Tutorial 2

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Topics Covered

This tutorial covers the following topics using Python and SymPy:

• Partial Derivatives: Using limit definition and rules.
• Chain Rule: For composite functions.
• Implicit Differentiation: For multivariable functions.
• Jacobians: For transformations.

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, z, s, t, r, theta, u, v, alpha # variables
from sympy.solvers import solve                   # Solving equations
from sympy import sqrt,tan,sin,cos,sec,pi,root,ln, E
from sympy import log,exp,atan,asinh,atanh,asin   # Import all required math function
from sympy import diff, idiff, Matrix                     # Solving differentiation
from sympy import simplify, Rational              # Simplifying expressions
sy.init_printing(use_latex=True)                  # Show it in natural display

Question 1

Find the partial derivative $\displaystyle\left(\frac{\partial f}{\partial y}\right)$ and $\displaystyle\left(\frac{\partial f}{\partial x}\right)$ of these functions using the limit definition

a. $f(x, y)=x^{2} y+2 x+y^{3}$

f = x**2*y + 2*x + y**3
display(diff(f, x))
display(diff(f, y))

b. $f(x, y)=x^{2}-4 x y+y^{2}$

f = x**2 - 4*x*y + y**2
display(diff(f, x))
display(diff(f, y))

c. $f(x, y)=2 x^{3}+3 x y-y^{2}$

f = 2*x**3 + 3*x*y - y**2
display(diff(f, x))
display(diff(f, y))

Question 2

Determine all the first and second order partial derivatives of the function

a. $f(x, y)=x^{2} y^{3}+3 y+x$

f = x**2*y**3 + 3*y + x
display(diff(f, x))
display(diff(f, y))
display(diff(f, x, x))
display(diff(f, x, y))
display(diff(f, y, x))
display(diff(f, y, y))

b. $f(x, y)=x^{4} \sin 3 y$

f = x**4*sin(3*y)
display(diff(f, x))
display(diff(f, y))
display(diff(f, x, x))
display(diff(f, x, y))
display(diff(f, y, x))
display(diff(f, y, y))

c. $f(x, y)=x^{2} y+\ln \left(y^{2}-x\right)$

f = x**2*y + ln(y**2 - x)
display(diff(f, x))
display(diff(f, y))
display(diff(f, x, x))
display(diff(f, x, y))
display(diff(f, y, x))
display(diff(f, y, y))

d. $f(x, y)=e^{x y}(2 x-y)$

f = exp(x*y)*(2*x - y)
display(diff(f, x))
display(diff(f, y))
display(diff(f, x, x))
display(diff(f, x, y))
display(diff(f, y, x))
display(diff(f, y, y))

Question 3

Find both partial derivatives for each of the following two variables functions

a. $g(x, y)=y e^{x+y}$

g = y*exp(x+y)
display(diff(g, x))
display(diff(g, y))

b. $h(x, y)=x \sin y-y \cos x$

h = x*sin(y) - y*cos(x)
display(diff(h, x))
display(diff(h, y))

c. $p(x, y)=x^{y}+y^{2}$

p = x**y + y**2
display(diff(p, x))
display(diff(p, y))

d. $U(x, y)=\frac{9 y^{3}}{x-y}$

U = (9*y**3)/(x-y)
display(diff(U, x))
display(diff(U, y))

Question 4

For $f(x, y, z)$, use the implicit function theorem to find $d y / d x$ and $d y / d z$

a. $f(x, y, z)=x^{2} y^{3}+z^{2}+x y z$

f = x**2*y**3 + z**2 + x*y*z
dydx = -diff(f, x) / diff(f, y)
dydz = -diff(f, z) / diff(f, y)
display(dydx)
display(dydz)

b. $f(x, y, z)=x^{3} z^{2}+y^{3}+4 x y z$

f = x**3*z**2 + y**3 + 4*x*y*z
dydx = -diff(f, x) / diff(f, y)
dydz = -diff(f, z) / diff(f, y)
display(dydx)
display(dydz)

c. $f(x, y, z)=3 x^{2} y^{3}+x z^{2} y^{2}+y^{3} z x^{4}+y^{2} z$

f = 3*x**2*y**3 + x*z**2*y**2 + y**3*z*x**4 + y**2*z
dydx = -diff(f, x) / diff(f, y)
dydz = -diff(f, z) / diff(f, y)
display(dydx)
display(dydz)

Question 5

Find $\frac{\partial F}{\partial s}$ and $\frac{\partial F}{\partial t}$, if applicable, for the following composite functions

a. $F=\sin (x+y)$ where $x=2st$ and $y=s^2+t^2$

F = sin(x+y)
x_s = 2*s*t
y_s = s**2 + t**2
F_s = F.subs([(x, x_s), (y, y_s)])
display(diff(F_s, s))
display(diff(F_s, t))

b. $F=\ln \left(x^{2}+y\right)$ where $x=\mathrm{e}^{(s+t 2)}$ and $y=s^{2}+t$

F = ln(x**2 + y)
x_st = exp(s+t**2)
y_st = s**2 + t
F_st = F.subs([(x, x_st), (y, y_st)])
display(diff(F_st, s))
display(diff(F_st, t))

c. $F=x^{2} y^{2}$ where $x=s \cos t$ and $y=s \sin t$

F = x**2*y**2
x_st = s*cos(t)
y_st = s*sin(t)
F_st = F.subs([(x, x_st), (y, y_st)])
display(diff(F_st, s))
display(diff(F_st, t))

d. $F=xy+y \mathrm{z}^{2}$ where $x=\mathrm{e}^{t}, y=\mathrm{e}^{t} \sin t$ and $\mathrm{z}=\mathrm{e}^{t} \cos t$

F = x*y + y*z**2
x_t = E**t
y_t = E**t*sin(t)
z_t = E**t*cos(t)
F_t = F.subs([(x, x_t), (y, y_t), (z, z_t)])
display(diff(F_t, t))

Question 6

Find $d y / d x$ and $d y / d z$ (if applicable) for each of the following

a. $7 x^{2}+2 x y^{2}+9 y^{4}=0$

f = 7*x**2 + 2*x*y**2 + 9*y**4
dydx = idiff(f, y, x)
display(dydx)

b. $x^{3} z^{2}+y^{3}+4 x y z=0$

f = x**3*z**2 + y**3 + 4*x*y*z
dydx = idiff(f, y, x)
dydz = idiff(f, y, z)
display(dydx)
display(dydz)

c. $3 x^{2} y^{3}+x z^{2} y^{2}+y^{3} z x^{4}+y^{2} z=0$

f = 3*x**2*y**3 + x*z**2*y**2 + y**3*z*x**4 + y**2*z
dydx = idiff(f, y, x)
dydz = idiff(f, y, z)
display(dydx)
display(dydz)

d. $y^{5}+x^{2} y^{3}=1+y \exp \left(x^{2}\right)$

f = y**5 + x**2*y**3 - (1 + y*exp(x**2))
dydx = idiff(f, y, x)
display(dydx)

Question 7

a. In polar coordinates, $x=r \cos \theta, y=r \sin \theta$, show that $\frac{\partial(x, y)}{\partial(r, \theta)}=r$

X = Matrix([r*cos(theta), r*sin(theta)])
Y = Matrix([r, theta])
J = X.jacobian(Y)
display(J)
display(J.det())

b. Obtain the Jacobian $J$ of the transformation $s=2 x+y, t=x-2 y$ and determine the inverse of the transformation $J_{1}$. Confirm that $J_{1}=J^{-1}$.

S = Matrix([2*x+y, x-2*y])
X = Matrix([x, y])
J = S.jacobian(X)
display(J)
display(J.det())

# Inverse
sol = solve([s - (2*x+y), t - (x-2*y)], (x, y))
x_st = sol[x]
y_st = sol[y]
X_st = Matrix([x_st, y_st])
S_st = Matrix([s, t])
J1 = X_st.jacobian(S_st)
display(J1)
display(J1.det())
display(J.inv())

c. Show that if $x+y=\mathrm{u}$ and $y=\mathrm{uv}$, then $\frac{\partial(x, y)}{\partial(u, v)}=u$.

# x = u - y => x = u - uv
x_uv = u - u*v
y_uv = u*v
X = Matrix([x_uv, y_uv])
Y = Matrix([u, v])
J = X.jacobian(Y)
display(J)
display(J.det())

d. Verify whether the functions $u=\frac{x+y}{1-x y}$ and $v=\tan ^{-1} x+\tan ^{-1} y$ are functionally dependent.

u_xy = (x+y)/(1-x*y)
v_xy = atan(x) + atan(y)
U = Matrix([u_xy, v_xy])
X = Matrix([x, y])
J = U.jacobian(X)
display(J)
display(J.det().simplify())

The Jacobian is 0, so the functions are functionally dependent.

e. If $x=uv, y=\frac{u+v}{u-v}$, find $\frac{\partial(u, v)}{\partial(x, y)}$.

jacobian = Matrix([[diff(u*v, u), diff(u*v, v)],
                   [diff((u+v)/(u-v), u), diff((u+v)/(u-v), v)]])
display(simplify(jacobian.det()))        # ∂(x,y)/∂(u,v) = 4uv/(u-v)²
display(simplify(1 / jacobian.det()))    # ∂(u,v)/∂(x,y) = (u-v)²/(4uv)

The question asks for $\frac{\partial(u,v)}{\partial(x,y)}$, which is the reciprocal of the Jacobian we just computed:

$\frac{\partial(u, v)}{\partial(x, y)}=\frac{1}{\frac{\partial(x, y)}{\partial(u, v)}}=\frac{(u-v)^{2}}{4 u v}$

note

Two sign/typo defects in the printed solution are avoided automatically by SymPy:

• Q2d — the printed mixed partial $f_{xy}=e^{xy}(2x^{3}y-xy^{2}+4x-2y)$ should be $e^{xy}(2x^{2}y-xy^{2}+4x-2y)$. diff(f, x, y) returns the correct expression, and it agrees with diff(f, y, x) as Clairaut's theorem requires.
• Q7e — $\frac{\partial y}{\partial v}=\frac{+2u}{(u-v)^{2}}$, not $\frac{-2u}{(u-v)^{2}}$. With the printed (wrong) sign the Jacobian determinant would be $0$ instead of $\frac{4uv}{(u-v)^{2}}$.

Question 8

Error and sensitivity analysis, using the increment (total differential) approximation $\Delta V\approx V_{r}\,\Delta r+V_{h}\,\Delta h$.

a. How sensitive is $V=\pi r^{2}h$ to small changes in $r$ and $h$ near $(r_{0},h_{0})=(1,3)$?

# Q3 above already used `g` and `h` as function names, so rebind them to plain symbols
# here — otherwise `V = pi*r**2*h` would silently contain that earlier expression.
h, l, g = sy.symbols('h l g')

V = pi*r**2*h
V_r, V_h = diff(V, r), diff(V, h)
display(V_r, V_h)                                        # 2*pi*h*r, pi*r**2

display(V_r.subs({r: 1, h: 3}), V_h.subs({r: 1, h: 3}))  # 6*pi, pi

$\Delta V\approx V_{r}(1,3)\,\Delta r+V_{h}(1,3)\,\Delta h=6\pi\,\Delta r+\pi\,\Delta h$

A one-unit change in $r$ shifts $V$ by about $6\pi$; a one-unit change in $h$ shifts it by about $\pi$. Near $(1,3)$ the volume is therefore roughly 6 times as sensitive to $r$ as it is to $h$.

Now reverse the pair, so that $r=3$ and $h=1$:

display(V_r.subs({r: 3, h: 1}), V_h.subs({r: 3, h: 1}))  # 6*pi, 9*pi

$\Delta V\approx 6\pi\,\Delta r+9\pi\,\Delta h$

note

The $r$-sensitivity is unchanged at $6\pi$ when the pair is swapped, because it depends only on the product $rh=3$; only $\pi r^{2}$ moves, and that is what makes the volume sensitive to $h$ instead. So the sensitivity to a change depends not only on the increment but on the relative size of $r$ and $h$.

b. If $r$ is measured to $\pm1\%$ and $h$ to $\pm0.5\%$, how accurately can $V$ be found?

Taking logarithms, $\log V=\log\pi+2\log r+\log h$, so $\dfrac{\Delta V}{V}\approx 2\dfrac{\Delta r}{r}+\dfrac{\Delta h}{h}$.

display(Rational(2, 100) + Rational(1, 200))            # 1/40
display((Rational(2, 100) + Rational(1, 200))*100)      # 5/2  ->  2.5 per cent

$\left|\frac{\Delta V}{V}\right|\le 2\left|\frac{\Delta r}{r}\right|+\left|\frac{\Delta h}{h}\right|=\frac{2}{100}+\frac{1}{200}=0.025$

so the maximum percentage error in $V$ is about $2.5\%$.

c. $T=2\pi\sqrt{\dfrac{l}{g}}$, with errors up to $1\%$ in $l$ and $2\%$ in $g$.

T = 2*pi*sqrt(l/g)
display(simplify(diff(T, l)*l/T))                       # 1/2
display(simplify(diff(T, g)*g/T))                       # -1/2

The elasticities are $+\frac{1}{2}$ for $l$ and $-\frac{1}{2}$ for $g$, so

display(Rational(1, 2)*Rational(1, 100) + Rational(1, 2)*Rational(2, 100))            # 3/200
display((Rational(1, 2)*Rational(1, 100) + Rational(1, 2)*Rational(2, 100))*100)      # 3/2

$\left|\frac{dT}{T}\right|\le\frac{1}{2}(0.01)+\frac{1}{2}(0.02)=0.015$

The maximum percentage error in $T$ is $1.5\%$.

d. $R=\dfrac{v^{2}\sin 2\alpha}{g}$, with a $1\%$ error in $v$ and a $0.5\%$ error in $\alpha$.

R = v**2*sin(2*alpha)/g
display(simplify(diff(R, v)*v/R))                       # 2
display(simplify(diff(R, alpha)*alpha/R))               # 2*alpha/tan(2*alpha)

Since $\log R=2\log v+\log\sin 2\alpha-\log g$,

$\frac{\delta R}{R}=2\frac{\delta v}{v}+2\cot(2\alpha)\,\delta\alpha$

so the percentage error is $2(1)+2\alpha\cot 2\alpha(0.5)=\left(2+\alpha\cot 2\alpha\right)\%$.

Question 9

Find the tangent plane and the normal line to each surface at the given point.

For a level surface $f(x,y,z)=c$, the gradient evaluated at the point is a normal vector, so the tangent plane is $f_{x}(x-x_{0})+f_{y}(y-y_{0})+f_{z}(z-z_{0})=0$ and the normal line has direction ratios $\left(f_{x},f_{y},f_{z}\right)$.

def grad(f):
    return Matrix([diff(f, x), diff(f, y), diff(f, z)])

def tangent_plane(f, pt):
    """Normal vector and tangent plane for the level surface f = 0 at pt."""
    n = grad(f).subs(pt)
    return n, sy.expand(n[0]*(x - pt[x]) + n[1]*(y - pt[y]) + n[2]*(z - pt[z]))

a. $x^{2}+2y^{2}+3z^{2}=6$ at $(1,1,1)$

n, plane = tangent_plane(x**2 + 2*y**2 + 3*z**2 - 6, {x: 1, y: 1, z: 1})
display(n)                                # <2, 4, 6>
display(plane)                            # 2*x + 4*y + 6*z - 12
display(simplify(plane/2))                # x + 2*y + 3*z - 6

Tangent plane $x+2y+3z=6$; normal line $\dfrac{x-1}{1}=\dfrac{y-1}{2}=\dfrac{z-1}{3}$.

b. $2x^{2}+y^{2}-z^{2}=-3$ at $(1,2,3)$

n, plane = tangent_plane(2*x**2 + y**2 - z**2 + 3, {x: 1, y: 2, z: 3})
display(n)                                # <4, 4, -6>
display(simplify(plane/2))                # 2*x + 2*y - 3*z + 3

Tangent plane $2x+2y-3z=-3$; normal line $\dfrac{x-1}{2}=\dfrac{y-2}{2}=\dfrac{z-3}{-3}$.

c. $x^{2}+y^{2}-z=1$ at $(1,2,4)$

n, plane = tangent_plane(x**2 + y**2 - z - 1, {x: 1, y: 2, z: 4})
display(n, plane)                         # <2, 4, -1>, 2*x + 4*y - z - 6

Tangent plane $2x+4y-z=6$; normal line $\dfrac{x-1}{2}=\dfrac{y-2}{4}=\dfrac{z-4}{-1}$.

d. $\ln\left(\dfrac{x}{y}\right)-z^{2}(x-2y)-3z=3$ at $(4,2,-1)$

f = ln(x/y) - z**2*(x - 2*y) - 3*z - 3
display(simplify(f.subs({x: 4, y: 2, z: -1})))    # log(2)  -> the point is NOT on f = 0

n, plane = tangent_plane(f, {x: 4, y: 2, z: -1})
display(n)                                # <-3/4, 3/2, -3>
display(plane)                            # -3*x/4 + 3*y/2 - 3*z - 3

caution

The given point does not satisfy the printed equation: $\ln\frac{4}{2}-(-1)^{2}(4-4)-3(-1)=\ln 2+3\approx3.693\neq3$. So no tangent plane exists at $(4,2,-1)$ for that level surface. The gradient and plane above belong to the surface that actually passes through the point (right-hand side $3+\ln 2$) — which is what the printed solution intended to compute.

e. $x^{3}z+z^{3}x-2yz=0$ at $(1,1,1)$

n, plane = tangent_plane(x**3*z + z**3*x - 2*y*z, {x: 1, y: 1, z: 1})
display(n)                                # <4, -2, 2>
display(simplify(plane/2))                # 2*x - y + z - 2

Tangent plane $2x-y+z=2$; normal line $\dfrac{x-1}{2}=\dfrac{y-1}{-1}=\dfrac{z-1}{1}$.

f. $z=5+(x-1)^{2}+(y+2)^{2}$ at $(2,0,10)$

n, plane = tangent_plane((x - 1)**2 + (y + 2)**2 - z + 5, {x: 2, y: 0, z: 10})
display(n, plane)                         # <2, 4, -1>, 2*x + 4*y - z + 6

Tangent plane $2x+4y-z=-6$; normal line $\dfrac{x-2}{2}=\dfrac{y}{4}=\dfrac{z-10}{-1}$.

g. $\dfrac{x^{2}}{12}+\dfrac{y^{2}}{6}+\dfrac{z^{2}}{4}=1$ at $(1,2,1)$

n, plane = tangent_plane(x**2/12 + y**2/6 + z**2/4 - 1, {x: 1, y: 2, z: 1})
display(n)                                # <1/6, 2/3, 1/2>
display(sy.expand(6*plane))               # x + 4*y + 3*z - 12

Tangent plane $x+4y+3z=12$; normal line $\dfrac{x-1}{1}=\dfrac{y-2}{4}=\dfrac{z-1}{3}$.

h. $ze^{x}+e^{z+1}+xy+y=3$ at $(0,3,-1)$

n, plane = tangent_plane(z*exp(x) + exp(z + 1) + x*y + y - 3, {x: 0, y: 3, z: -1})
display(n, plane)                         # <2, 1, 2>, 2*x + y + 2*z - 1

Tangent plane $2x+y+2z=1$; normal line $\dfrac{x}{2}=\dfrac{y-3}{1}=\dfrac{z+1}{2}$.