₦airaland Forum

Cymbal:
nah

ok. Just google "intersecting chord theorem" because i can't draw the diagram here

Cymbal:
it helps buh please kindly give me full solution and explain how k=8/3.THANKS.

heard of Newton's theorem for the radius of curvature of a line?

agentofchange1:
How will I go about this plz..?.
The function f(x,y) satisfies
1) f(0,y)=y+1
2)f(x+1,0)=f(x,1)
3)f(x+1,y+1)=f(x,f(x+1,y)) for all non negative integers x,y.
determine f(4,1981)

startin wit the third, set x=0
then;
f(1,y+1)=f(0,f(1,y)), now apply rule 2 to the RHS to get
f(1,y+1)=1+f(1,y). And this is all we need, we can apply this recurrence y times to get;
f(1,y+1)=1+f(1,y)=2+f(1,y-1)=...=y+1+f(1,0).
Hence,
f(1,y)=y+f(1,0)............eqn4
from eqn2, put x=0, then;
f(1,0)=f(0,1).
from rule1,
f(0,1)=1+1=2.
Hence, f(1,y)=y+2..............eqn5

so eqn1 and eqn5 give us;
f(0,y)=y+1 and
f(1,y)=y+2.

Next, set x=1 in eqn3 and use eqn5. Repeat the process until u get to f(4,y) and set y=1981 to get your result. I will not solve it completely cos i don't have the time.

[quote author=jackpot post=32654806][/quote]hi jackpot.
Question 3 is analogous to a ball rolled off a cliff with initial velocity equal to that of the space craft

Cymbal:
Help me with this : Find the equation of the circle which touches the axis of y at a distance +4 from the origin and cuts off an intercept 6 from the axis of x.
TAGS: agentofchange1 ,jackpot,laplacian.

at least three points are required to specify a circle, so we must determine the second intercept on the x-axis; if the second intercept is k, then; k=4²/6=8/3
so the three points are; (0,4), (6,0), (8/3,0). Hope it helps!!

Nature130:
Help me with these questions ; find the values of x and y in ; x^y+y^x=17 and x+y=5. (2) find the value of x in 4^x=8x.

4^x=8x or 2^2x-3=x
this clearly shows that x must be some power of 2, i.e x=2^t for some positive integer t. Hence,
2^{2^(t+1)-3}=2^t, or
2^t+1-3=t, now t cannot divide 2 otherwise it divides 3, so using Fermat's Little Theorem, we have;
2^t+1-3 # 0 (mod t)
or
2*2^t-3 # 0 (mod t)
or
2*2-3 # 0 (mod t)
or
4-3 # 0 (mod t)
or
1 # 0 (mod t)
hence, t is a positive divisor of 1. Hence, t=1, but x=2^t. Hence, x=2

agentofchange1:
nice try sir, av known of this method too but the problem is , its still an approximate, since we truncate @ the 3rd term of expansion , why not 4th or 5th or higher ?. that's because we already know the answer to be 2 by trial -by-error , which is not always the ideal way of solving mathematical problems .


Now can it work for this ?

4^x = x^4 .?

or 3x = 8^x

x⁴=4^x
first; (x⁴)^1/x=4, since the RHS is an integer, x must divide 4;
similarly;
x=(4^x)^1/4, again since LHS is an integer, 4 must divide x.
Hence, if x divides 4 and 4 divide x, then x=4.

For; 3x=8^x, since 3 cannot divide 8, no integer x can satisfy that equation

Nature130:
Help me with these questions ; find the values of x and y in ; x^y+y^x=17 and x+y=5. (2) find the value of x in 4^x=8x.

x^y+y^x=17 and x+y=5, using modular arithmetic, and letting # denote "congruent to" we have;
x^y # 17 (mod y) and
x # 5 (mod y)
hence; 5^y # 17 (mod y). Since y is relatively prime to 5 and 17, we have, by Fermat's Little Theorem,
5^y # 5 # 17 (mod y), hence, 12 # 0 (mod y), so y is a positive divisor of 12 and is less than 5, the only candidate are
y= 2 or 3 or 4, same argument goes for x since the equation is symetrical. So,
x= 2, or 3, or 4 and
y=2, or 3, or 4.
Suppose x<y, then, 2x<x+y=5 hence x<2.5. If x is less than x then x=2 and hence, from y=5-x, y=3

Find the differential equation corresponding to;
y=Ae^3x+Be^x.............(1)

Solution:
since there are two unknown constants (A & B), we need two differential equations.
y'=3Ae^3x+Be^x..............(2.)

y"=9Ae^3x+Be^x................(3.)

if, from eqn(1), we substitute for Be^x in (2) and (3) we get;
y'=2Ae^3x+y...................(4)

y"=8Ae^3x+y...................(5)

now, y"-4*eqn(4) gives
y"-4y'+3y=0

AlphaMaximus:
Good morning, Math Clinic! Kudos to all that have been thoroughly active on this thread, its not easy. I seem to be in a fix as regards Complex Analysis: Non-Linear Transformations of the form w=(az+b)/(cz+d), which can be found in page 927 of Advanced Engineering Math By K.A Stroud......its specifically involves mapping a circle of modulus, |z|=2 onto the w- plane by a particular transformation equation of w= (z+2j)/(z+j). And the question demands that i determine the centre and radius of the resulting circle in the w-plane which normally is easy but for an impedement i encountered which i shall expatiate on.

w= (z+2j)/(z+j), w(z+j)=z+2j or

z=(2j+wj)/(w-1)=[(u+2)j-v]/[(u-1)+vj] or
taking conjugates;
z=[(u+2)j-v]*[(u-1)-vj]/[(u-1)^2+v^2]

={[(u-1)(u+2)+v^2]j+[-uv+v(u+2)]}/[(u-1)^2+v^2]
so
x=[-uv+v(u+2)]/[(u-1)^2+v^2]
and
y=[(u-1)(u+2)+v^2]/[(u-1)^2+v^2]

if u make this substitution u 'll get the centre

agentofchange1:
What are the next three numbers in this series
4,6,12,18,30,42,60,72,102,108,?,?,?

(the bolded)
i don't have the table of primes but i 'll give u the general rule:
1.) write the pairs of twin primes in ascending order
2.) write the greater of each pair in ascending order
3.) subtract 1 (one) from each term of sequence (2.) above.
Observation; it turns quite amazing that each term (aside the first) turns out to be divisible by 3. I'll leave it to the author of this post to offer a proof if he can!!

Determine the two values of c for which the line 3x+4y+c=0 is a tangent to the circle x^2+y^2-6x-2y-15=0.

The centre of the circle is Q=(3,1). If the given point is actually a tangent, then the normal thru the point @ which it is tangent, i.e the normal passes thru Q=(3,1). Now, since the slope of the tangent is -3/4, the slope of the normal must be 4/3, hence, the equation of the normal is; (y-1)/(x-3)=4/3 or y+3=4x/3, substitute this for y in the equation of the circle,
x^2+(-3+4x/3)^2-6x-2(-3+4x/3)-15=0
or
25x^2/9-38x/3=0
i.e x=0 or 114/25, correspondingly, y=-3 or 77/25,
the two points where the normal cut the circle (or the two points where the tangents touch the circle) are; A=(0,-3) and B=(114/25,77/25), substitute each cordinate in the equation of the straight line to get c.
I.e c=12 and c=-26.
(cross check, i have no calculator)

Prove that the line 3x +4y=13 is a tangent to the circle x^2+y^2-2x-3=0 and find the equation of the two tangents perpendicular to this.

If the line is a tangent, then, eliminating y should give two real but equal values of x. Hence, 16x^2+(4y)^2-32x-48=0 or
16x^2+(13-3x)^2-32x-48=0 or
25x^2-110x+121=0
if b^2=4ac, then we are done (pls verify that).
To find the two tangents perpendicular to the first, we take a line parallel to the first and passes thru the center. The center Q=(1,0) the line is 3x+4y=c but iit passes thru the center, so c=3, hence 3x+4y=3, we now find where this line cut the circle, 16x^2+(3-3x)^2-32x-48=0 or
25x^2-50x-39=0,
get the two points x1 and x2, from 3x+4y=3 get y1 and y2, so the points of intersection are A=(x1,y1) and B=(x2,y2). The slope of the perpendicular must be 4/3 so its equation must be 3y-4x=k, subtitute the cordinate of A into the line to get k, do the same B. And that gives the equation of the two perpendiculars.

y=§cos²@xsin²@d@

=§(cos@xsin@)²d@

=§[(1/2)xsin2@]²d@

=§1/4xsin²2@d@

=§1/8x2sin²2@d@

=§1/8x(1-cos4@)d@

=1/8x(@-sin4@/4)+C

NOTE; 1-2sin²x=cos2x

emmyeuler1:
it is not second circle oooo......it is circumcircle i meant

k. Solve three equations choosing two at a time and u'll get the points of intersections. Substitute each points of intersection in the general equation of the circle; x^2+y^2+2ax+2by+c=0, and solve the three simultaneous equations for a, b and c.....its too bulky 2 solve without calculator...

Find the equation and radius of d second circle of the triangle formed by d three lines; 2y-9x+26=0, 9y+2x+32=0, 11y-7x-27=0.
I dont understand this questio. Is it the circumcircle we are lookin 4? or the inscribed circle?

D length of d tangent from d point (3,2) to d circle x^2 +y^2-2x-3y+k=0 is 9 units, find k.

Centre of d circle, C=(1,3/2).
Length of of given point from center L^2=(3-1)^2+(2-3/2)^2=17/4.
Since d the system forms a pythagorean triangle, the radius is found from;
r^2=17/4-9^2.
This shows that the circle does not have a REAL radius, and so, k cannot be real

Find the eqns of d sides of d triangle ABC where A, B, C are d points (5,7), (3,3), (7,1) respectively. Hence, show that the triangles ABC has angles of 90, 45 and 45. Verify this results by finding the lengths of d triangle.

Slope of line AB is (3-7)/(3-5)=2
eqn of line AB is y=2x-3

slope of line BC is (1-3)/(7-3)=-1/2
eqn of line BC is y=-x/2+9/2

slope of line AC is (1-7)/(7-5)=-3
eqn of line AC is y=-3x+22

now, two lines are perpendicular if the product of their slope is -1.
Multiplyin slope of AB and BC we get; 2*-1/2=-1.
Now, d angle between two lines is found from;
tan@=(m1-m2)/(1+m1*m2), where m1 and m2 are the slope of d lines. I dont have calculator, complete the rest.

What must be d value of k in order that; 2x+y-3=0, kx+3y+1=0, x+y+7=0 may meet in a point. Discuss d cases when k=3 and k=6

d point of intersection of d first and last lines are obtained first. Subtractin their eqns give; x=10 and y=-17. For the second line to satisfy this point, we must have;
10k-3(17)+1=0, or k=52/10.
(i)caseI; if k=3, we have, 3x+3y+1=0 or by dividing thru by 3; x+y+1/3=0. Hence the line is parallel to the third line (since their slope is the same) and is higher the third line (and also the point of intersection) since the intercept is greater.
(ii) caseII; if k=6, then, 6x+3y+1=0 or by dividing by 3, 2x+y+1/3=0. Again, this line is parallel to the first line and is lower, below it.

Find d eqn of d cicle which passes thru d point (1,1), half a radius of sqr(10)/2 and whose centre lies on d line y=3x-7.
If d centre passes thru d given line, then d co-ordinate of d centre must satisfy d line.
Let d coordinate of d centre b (a,b), then; b=3a-7, the radius r=sqr(10) and d eqn of d circle is;
(a-1)^2+(b-1)^2=10
or substitutin for b gives
(a-1)^2+(3a - 8 ) ^2=10
put a-1=z, then,
z^2+(3z-5)^2=10
or
10z^2-30z+15=0
or
2z^2-6z+3=0
z=[3(+ or -)sqr(3)]/2.
From z=a-1, get a & from b=3a-7, get b.

Substitute for a and b in d eqn of d circle above

Find d area of d region between d curves; y=x^4 & y=2x-x^2.

First we find their point of intersection and ensure that none of d two curves crosses d x-axis bewteen dis pionts of intersection.
x^4=2x-x^2 or
(x^3+x-2)x=0.
Hence, x=0, x=1 are d only real solutions. Equatin each of d given eqns to zero, we obtain; x^4=0 i.e x=0. Then, 2x-x^2=0, x=0 or 2. Hence, none of d two equations crosses d x-axis between x=0 and x=1. So,
area bounded by y=x^4 is;
A1=§x^4dx=x^5/5 limit is from 0 to 1. So, A1=1/5.
Area bounded by y=2x-x^2 is,
A2=§(2x-x^2)dx=x^2-x^3/3 limit is from 0 to 1. Hence, A2=2/3.
So area between d two curves; A=A2-A1=2/3-1/5=7/15 units

Show that; If 2ⁿ+1 is a prime, then, n is a power of 2

jackpot:
In an hour,
Mr A must have done (1/t_1) times the work
Mr B - (1/t_2) times the work
so, in an hour, both of them working together must have done (1/t_1+1/t_2) times the work
so, working together, they completes the work in 1 / (1/t_1 + 1/t_2) hours. This is the same as t₁t₂/(t₁+t₂) hours.

@Laplacian and others, please feel free to criticize.

that's pretty!! (i didn't think of dis approach)

Scrypt:
Let the farm work done be 1 farmwork. Hence the rate at which Mr. A works is 1/t₁ (Work/hrs). Similarly Mr. B's rate since the quantity of work is similar (1 farm work) is 1/t₂.

Hence both of them working together would accomplish 1/t₃ rate.

= 1/t₁ + 1/t₂ = 1/t₃

I'm sure you can take it from there.

hahahaha...u pay me back in my coin? (nice work!!)

Mr. A can complete a farm work in t₁ hours. Mr. B can finish the same work in t₂ hours. How would it take them if they worked together?

Arithmetic:
If tan²x = 1 + 2tan²y. Show cos2x + sin²y = 0. Thanks…

recall, sec²y=1+tan²y.
From the LHS,
sec²x-1=1+2(sec²y-1) or
sec²x=2sec²y or
cos²y=2cos²x or
cos²y=(2cos²x-1)+1, does the expression in the bracket look familiar? *****





That's cos2x.
So, cos²y=cos2x+1
or
cos²y-1=cos2x
can u finish it from here? Hop it helps!

jackpot:
There's no theorem saying that if you differentiate a DE, the degree will increase or decrease. In other words, there is no theorem comparing the degree of a DE with that of the differentiated DE. So, you shouldn't compare both degrees.

Maybe you're trying to draw analogy from the fact that the difference between the order of a DE and with that of the differentiated DE is 1.

Catch my drift, Sir?

i catch ur drift ma. But, with due respect, just because u've not seen a theorem on that does not erase d fact that it does exist. And do i always need a theorem to confirm what i know?

jackpot:
since y'+y=0 is equivalent (you said so) to y''+y'=0, it is equivalent to y''+y'-(y'+y)=0, which is the same thing as y''-y=0.

The DE y''-y=0 now has the solution y=C₁e^x+C₂e^-x.

but the solution of y'+y=0 always satisfies any of its extended D.E doesn't it? But the matter of the extended D.E's solution (y"-y=0) satisfying the original D.E (either y'+y=0 or y'-y=0) is decided by initial value. The values on e^x are not on e^-x, so with d initial value inputed in the solution of the extended D.E, either C₁ or C₂=0 as appropriate. Turning tables around, i ask u; if the D.E y'+sin(y" )=0 is of infinite degree, why should the D.E obtained from it have degree 2?

jackpot:
using your logic,
y=Ce-^x is equivalent to y=C₁e^x+C₂e^-x which is (again, by your logic) equivalent to y=Ce^x.

Then, by transitivity of an equivalence relation, y=Ce^-x is equivalent to y=Ce^x.

Do you really believe that both solutions are equivalent?

what extension of the D.E for y=Ce-^x did u get the solution y=C₁e^x+C₂e^-x if u want to play by my logic?

jackpot:
hmmm.

Suppose for contradiction that y'+y=0 is equivalent to y''+y'=0. By differentiating once more, we must have that y''+y'=0 is also equivalent to y'''+y''=0.

But these guys are an equivalence relation, so by transitivity,
y'+y=0 is equivalent to y''+y'=0 which is equivalent to y'''+y''=0 implies that y'+y=0 is equivalent to y'''+y''=0.
Now, y'+y=0 has the solution y=C^-x, but y'''+y''=0 has the solution y=C₁+C₂x+C₃e^-x.
Are these two solutions equivalent?

yes!
A simple substitution C₁=C₂=0 will pay the price. It stil boils down to initial value.

jackpot:
maybe in the world of cartoons(where things happen the way we want), we may say that y'+y=C is equivalent to y'+y=0 for all values of the constant C.

everyone is entitled to his opinion

jackpot:
Both differential equations won't even have the same number of initial conditions. The one you differentiated will need extra one initial condition.

For example, y'+y=0 needs only one initial condition before you obtain the particular solution free from arbitrary constants, whereas y''+y'=0 needs two initial conditions. So, i am thinking it makes them even far from being equivalent?

C₁ came as a result of an integration (see above). So u are saying in essence that, differentiating a function and integratin the result will not yield the initial function.
Because i differentiate; y'+y=0 to get y"+y'=0 and i integrate this to get y'+y=C₁

jackpot:
hmmm. I can see the mathematics. But. . .

Let's take another look.

Given y'+y=0, differentiating both sides w.r.t. x gives a new differential equation
y''+y'=0.

It is really a heresy to say that both differential equations are equivalent since y'+y=0 has the solution y=C e^-x, whereas y''+y'=0 has the solution y=C₁+C₂e^-x.


Because y'+y=0 is easier to handle and it's solution satisfies y''+y'=0, I can not conclude that both differential equations has the same (general) solution.

So, I will say that the two differential equations you compared are not equivalent.

Or, what do you think?

y should they not be equivalent?
Integration as we know is d inverse of differentiation. Does it not imply that if one differentiates a function and integrates it he should still have the same general solution? Let's use ur example.
Given
y'+y=0 after differentiating, y"+y'=0, should we not have same function (in a more general form) if we integrate? §y"dt+§y'dt=§0xdt so that y'+y=C₁ and this is slightly more general and does no harm to the original function (in terms of generality). Let's solve the two equations; from dy/dt=C₁-y
we get
dy/(C₁-y)=dt so that
-In(C₁-y)=t+K
or
C₁-y=e^-t-K
or
y=C₁-e^-Kxe^-t
so that y=C₁+C₂e^-t. This tells u how C₁ came into the picture. And as i have explained above, the are one and the same. Questioning its validity amounts to doubtin if the following is true; given y=t²+3........(1.)
differentiate, y'=2t, now let's integrat d last to get y=t²+C............(2.)
and u say (1.) and (2.) are not the same? Leave it or take it INITIAL VALUE SETTLES EVERYTHING MY DEAR.
So on what ground do u accuse me of heresy, and what is the FOUNDATION of ur claim of inequivalence?

agentofchange1:
greetings .....


happy new year

please help out with this. integral of ln(sinx) dx = ?

hi ben!
Let I=§In(sinz)dz,

let y=In(sinz).........(1.), dy/dz=cosz/sinz, so ,

I=§ytanz.dy............(2.) from (1.)

e^y=sinz or e^-2y=cosec²z

so that (e^-2y-1)^1/2=cotz so that (2.) becomes;

I=§y/(e^-2y-1)^1/2dy or

I=§ ye^y/(1-e^2y)^1/2dy

if we integrate by part by keepin y constant first we get;

I=-y(1-e^2y)^1/2+§(1-e^2y)^1/2dy

now let
H=§(1-e^2y)^1/2dy

let u=e^y so that du/dy=u

Hence

H=§(1-u²)^1/2du/u

Or

H=§(u^-2-1)^1/2du

now let u=cosw then du/dw=-sinw, Hence,
H=§tanw.-sinwdw

or

H=§-sin²w/cosw dw

=§(cos²w-1)/cosw dw

=§(cosw - 1/cosw) dw

=sinw-§1/cosw dw

let G=§1/cosw dw

=§cosw/cos²w dw

let k=sinw then dk/dw=cosw
hence
G=§1/(1-k²) dk

using partial fraction and resolvin we get

G=1/2 In[(1+k)/(1-k)]

= 1/2 In[(1+sinw)/(1-sinw)]

therefore
H=sinw-1/2 In[(1+sinw)/(1-sinw)]
but
I=-y(1-e^2y)^1/2+H
hence

I=-y(1-e^2y)^1/2+sinw-1/2 In[(1+sinw)/(1-sinw)]+Constant

where w=cos^-1(e^y)
and y=In(sinz)

simplifying further if u wish,

I=[1-In(sinz)]cosz +1/2 In[(1+cosz)/(1-cosz)]+constant