Some calculus questions (1 Viewer)

Ragerunner

Your friendly HSC guide
Joined
Apr 12, 2003
Messages
5,472
Location
UNSW
Gender
Male
HSC
2003
A bit stuck here..

Find the equation of the tangent to x^3 + y^3 - xy^2 = 1 at (1,1)


Also, what is the easiest way to find out whether a function is continuous or differentiable when given a function equation?



Thanks!
 

:: ck ::

Actuarial Boy
Joined
Jan 1, 2003
Messages
2,414
Gender
Male
HSC
2004
mmm hint...

implicit differentiation... then plug 1,1 in....
---------

and to tell if a function is continuous at a given point, wot u need to do is use limits, say at the point where x = a

lim(x->a<sup>-</sup>)f(x) = f(a) = lim(x->a<sup>+</sup>)f(x)

in terms of differntiability at the given point (x=a again), it needs to be
1) continuous at x=a and
2) lim(x->a<sup>-</sup>)f'(x) = f'(a) = lim(x->a<sup>+</sup>)f'(x)
 
Last edited:

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
Find the equation of the tangent to x^3 + y^3 - xy^2 = 1 at (1,1)
Implicit differentiation is the way to go, differentiate both sides with respect to x:

x^3 + y^3 - xy^2 = 1
3x^2 + 3y^2*dy/dx - (y^2 + dy/dx*2xy) = 0
dy/dx * (3y^2 - 2xy) = y^2 - 3x^2
dy/dx = (y^2 - 3x^2)/(3y^2 - 2xy)
subbing (x,y)=(1,1),
dy/dx = -2

so the gradient of the line is -2. Seeing as it goes through (1,1):
y = -2x + 3
 
Last edited:

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
Also, what is the easiest way to find out whether a function is continuous or differentiable when given a function equation?
From the fundamental definition of the continuity:

A function f is continuous at x=a iff:

for every e>0 there exists a d>0 such that:

whenever |x - a| < d,
then |f(x) - f(a)| < e

So for example if you're asked to prove from the fundamental definition that y=x^2 is continuous at x=0:

whenever |x - 0| = |x| < d,
we require |x^2 - 0^2| = x^2 < e

since |x| >= x^2 for x near 0,
it will suffice to require that |x| < e, in which case we can choose:
d = e

therefore y=x^2 is continuous at x=0.
-----------------------------------------
Obviously this isn't the easiest way to work out if a function is continuous, you only use this method if you're asked to prove a function is continuous from the fundamental definition. Otherwise see if it is composed of functions you know that are continuous.

To prove a function f is differentiable at x=a, it is quite easier.

f is differentiable at x=a if:

lim (h->0) {f(a+h) - f(a)}/ h exists
 

McLake

The Perfect Nerd
Joined
Aug 14, 2002
Messages
4,187
Location
The Shire
Gender
Male
HSC
2002
Can you please post your uni questions in the extra-cirircular topic from now on, thanks ...
 

:: ck ::

Actuarial Boy
Joined
Jan 1, 2003
Messages
2,414
Gender
Male
HSC
2004
i thought we needed 2 know that as part of curve sketching in 4unit?
 

McLake

The Perfect Nerd
Joined
Aug 14, 2002
Messages
4,187
Location
The Shire
Gender
Male
HSC
2002
Originally posted by :: ryan.cck ::
i thought we needed 2 know that as part of curve sketching in 4unit?
Part i) Yes
Part ii) I don't think so ...
 

:: ck ::

Actuarial Boy
Joined
Jan 1, 2003
Messages
2,414
Gender
Male
HSC
2004
ah ok well we covered it in tutoring briefly so yeh.. =\

oh well doesn't hurt to know it for hsc :)
 

McLake

The Perfect Nerd
Joined
Aug 14, 2002
Messages
4,187
Location
The Shire
Gender
Male
HSC
2002
Originally posted by :: ryan.cck ::
ah ok well we covered it in tutoring briefly so yeh.. =\

oh well doesn't hurt to know it for hsc :)
And it may help at uni ...
 

Ragerunner

Your friendly HSC guide
Joined
Apr 12, 2003
Messages
5,472
Location
UNSW
Gender
Male
HSC
2003
Thanks for the replies. Understand it now :)

I got few more questions a bit confused with.

1) Determine the limit as x -> infinity of : sqroot(4x^2+7x-2) - (2x+1)


2) Determine the limit as x -> 0 of : (x^2)*sin(1/x)


3) f(x) = (x^3 - 6x^2 + 11x - 6) / (x-a)

This is not continuous at x = a. For which values of a is the discontinuity removable?
 

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
1) Determine the limit as x -> infinity of : sqroot(4x^2+7x-2) - (2x+1)
If you see something like this in the form of lim {x -> ?} (a - b), where both a and b go to infinity, the following is often a handy trick to use:

lim {x -> ?} a - b
= lim {x -> ?} (a^2 - b^2)/(a+b)

so
lim {x -> infinity} sqrt(4x^2 + 7x - 2) - (2x+1)
= lim {x -> infinity} (4x^2 + 7x - 2 - (2x + 1)^2)/(sqrt(4x^2 + 7x - 2) + 2x + 1)
= ...
= 3/4

2) Determine the limit as x -> 0 of : (x^2)*sin(1/x)
since -1 <= sin(1/x) <= 1 for all real x,
-x^2 <= x^2 * sin(1/x) <= x^2 for all real x,
so lim{x->0} -x^2<= lim{x->0} x^2 * sin(1/x) <= lim{x->0} x^2
0 <= lim{x->0} x^2 * sin(1/x) <= 0
so by the pinching theorem,
lim{x->0} x^2 * sin(1/x) = 0

f(x) = (x^3 - 6x^2 + 11x - 6) / (x-a)
In order for there to be a removable discontinuity at x=a, (x-a) must be a factor of the polynomial x^3 - 6x^2 + 11x - 6 (otherwise the graph will asymptote to +- infinity)

so you need to solve the polynomial
a^3 - 6a^2 + 11a - 6 = 0, using techniques from 3U HSC maths.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top