Saturday

July 26, 2014

July 26, 2014

Total # Posts: 50

**Calculus**

Thank you very much for this valuable guidance.

**Calculus**

Thanks for reply. Sorry for confusion. Second and II DC are same and only mean second diff. coeff. or derivative d^2y/dx^2. Nr is numerator and A is dy/dx as shown as equation(A)in the first line of answer. What I meant was that for calculating second DC we have to apply quoti...

**Calculus**

Q: If y=sinx/(1+tanx), find value of x not greater than pi, corresponding to maxima or minima value of y. I have proceeded thus- Equating dy/dx=0 we get{ (1+tanx)cosx-sinx.sec^2 x}/(1+tanx)^2=0……..(A) Or cosx+sinx=sinx.sec^2 x or cosx= sinx.sec^2-sinx Or cosx=sinx.ta...

**Calculus**

Thanks Mr. Steve for the guidance.

**Calculus**

Find the area cut off by x+y=3 from xy=2. I have proceeded as under: y=x/2. Substituting this value we get x+x/2=3 Or x+x/2-3=0 Or x^2-3x+2=0 Or (x-1)(x-2)=0, hence x=1 and x=2 are the points of intersection of the curve xy=2 and the line x+y=3. Area under curve above X axis f...

**Calculus**

Thanks for the advice. I checked the problem statment and answer several times and got the same result. I also suspect it to be a print mistake in the book.

**Calculus**

Find the area cut off by x=4 from the hyperbola x^2/9-y^2/4=1. Answer is 4.982 in the book. I have proceeded as under: Y=2/3*sqrt(x^2-9) and rhe reqd. area is double of integral 2/3*sqrt(x^2-9) from 3 to 4. Int= 2/3*[xsqrt(x^2-9)/2 – 9/2*log{x+sqrt(x^2-9)}] from 3 to 4 =x...

**math**

Differentiate cost function and equate it to 0, i.e. 0.1(2x)-1.2=0 or x=1.2/0.2=6; Second derivative is0.2which is+ve hence it corresponds to minimum cost level. Hence x=6,i.e. 600 bikes. Cost/bike for 600 bikes will be =0.1x36-1.2x6+4.993=1.393 hunderd $. For 590 and 610 bike...

**Calculus**

Thank you very much, Mr. Steve. It was actually x^2/2*logx but I feel it won't matter.

**Calculus**

Following 2 questions are from a book at a point where L’Hopital’s Rule, Squeeze Theorem etc. have not been discussed and limits (A) and (B) as given below are to be evaluated by simple methods like algebraic simplification etc. 1. Int. of (xlogx)dx from 0 to 1. Inde...

**Calculus**

Following 2 questions are from a book at a point where L’Hopital’s Rule, Squeeze Theorem etc. have not been discussed and limits (A) and (B) as given below are to be evaluated by simple methods like algebraic simplification etc. 1. Int. of (xlogx)dx from 0 to 1. Inde...

**Calculus**

Integral of dx/{(x+1)sqrt(x^2+4x+2)} is given as -arcsinh{(x+2)/(xsqrt3)}, but I am not getting it.? The hint given is 'Put 1/(x+1)=t' Using it, the expression reduces to Int. -1/sqrt(1+2t-t^2). How to proceed further to get the desired result?

**Calculus**

Integrate x dx/(1-x). I have proceeded thus- Int xdx/(1-x)=int -(x-1+1)/(x-1) =-Int[1+ 1/(x-1)]dx =-Int dx-Int dx/(x-1) =-x-log(x-1). On differentiating, we get original expression- d/dx[-x-log(x-1)]=-1-1/(x-1)=-x/(x-1)=x/(1-x). However, the answer in the book is -x-log(1-x)an...

**Calculus**

S=Integral xdx/sqrt(x-1). I have proceeded thus- Put sqrt(x-1)=u then x=u^2+1 and dx/sqrt(x-1)=2du. S=(u^2+1)2du/u =(2u+2/u)du=u^2+2 log u +C =(x-1)+ 2 log sqrt(x-1)=(x-1)+log(x-1)+C Required answer is 2/3*(x+2)sqrt(x-1)+C Have I proceeded correctly and if so can these 2 resul...

**Calculus**

How to integrate dx/(4-5 sin x) using t-substitution method(i.e. taking tan x/2=t)?

**Calculus**

Thanks. I checked and really got them as same. But can we analytically show them to be so? I tried but could not make it. Please guide.

**Calculus**

s=çdx/(4+5cos x). By using t-substitution, i.e. t=tan(x/2) we get cos x=(1-t^2)/(1+t^2) and dx=2dt/(1+t^2). Substituting in s and simplifying, we get s= 2çdt/4(1+t^2)+5(1-t^2)=2çdt/(9-t^2). Using standard result çdx/(a^2-x^2)...

**Calculus**

Book works out integral of sqtr(a^2-x^2)dx as a^2/2*arcsin(x/a)+x/2*sqrt(a^2-x^2) by taking x=a sin theta, and advises to work it out using x=a cos theta. I did and got the first term as -a^2/2*arccos(x/a)and the same second term. This means arcsin x/a=-arccos x/a which can be...

**algebra**

30x2+Ax1=(30+A)1.6 60+A=48+1.6A or 0.6A=12 or A=20. Total mix=50oz and cost=30*2+20=80$ Or 80/50=1.6$/oz.

**Maths**

3x-y=2 Multiplt it by 3 9x-3y=6 Add to it the other eq. 9x+3y=4 You get 18x=10 or x=5/9 Substitute in any eq. 3*5/9-y=2 or y=15/9-2=5/3 -2 or y=(5-6)/2=-1/6. Check: 3*5/9-(-1/6)=2 or 5/3+1/6=2 or (10+2)/6=2 ok. 9*5/9-3*1/3=4 or 5-1=4 ok.

**Calculus**

Thank yoy very much Mr. Steve for making it so clear and easy.

**Calculus**

I have done Int(1/ax)dx=1/a*Int(1/x)dx =[(log x)/a] + C. The working in the book is Int(1/ax)dx=1/a*Int(a/ax)dx =[(log ax)/a]+C. Which one is correct and where is the error in my working, if so?

**Physics**

v^2=u^2+2fs or f=(v^2-u^2)/2s =(7.4^2-79^2)/(2*710) =(54.76-6241)/1420=-4.36m/s^2 Negative acceleration (i.e. against the direction of motion)of 4.36m/s^2

**Physics**

That will be sec^2 per m but that is most unlikely to be required as generally time is taken as independent variable and variations are measured per unit time i.e. time rate of variation.

**Physics**

If time is on X axis and distance on Y axis then units will be m/sec^2.

**physics l**

Angle will be arctan 15/100 =arctan 0.15=8.53 degrees.

**physics**

23.5kmph=23500/3600=6.53m/s v=u+ft=6.53+0.98*4=6.53+3.92=10.45m/s

**physics**

v=u+ft or f=(v-u)/t=(10.6-13)/4=-2.4/4 =-0.6m/s^2. Now the u is 10.6m/s and f=-0.6m/s^2 v=u+ft=10.6-0.6*2=10.6-1.2=9.4 m/s^2

**calculus**

Thank you very much.

**calculus**

Mr. Graham, Thank you very much for elaborate explanation. Kindly also let me know if it is ok to find at what values of theta the r becomes 0 and take that as period of curve? If so, plotting the complete curve every time may not be required.

**calculus**

Thanks a lot for guiding. Kindly let me know if it should be a routine procedure to plot the complete curve everytime or can it be conveniently worked out analytically? I used this formula for integrating Int sin^2x dx=1/2*(x-1/2*sin2x)which is a std formula. Why the result di...

**calculus**

Find complete length of curve r=a sin^3(theta/3). I have gone thus- (theta written as t) r^2= a^2 sin^6 t/3 and (dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3) s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt =a Int Sqrt[sin^4(t/3){(sin^2(t/3)+cos^2(t/3)}]dt=a Int Sqrt[sin^4(t/3)dt =a ...

**calculus**

Thanks a lot for guiding,please.

**calculus**

Sorry, slight typo. Please reasd it as Int (0 to 2pi)2a*Int d theta=2a(2pi-0)=4a*pi

**calculus**

Find circumference of the circle r=2acos theta. s= Int (0 to 2pi) of Sqrt(4a^2cos^2 theta+4a^2sin^2 theta)d theta =Int (0 to 2pi)2a*Int theta d theta =2a(2pi-0)=4a*pi Book shows 2a*pi. Am I wrong somewhere?

**Calculus**

I reached upto Int csc u sec^2 u du from u=arctan 1 to 2 but am not clear how to go further.

**Calculus**

Find arc length of y=logx from x=1 to x=2. dy/dx)^2=1/x^2 arc length=Int of [sqrt(1+1/x^2)]dx =Int of [sqrt(1+x^2)/x^2] =Int of [sqrt(1+x^2)]/x from x=1 to x=2. How to proceed further to integrate?

**MATHS**

Find arc length of y=logx from x=1 to x=2. dy/dx)^2=1/x^2 arc length=Int of [sqrt(1+1/x^2)]dx =Int of [sqrt(1+x^2)/x^2] =Int of [sqrt(1+x^2)]/x from x=1 to x=2. How to proceed further to integrate?

**Maths**

Does it indicate that the answer 'b/a' given in the book may be wrong? I tried many times but could not get it.

**Maths**

Eq of curve is y=b sin^2(pi.x/a). Find mean value for part of curve where x lies between b and a. I have gone thus far- y=b[1-cos(2pi x/a)/2]/2 Integral y from a to b=b/2(b-a)-ab/4pi[sin(2pi b/a)-sin2pi) MV=b/2-[ab sin(2pi b/a)]/(b-a) Ans given is b/a. I am not getting further.

**chemistry**

When heated, elemental phosphorus, P4, produces phosphine, PH3, and phosphoric acid, H3PO4. How many grams of phosphine are produced if 56 grams of P4 have react

**MSJH**

When heated, elemental phosphorus, P4, produces phosphine, PH3, and phosphoric acid, H3PO4. How many grams of phosphine are produced if 56 grams of P4 have reacted?

**plzz help physics**

An experiment makes use of a water manometer attached to a flask. Initially the two columns in the gas manometer are at the same level and the air pressure in the flask and both sides of manometr is 1 atm. The experiment is set up when the air pressure is 7 degree Celcius. Th...

**physics**

An experiment makes use of a water manometer attached to a flask. Initially the two columns in the gas manometer are at the same level and the air pressure in the flask and both sides of manometr is 1 atm. The experiment is set up when the air pressure is 7 degree Celcius. The...

**physics**

An experiment makes use of a water manometer attached to a flask. Initially the two columns in the gas manometer are at the same level and the air pressure in the flask and both sides of manometr is 1 atm. The experiment is set up when the air pressure is 7 degree Celcius. The...

**chemistry**

no i think we dont need to use 150ul.. can u solve it without using 150ul?

**chemistry**

a solution has gluten concentration of 2000ug/mL at 150uL. if 2uLof that solution is taken out what is the quantity of gluten in ppm?

**bcc**

One canned juice drink is 30% orange juice, another is 10% orange juice. How many liters of each should be mixed together in order to get 20L that is 18% orange juice?

**physics**

A child operating a radio-controlled model car on a dock accidentally steers it off the edge. The car's displacement 1.0 s after leaving the dock has a magnitude of 12.8 m. What is the car's speed at the instant it drives off the edge of the dock?

**chemistry**

1.30 M = 1.30mols of solute/L of solvent mass of Cacl2 = 1.30 mols CaCl2 x 111g/mol CaCl2(molar mass)= 144.3g of CaCl2 (solute) Solution = 1.11g/ml mass of solution = 1.11g/ml x 1000ml/L x1L = 1110g (solution) mass of solvent = 1110g (solution) - 144.3g (solute)= 965.7g of sol...

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