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Re: Nairaland Mathematics Clinic by Nobody: 1:59pm On Nov 18, 2013
rhydex 247:
Solution.
let y=xlog~e 3x
let u=x. du/dx=1
let v=log~e 3x. dv/dx=1/x.
usin product rule.
dy/dx=udv/dx+vdu/dx.
dy/dx=x(1/x)+log~e 3x(1).
dy/dx=1+log~e 3x. All is well. Similarly to ur no 1 questn apply d same method.
,..9c 1
log _e (u) = ln u
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 4:11pm On Nov 18, 2013
echibuzor:

Show your work and we can figure it out together, although questions of this sort aint my strong suit..
@ d first 1 i did something like:
(1+3)^x=8x.
and i applied (1+x)^n method but my answers deviated thoroughly.
d second 1 hmm.....i no even know hw to start am
Re: Nairaland Mathematics Clinic by Laplacian(m): 6:02pm On Nov 18, 2013
Mbahchiboy: plz u guys should help solve dis using "d citizen" approach.i tried doing it but i dont know my mistake.
(1)if 4^x=8x find x.
(2) find x if (4/x)^x=2x
2^2x=8x, or 2^(2x-3)=x or

2^(2x-3)/4^(2x-3)=x/4^(2x-3)

(1/2)^(2x-3)=x*(1/4)^(2x-3)

(0.5)^(2x-3)=x*(0.25)^(2x-3)

(1-0.5)^(2x-3)=x*(1-0.75)^(2x-3)

by binomial approx.

1-(2x-3)*0.5...=

x*[1-(2x-3)*0.75+...]

re-arrangin gives
1-x+3/2=x-3x/4*(2x-3)
or
4-4x+6=4x-6x^2+9x
or
6x^2-17x+6=0

for d second question;
(4/x)^x=2x or 4^x=2x^(x+1) or

2^(2x-1)=x^(x+1)...i hope u can go on from there?

1 Like

Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 8:41pm On Nov 18, 2013
Laplacian:
2^2x=8x, or 2^(2x-3)=x or

2^(2x-3)/4^(2x-3)=x/4^(2x-3)

(1/2)^(2x-3)=x*(1/4)^(2x-3)

(0.5)^(2x-3)=x*(0.25)^(2x-3)

(1-0.5)^(2x-3)=x*(1-0.75)^(2x-3)

by binomial approx.

1-(2x-3)*0.5...=

x*[1-(2x-3)*0.75+...]

re-arrangin gives
1-x+3/2=x-3x/4*(2x-3)
or
4-4x+6=4x-6x^2+9x
or
6x^2-17x+6=0

for d second question;
(4/x)^x=2x or 4^x=2x^(x+1) or

2^(2x-1)=x^(x+1)...i hope u can go on from there?
THANKU VERY MUCH SIR 4 D HIGHLIGHT.....BUT PLZ COMPLETE D NO 2 I DON PAUSE
Re: Nairaland Mathematics Clinic by Laplacian(m): 10:52pm On Nov 18, 2013
Mbahchiboy: THANKU VERY MUCH SIR 4 D HIGHLIGHT.....BUT PLZ COMPLETE D NO 2 I DON PAUSE
2^(2x-1)=x^(x+1).....(eqn1)

let y=x^(x+1).........(eqn2)

substitut (2) in (1) to get

2^(2x-1)=y........(eqn3)

from (3),
2^(2x-1)/4^(2x-1)=y/4^(2x-1)
or
(0.5)^(2x-1)=y*(0.25)^(2x-1)
or
(1-0.5)^(2x-1)=y*(1-0.75)^(2x-1)
or
1-0.5(2x-1)=y[1-0.75(2x-1)]
.......(eqn4)

from (2), y=x^(x+1)
or
y*[1/(4x)]^(x+1)=(x/4x)^(x+1)
or
y*[1+1/(4x)-1]^(x+1)=(0.25)^(x+1)
or
y[1+(x+1){1/(4x)-1}+...]=

(1-0.75)^(x+1)
or
y[1+(x+1)(1-4x)/4x]=1+0.75(x+1)....(eqn5)

eliminat y from eqn4 and eqn5 and solve d cubic to obtain x...both equtaion can b easier solved using logarithmic approximation...but obviously dis thread is not a convenient place to use logarithmic operations...
Re: Nairaland Mathematics Clinic by rhydex247(m): 9:09am On Nov 19, 2013
Question 1.
Show that the four points whose position vectors are 3i-2j+4k, 6i+3j+k, 5i+7j+3k and 2i+2j+6k are coplanar.
Question 2.
If a ^ r=b+¥a and a.r=3 where a=2i+j-k and b=-i-2j+k. Then find r and ¥.
Question 3.
Prove the identity. (a ^ b).(c ^d)=(a.c)(b.d)-(a.d)(b.c). Where ^ means cap or cross
Re: Nairaland Mathematics Clinic by Nobody: 10:04am On Nov 19, 2013
An error of P% is made in measuring the radius of a circle. Calculate the approximate percentage error in the measurement of the circumference.
Re: Nairaland Mathematics Clinic by echibuzor: 10:21am On Nov 19, 2013
smurfy: An error of P% is made in measuring the radius of a circle. Calculate the approximate percentage error in the measurement of the circumference.
(P2 - 2rP) / ^r2 x 100
Re: Nairaland Mathematics Clinic by factorial1(m): 10:29am On Nov 19, 2013
How many odd numbers with three digits can be formed from the digits 1, 2, 3, 4, 5 if you can use any of the digits more than one?
Re: Nairaland Mathematics Clinic by factorial1(m): 10:30am On Nov 19, 2013
explain hw you arrived @that bro...?
echibuzor:
(P2 - 2rP) / ^r2 x 100
Re: Nairaland Mathematics Clinic by echibuzor: 10:51am On Nov 19, 2013
factorial1: explain hw you arrived @that bro...?
I sort of read the question wrong, my previous answer is for the percentage error in area. The circumference is p/r x 100
Re: Nairaland Mathematics Clinic by Nobody: 12:00pm On Nov 19, 2013
factorial1: How many odd numbers with three digits can be formed from the digits 1, 2, 3, 4, 5 if you can use any of the digits more than one?

Here goes...

The last digit can be arranged in 3 ways, since it must end with 1, 3 or 5. Each of the first two digits can be arranged in 5 ways, since each digit can be used more than once. The answer therefore is 5 * 5 * 3 = 75 ways.
Re: Nairaland Mathematics Clinic by Nobody: 12:05pm On Nov 19, 2013
echibuzor:
I sort of read the question wrong, my previous answer is for the percentage error in area. The circumference is p/r x 100

If the above is your answer, then it's wrong. The options are

A. (1/2)P%
B. P%
C. (3/2)P%
D. 2P%

Note: Clear working'll be appreciated.
Re: Nairaland Mathematics Clinic by echibuzor: 12:42pm On Nov 19, 2013
smurfy:

If the above is your answer, then it's wrong. The options are

A. (1/2)P%
B. P%
C. (3/2)P%
D. 2P%

Note: Clear working ll be appreciated.
Correct Measurement gives you 2^r (note that '^' means pi)
measurement with the error in it gives 2^(r+p) == 2^r + 2^p
Error in measurement is 2^p (in whatever unit it is m, cm, km)
percentage is (2^p / 2^r) x 100
So you see how I arrived at my answer...
I used Linear Approximation and got the same thing.. So it's either I am hungry (which I am BTW) or the Option is just not there or there is something I am not seeing (This one is most likely)
Re: Nairaland Mathematics Clinic by factorial1(m): 12:57pm On Nov 19, 2013
Correct!
smurfy:

Here goes...

The last digit can be arranged in 3 ways, since it must end with 1, 3 or 5. Each of the first two digits can be arranged in 5 ways, since each digit can be used more than once. The answer therefore is 5 * 5 * 3 = 75 ways.
Re: Nairaland Mathematics Clinic by factorial1(m): 12:59pm On Nov 19, 2013
Find the equation of the tangent(s) where x = 2 on the curve x^2 + y^2 - 2x + y = 6.
Re: Nairaland Mathematics Clinic by Nobody: 1:16pm On Nov 19, 2013
smurfy: An error of P% is made in measuring the radius of a circle. Calculate the approximate percentage error in the measurement of the circumference.

The only way I could solve this is by using numbers.

Here goes...

Let True Radius = 20cm
Let Error Radius = 18cm.
PE = (TR - ER)/TR * 100%
This gives 10% (My P is therefore 10. Now let's find PE for the circumference. Whatever I get is directly compared with the value of P above.)

Now, C = 2 * pi * r
True C = 2 * pi * 20
= 40 * pi cm
Error C = 2 * pi * 18
= 36 * pi cm
PE = (TC - EC)/TC * 100%.
This gives (4pi/40pi) * 100% = 10%.

This shows that if there's a 10% error in the measurement of the radius of a circle, there'll be a 10% corresponding error in the measurement of its circumference. Hence, the correct answer is P%.

The issue is whether this can be solved neatly, algebraically without using this primary school method. That, I'd like to know.
Re: Nairaland Mathematics Clinic by echibuzor: 1:24pm On Nov 19, 2013
smurfy:

The only way I could solve this is by using numbers.

Here goes...

Let True Radius = 20cm
Let Error Radius = 18cm.
PE = (TR - ER)/TR * 100%
This gives 10% (My P is therefore 10. Now let's find PE for the circumference. Whatever I get is directly compared with the value of P above.)

Now, C = 2 * pi * r
True C = 2 * pi * 20
= 40 * pi cm
Error C = 2 * pi * 18
= 36 * pi cm
PE = (TC - EC)/TC * 100%.
This gives (4pi/40pi) * 100% = 10%.

This shows that if there's a 10% error in the measurement of the radius of a circle, there'll be a 10% corresponding error in the measurement of its circumference. Hence, the correct answer is P%.

The issue is whether this can be solved neatly, algebraically without using this primary school method. That, I'd like to know.
This is literally what I showed in my workings. My answer is right, it just lacked the conclusion. P% in the question is gotten by error p / actual r x 100.
which is repeated in the final answer..
Re: Nairaland Mathematics Clinic by Nobody: 1:25pm On Nov 19, 2013
factorial1: Find the equation of the tangent(s) where x = 2 on the curve x^2 + y^2 - 2x + y = 6.

Differentiate the equation of the circle implicitly. Find y when x =2. Make dy/dx the subject, slot in x = 2, y = ... to find the gradient. Then use y - y1 = m(x - x1) to find the required equation.

Note: x1 = 2, y1 = -3 or 2
Re: Nairaland Mathematics Clinic by rhydex247(m): 1:33pm On Nov 19, 2013
rhydex 247: Question 1.
Show that the four points whose position vectors are 3i-2j+4k, 6i+3j+k, 5i+7j+3k and 2i+2j+6k are coplanar.
Question 2.
If a ^ r=b+¥a and a.r=3 where a=2i+j-k and b=-i-2j+k. Then find r and ¥.
Question 3.
Prove the identity. (a ^ b).(c ^d)=(a.c)(b.d)-(a.d)(b.c). Where ^ means cap or cross
Re: Nairaland Mathematics Clinic by ameer2: 2:27pm On Nov 19, 2013
Prove dat
cosecX-1/cosecX+1=[1-tan(x/2)/1+tan(x/2)]^2. . .any1 dat cn prove dis will have a special reward.I promise.
Re: Nairaland Mathematics Clinic by Laplacian(m): 3:43pm On Nov 19, 2013
smurfy: An error of P% is made in measuring the radius of a circle. Calculate the approximate percentage error in the measurement of the circumference.
let C denote d circumference and let # denot pi
then, C=2#r, now differentiat C wit respet to r to get
dC/dr=2#
or
the error in the circumference;
dC=2#*dr,
now divide thru by C=2#r to get
dC/C=2#*dr/(2#r)
or
dC/C=dr/r=p%
showin dat d percentag error in measurin d circumference must be d same as dat made in measurin d radius
Re: Nairaland Mathematics Clinic by Laplacian(m): 3:52pm On Nov 19, 2013
ameer!:
Prove dat
cosecX-1/cosecX+1=[1-tan(x/2)/1+tan(x/2)]^2. . .any1 dat cn prove dis will have a special reward.I promise.
...i suggest u use brackets apprioprately to for ambiguity (for those who 'll solv d question)...
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 3:55pm On Nov 19, 2013
factorial1: Find the equation of the tangent(s) where x = 2 on the curve x^2 + y^2 - 2x + y = 6.
BEFORE I START PLZ cross check d ques i think u omitted d value of y
Re: Nairaland Mathematics Clinic by Nobody: 6:01pm On Nov 19, 2013
Mbahchiboy: BEFORE I START PLZ cross check d ques i think u omitted d value of y

Yeah. I think you're right. I got two values for y, thereby giving two different values for dy/dx, -2/5 and 2/5.

Equation of tangent is 2x + 5y - 14 = 0 or
-2x + 5y + 19 = 0

I don't think we're meant to get two answers. He probably missed something.
Re: Nairaland Mathematics Clinic by Mbahchiboy(m): 6:55pm On Nov 19, 2013
[quote author=Laplacian]
2^(2x-1)=x^(x+1).....(eqn1)
let y=x^(x+1)...(eqn2)
substitut (2) in (1) to get
2^(2x-1)=y(eqn3
from (3),
2^(2x-1)/4^(2x-1)=y/4^(2x-1)
or
(0.5)^(2x-1)=y*(0.25)^(2x-1)
or
(1-0.5)^(2x-1)=y*(1-0.75)^(2x-1)
or
1-0.5(2x-1)=y[1-0.75(2x-1)]
from (2), y=x^(x+1)
or
y*[1/(4x)]^(x+1)=(x/4x)^(x+1)
or
y*[1+1/(4x)-1]^(x+1)=(0.25)^(x+1)
or
y[1+(x+1){1/(4x)-1}+...]=
(1-0.75)^(x+1)
or
y[1+(x+1)(1-4x)/4x]=1+0.75(x+1[\quote/]thanks sir i think i'l try it 4 better understandin
Re: Nairaland Mathematics Clinic by Nobody: 7:47pm On Nov 19, 2013
Laplacian:
let C denote d circumference and let # denot pi
then, C=2#r, now differentiat C wit respet to r to get
dC/dr=2#
or
the error in the circumference;
dC=2#*dr,
now divide thru by C=2#r to get
dC/C=2#*dr/(2#r)
or
dC/C=dr/r=p%
showin dat d percentag error in measurin d circumference must be d same as dat made in measurin d radius

Thanks Laplacian. Thanks for this method.

Tell me, assuming you got something like dC/C = 1/2(dr/r), what would you have said about P%?
Re: Nairaland Mathematics Clinic by factorial1(m): 9:58pm On Nov 19, 2013
I omitted nothing der, the qt is correct n' accurate!
Mbahchiboy: BEFORE I START PLZ cross check d ques i think u omitted d value of y
Re: Nairaland Mathematics Clinic by Calculusfx: 10:13pm On Nov 19, 2013
Hmmmm
Re: Nairaland Mathematics Clinic by Calculusfx: 10:21pm On Nov 19, 2013
I have not been to the thread since morning...kudos to master laplacian,master echibuzor,master factorial(who used logic to abuse me on the quiz thread),my main oga rhydex,master chiboy and master smurfy...for not allowing the clinic to be silent as mortuary...i respect you all.

1 Like

Re: Nairaland Mathematics Clinic by Calculusfx: 10:24pm On Nov 19, 2013
ameer!:
Prove dat
cosecX-1/cosecX+1=[1-tan(x/2)/1+tan(x/2)]^2. . .any1 dat cn prove dis will have a special reward.I promise.
...hmmm.from rhs {(1-tan(1/2)x)/(1+tan(1/2)x)}^2 since tan(1/2)x=sin(1/2)x/cos(1/2)x...then {[1-sin(1/2)x/cos(1/2)x]/{1+sin(1/2)x/cos(1/2)x}^2...{LET'S DEAL WITH THE ONE IN THE BRACKET FIRST)multiply the numerator and denominator by cos(1/2)x which gives {(cos(1/2)x-sin(1/2)x)/(cos(1/2)x+sin(1/2)x)}^2...then square to give (sin^2(1/2)x-2sin(1/2)xcos(1/2)x+cos^2(1/2)x)/(sin^2(1/2)x+2sin(1/2)xcos(1/2)x+cos^2(1/2)x)...don't forget from trig. that cos^2(1/2)x+sin^2(1/2)x=1 and 2sin(1/2)xcos(1/2)x=sinx...substitute those to give...(1-sinx)/(1+sinx)...multiply the numerator and the denominator by 1/sinx...to give {1/sinx(1-sinx)/1/sinx(1+sinx)=(1/sinx-1)/(1/sinx+1)...don't forget 1/sinx=cosecx...substitute that to give cosecx-1/cosecx+1
QED
Re: Nairaland Mathematics Clinic by statusnet(m): 12:23am On Nov 20, 2013
Integrate (tan^2) x / tan2x dx

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