve in the plane is defined parametrically by the equations $x=t^2-1$ and $y=2e^t$. Find $\frac{dy}{dx}$

dydt = 2*exp(t).diff(t)
dxdt = (t**2-1).diff(t)
dydt/dxdt

Question 7

Find $y'$ for each of the following:

a. $x^2\tan(y)+y^{10}\sec(x)=2x$

eqn=x**2*tan(y)+y**10*sec(x)-2*x
idiff(eqn,y,x)

b. $x^3y^5+3x=8y^3+1$

eqn=x**3*y**5+3*x-8*y**3-1
idiff(eqn,y,x) 

c. $e^{2x+3y}=x^2-\ln(xy^3)$

eqn=exp(2*x+3*y)-x**2+ln(x*y**3)
idiff(eqn,y,x)

Question 8

Solve for $y'$ if $y=\ln(\cos x^2)$

diff(ln(cos(x**2)))

Question 9

Find $y'$ for $10e^{2xy}=e^{15y}+e^{13x}$

eqn=10*exp(2*x*y)-exp(15*y)-exp(13*x)
idiff(eqn,y,x)

Question 10

Solve $f'(x)$ if $f(x)=2x(\arctan 5x)^2+6\tan(\cos 6x)$

diff(2*x*(atan(5*x))**2+6*tan(cos(6*x)))

Question 11

Solve $y'$ if $y=4x\sinh^{-1}(\frac{x}{6})+\tanh^{-1}(\cos 10x)$

diff(4*x*asinh(x/6)+atanh(cos(10*x)))

Question 12

A closed-ended pipe must hold $200$ litres of fluid. Determine the radius and height (in cm) that minimise the amount of material used.

The volume of a closed cylinder is $\pi r^2 h = 200{,}000\ \text{cm}^3$, so $h=\frac{200000}{\pi r^2}$. Substituting into the surface area gives $A(r)=2\pi r^2+\frac{400000}{r}$. Because this is an applied problem, we declare r as a positive symbol so that solve only returns the physically meaningful root.

r = sy.Symbol('r', positive=True)      # radius must be positive
A = 2*pi*r**2 + 400000/r               # A(r) after eliminating h
display(A.diff(r))

r_crit = solve(A.diff(r), r)           # critical number: A'(r) = 0
display(r_crit, [float(s) for s in r_crit])

# Second derivative test: A''(r) > 0 means concave up, i.e. a minimum
display(A.diff(r, 2).subs(r, r_crit[0]))

h = 200000/(pi*r_crit[0]**2)           # back-substitute to find the height
display(h, float(h))

So $r \approx 31.69\ \text{cm}$ and $h \approx 63.38\ \text{cm}$. Since $A''(r)>0$, the surface is concave upward and the critical point is indeed a minimum.

Question 13

Three copper cones have a total volume of $15\ \text{m}^3$, so each cone has volume $5\ \text{m}^3$. Only the lateral surface $A=\pi r s$ is to be minimised, where $s=\sqrt{r^2+h^2}$ and $h=\frac{15}{\pi r^2}$.

r = sy.Symbol('r', positive=True)
h = 15/(pi*r**2)                       # height from the volume constraint
A = pi*r*sqrt(r**2 + h**2)             # lateral surface, A = pi*r*s

r_crit = solve(sy.simplify(A.diff(r)), r)
r_star = [s for s in r_crit if s.is_real and s > 0][0]
display(sy.simplify(r_star), float(r_star))

h_star = 15/(pi*r_star**2)
s_star = sqrt(r_star**2 + h_star**2)
display(float(h_star), float(s_star))

Hence the optimal dimensions of each cone (for the requested lateral area) are

• Radius, $r \approx 1.5002\ \text{m}$
• Height, $h \approx 2.1216\ \text{m}$
• Slant length, $s \approx 2.5984\ \text{m}$

note

The calculation minimises the lateral surface only, matching the final sentence of the question. If the flat base is also counted ($A=\pi r^2+\pi r s$), replace the area above with A = pi*r**2 + pi*r*sqrt(r**2 + h**2); the optimum is then approximately $r=1.1907\ \text{m}$, $h=3.3678\ \text{m}$, and $s=3.5721\ \text{m}$.

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Extra Learning Resources

Key Concepts to Master

1. Limit Laws: Sum, product, quotient rules for limits
2. Differentiation Rules: Power rule, product rule, quotient rule, chain rule
3. Trigonometric Derivatives: Derivatives of sin, cos, tan, sec, etc.
4. Implicit Differentiation: Finding derivatives without solving for y explicitly
5. Parametric Derivatives: Using $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

SymPy Resources

• SymPy Calculus Tutorial - Official tutorial on limits and derivatives
• SymPy Limits Documentation - Detailed limit computation methods
• SymPy Differentiation Guide - Complete guide to derivatives in SymPy

Practice Problems

• Try solving the same problems using different methods (e.g., algebraic simplification before limits)
• Experiment with display() function to verify your input expressions
• Use simplify() to clean up complex derivative expressions
• Practice with one-sided limits using limit(f, x, a, '+') or limit(f, x, a, '-')

Common Pitfalls

• Remember to use ** for exponents, not ^
• Import necessary functions from SymPy (sin, cos, log, etc.)
• For implicit differentiation, use idiff(equation, y, x) not diff()
• Logarithm base: log(x, base) - e.g., log(x, 10) for log₁₀(x)