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Re: Nairaland Mathematics Clinic by jackpot(f): 6:13am On Apr 08, 2015 |
tohero:well, I think for each hit, there are two firing angles (as long as the firing angle isnt 45degrees) namely theta and 90-theta. I also think the question was asking for the lesser of the two since the greater of the two will make the particle spend longer time in the air. tohero:why did you set v=0? Are you assuming the height to be the maximum height? It wasn't given like that nah. What do you think? |
Re: Nairaland Mathematics Clinic by jackpot(f): 6:15am On Apr 08, 2015 |
benji93:are you using pythagora's? I think it would work like that if there's no acceleration due to gravity acting towards the surface of the earth 1 Like |
Re: Nairaland Mathematics Clinic by tohero(m): 12:41pm On Apr 08, 2015 |
jackpot:We are looking for the initial velocity which means the object has been stationary, hence, it implies that velocity=zero. jackpot:We are already given the angle as 75o. What other angles are you suggesting for ![]() Our aim is to find the initial velocity not --the lesser of two angles. That's my solution anyway. Bring forth your possible solution probably we may get convinced. |
Re: Nairaland Mathematics Clinic by benji93: 2:09pm On Apr 08, 2015 |
jackpot:rit if gravity is taken into consideration then s= ut + (1/2)gt^2,but since it is shot at a component of the velocity would be acting vertically upwards so we can find theta first using the sides of the right angled triangle and then find t since we know u,g and we have found s=x=the hypotenuse. or probably s = usin theta + (1/2)gt^2,where s is the vertical of the right angled triangle, |
Re: Nairaland Mathematics Clinic by Miscellaneous(m): 7:24pm On Apr 08, 2015 |
jackpot: it would be easier to solve all this ur projectile if u consider just the vertical distance, horizontal & tan of the angle at first. They are cases of projectiles of say a footballer hitting a ball at angle theta & the ball landing on a roof ymetres from the ground at a distance xmetres from the footballer. If you do not follow the approach I gave, you will probably bask in the euphoria that the answer you got was correct when it isn't. |
Re: Nairaland Mathematics Clinic by jackpot(f): 10:02pm On Apr 08, 2015 |
Miscellaneous:alright, Sir. solve one please. |
Re: Nairaland Mathematics Clinic by jackpot(f): 10:10pm On Apr 08, 2015 |
tohero:thanks, it was a mix-up. I was referring to a similar question in my text. |
Re: Nairaland Mathematics Clinic by Miscellaneous(m): 10:24pm On Apr 08, 2015 |
jackpot: lol……… I'm relaxing & I'm not with calc. here use, y= (xtan©) - { (gx²)/(2u²cos²©)} where; y=vertical distance x=horizontal distance © = given angle u= initial velocity also put R= Ux • T into use R= range Ux = ucos© T= flight time also put say, Vy = usin© - gT into use; Vy = vertical component of velocity try it ……… u na scholar na! |
Re: Nairaland Mathematics Clinic by benji93: 11:26pm On Apr 08, 2015 |
jackpot:angle = 48.743 |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 2:53pm On Apr 09, 2015 |
hmmmm. see math scholars abeg . busy solving with passion , greets you all guys @ jackpot sorry couldn't post solutions, but am glad. guys are already doing so , really gat tight schedules with school tinz & other engagements . Nice solving #shalom |
Re: Nairaland Mathematics Clinic by jackpot(f): 6:15am On Apr 10, 2015 |
Miscellaneous:lol. You too na super-scholar. Take five! ![]() 1 Like |
Re: Nairaland Mathematics Clinic by emmyeuler1: 8:23am On Apr 10, 2015 |
jackpot:lol...........ur own na just to make fun of guyz here.......loool |
Re: Nairaland Mathematics Clinic by Nature130: 9:17am On Apr 10, 2015 |
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. |
Re: Nairaland Mathematics Clinic by Nature130: 9:18am On Apr 10, 2015 |
Richiez: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. |
Re: Nairaland Mathematics Clinic by Admissnandjobs(m): 9:40am On Apr 10, 2015 |
hi |
Re: Nairaland Mathematics Clinic by Admissnandjobs(m): 10:11am On Apr 10, 2015 |
hi |
Re: Nairaland Mathematics Clinic by jackpot(f): 3:23pm On Apr 10, 2015 |
benji93:show steps, Sir |
Re: Nairaland Mathematics Clinic by benji93: 4:11pm On Apr 10, 2015 |
jackpot:tan theta = 28.5/25 theta = 48.743degrees |
Re: Nairaland Mathematics Clinic by jackpot(f): 3:33pm On Apr 11, 2015 |
benji93:you're using SOHCAHTOA? ![]() 3 Likes |
Re: Nairaland Mathematics Clinic by Nwiboazubuike(m): 7:56pm On Apr 11, 2015 |
Differentiate X^x^x |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 10:27pm On Apr 11, 2015 |
Nwiboazubuike: x^x^x[ x^x(lnx+1)lnx + x^(x-1) ] don't ask 4 workings except u wanna learn not test our intelligence. |
Re: Nairaland Mathematics Clinic by factorial1(m): 10:28pm On Apr 11, 2015 |
Nwiboazubuike:Ok, here is the solution. Let Y = X^x^x firstly... let U = x^x so that... Y = X^U Now... dy/dx will be equal to dY/dU x dU/dx Considering U = x^x first. Applying natural logarithm to the base of exponential to both sides... we have lnU = lnx^x which can be written as lnU = xlnx, Now, the derivative of the equation gives (1/U)dU/dx = x(1/x) + lnx(1)... Using product rule. And don't forget that, U is differentiated implicitly with respected to x. (1/U)dU/dx = 1 + lnx Therefore dU/dx = U(1 + lnx). Since U = x^x, therefore dU/dx = x^x(1 + lnx). Alright, considering Y = X^U also. Applying log into base of e to both sides... we have lnY = UlnX... Also, following the same rule for this also... we have (1/Y)dY/dU = U(1/x) + lnx(x^x + x^xlnx)... since du/dx = x^x + x^xlnx (1/Y)dY/dU = x^x(1/x) + x^xlnx + x^x(lnx^2) dY/dU = x^(x+1) + x^xlnx + x^x(lnx^2) So therefore dY/dx = dY/dU x dU/dx which is equal to x^(x+1) + x^xlnx + x^x(lnx^2) x x^x(1 + lnx). Finally dY/dx = [x^(x+1) + x^xlnx + x^x(lnx^2)] X [x^x(1 + lnx)]. You can expand further by opening the bracket. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 10:34pm On Apr 11, 2015 |
@ factorial1 u try sir , differential calculus is quite cheep , but even though integral calculus is the reverse of it , its not usually funny to do so . Now in this regards , how can we then obtain back the integrand .? |
Re: Nairaland Mathematics Clinic by factorial1(m): 10:39pm On Apr 11, 2015 |
agentofchange1:Lol... By finding the integral of course, which I'm sure won't be quite easy. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 10:44pm On Apr 11, 2015 |
Nature130: there exist NO analytical solutions to such equations yet , except approximate or graphical though 1) (x,y) =(2,3)=(3,2) 2) x=2 you could be the first mathematician to develop that , happy trying . 1 Like |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 10:46pm On Apr 11, 2015 |
factorial1: sure , that's the point . |
Re: Nairaland Mathematics Clinic by factorial1(m): 10:47pm On Apr 11, 2015 |
Nature130:May not be able to type the solutions tonight but lemme give an Hint to solving question 2. Since 4^x = 8x... Divide both sides by 4 to get 4^(x-1) = 2x. Don't forget 4^(x-1) can be written as (1 + 3)^(x-1)... equating that to 2x gives (1 + 3)^(x-1) = 2x... Then solve using Binomial Expansion . The answer should be 2 and 0 cause it will surely lead to quadratic equation. Note that you only need the first 3 terms of the expansion. |
Re: Nairaland Mathematics Clinic by factorial1(m): 10:51pm On Apr 11, 2015 |
agentofchange1:Question 2 is solvable bro.. Try it out. Gonna lead to you using Binomial expansion. Can't really be typing tonight... Have got a lot. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 10:57pm On Apr 11, 2015 |
factorial1: 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 |
Re: Nairaland Mathematics Clinic by factorial1(m): 11:05pm On Apr 11, 2015 |
agentofchange1: Solving that question doesn't require using all the terms...as a matter of fact... not for only this question. There are some maths question that one will just have to make use of some part of the equation. |
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