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Re: Nairaland Mathematics Clinic by johnpaul1101(m): 1:33pm On Jan 19, 2013 |
biolabee:general biolabee and richiez, if you truly want to see my work on this question, u must know that your mathematical knowledge is more than mine, so plz however premature my ideas, solution, or logic may b to u, dont make me look like a mathematical fool, and to tell u the truth, i like this idea of posting my work, nd if not correct, u help me |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 1:38pm On Jan 19, 2013 |
for the question, this was what i did: let the running speed be x let the walking speed be y, from the fist statement, (8km/50min)x + (2km/50min)y=10km/50min..........eq 1 from the second statement (4km/75min)x + (6km/75min)y=10km/75...........eq 2 now, the parameter(minute) need to b converted to hour to give the correct unit(km/hr).....in doing that., |
Re: Nairaland Mathematics Clinic by Richiez(m): 1:47pm On Jan 19, 2013 |
johnpaul1101: back to simultaneous, plz help me.....i've been trying this since this evening, but i keep getting confused: well for the sake of others who wants to learn, here's the solution. first of all, we must recall that Time = Distance/Speed .........(1) lets convert unit of time to hours only 50mins = 50/60 =5/6 hrs 1hr 15mins =75mins = 75/60 = 5/4 hrs let running speed = x and walking speed = y now applying eqn(1) we get 5/6 = 8/x + 2/y ......................(2) 5/6 = (8y + 2x)/xy 5xy = 6(8y + 2x) 5xy = 48y + 12x x(5y-12) = 48y x = 48y/(5y-12)................(3) also; 5/4 = 6/x + 4/y ...............(4) putting value of x obtained in eqn(3) into eqn(4) we get; 5/4 = 6(5y-12)/48y + 4/y multiply thru by 'y' 5y =4[6(5y-12)/48 + 4] 5y= 5y/2 -6 +16 5y/2 = 10 y =20/5 = 4 from eqn(3) x= 48y/(5y-12) x= 48(4)/[5(4) -12)] x= 192/8 =24 therefore running speed= x = 24km/hr and walking speed = y = 4km/hr |
Re: Nairaland Mathematics Clinic by johnpaul1101(m): 2:01pm On Jan 19, 2013 |
Richiez:but the answer here says 16km/hr and 6km/hr |
Re: Nairaland Mathematics Clinic by Richiez(m): 2:19pm On Jan 19, 2013 |
johnpaul1101: for the question, this was what i did: now you anticipated the solution very well, but i must tell you, knowing how to play with your distance, speed and time relationships give you an edge over how to generate your simultaneous equations. 1. distance = speed * time 2. speed = distance/time 3. time = distance/speed now i did not choose the first and the second one because the time given was for distance covered after running and walking, and not for either running or walking alone. for example it will be a big error to say 8km = running speed/(5/6)hrs note 5/6 hrs = 50mins this is 50 mins is not for running alone, it was time taken after running and walking. i chose the 3rd relationship because it keeps total time on one side of the eqn and the i can equate it to the addition of the time after running and the time after walking for example 5/6 hrs = 8km/running speed + 2km/walking speed note; time=distance/speed 1 Like |
Re: Nairaland Mathematics Clinic by SOBSISRAEL(m): 2:28pm On Jan 19, 2013 |
[quote author=Richiez] nice try bro, although you made some some mistakes especially at the point where you took log of both sides. e.g it ought to be log(x+y)= log5 and not logx + logy = log5. but i must admit, you did a very good job. here's another approach; x + y = 5.......(1) x^x + y^y =31.......(2) from eqn(1), x=5-y, hence we substitute this value for x in eqn(2). (5-y)^(5-y) + y^y = 31.......(3) now, we have reduced the problem to what PLS AM NEW HERE HELP ME IN MATH |
Re: Nairaland Mathematics Clinic by Richiez(m): 2:36pm On Jan 19, 2013 |
johnpaul1101: Richiez: this goes to show that even the minutest mistakes could cost alot in mathematics |
Re: Nairaland Mathematics Clinic by Richiez(m): 5:23pm On Jan 19, 2013 |
SOBSISRAEL: now, we have reduced the problem to what PLS AM NEW HERE HELP ME IN MATH you're most welcome. how do you want us to help you? please feel free to express yourself clearly |
Re: Nairaland Mathematics Clinic by biolabee(m): 6:28pm On Jan 19, 2013 |
johnpaul, this is not to make fun of you but use the opportunity to learn you used km and mins which are not the common units of speed and as richie showed u, you have to take steps and recheck More questions but please put whAT YOU have done so far before we reply |
Re: Nairaland Mathematics Clinic by kasbeats(m): 7:12pm On Jan 19, 2013 |
i salute all ma 5* generals for here lo........abeg its this question i would like sum1 to solve oo.....i knw its gonna b too simple for ppl like Richiez,ositadima,biolabee and d likes........find §^2V/§y^2 wen V=tan`(y/x)........§ means partial differentiation,tan` means arc tan......tanks bros |
Re: Nairaland Mathematics Clinic by Nobody: 5:47am On Jan 20, 2013 |
kasbeats: i salute all ma 5* generals for here lo........abeg its this question i would like sum1 to solve oo.....i knw its gonna b too simple for ppl like Richiez,ositadima,biolabee and d likes........find §2V/§y2 wen V=tan`(y/x)........§ means partial differentiation,tan` means arc tan......tanks bros I want to believe your question is, find ∂2v/∂y2 if v = tan^(-1) (y/x)? If it is then here is the solution. To solve this we have to use the derivative of tan(-1)A or arcTanA which is 1/(A2 + 1). Therefore, if A represents => y/x, the first derivative of the entire expression can be expressed using chain rule => ∂[tan^(-1) (y/x)]/∂y => ∂v/∂y = {1/[(y/x)2 +1]}.∂(y/x)/∂y ∂v/∂y = {1/[(y/x)2 +1]}.1/x ∂v/∂y = x/(y2 + x2) Having computed the first derivative of function wrt y, we can now differentiate the result wrt y for the second derivative of the given function => ∂2v/∂y2. x/(y2 + x2) Lets put t = y2 + x2, ∂t/∂y = 2y, ∂[x]/∂t = 0 Then the function becomes = x/t and using quotient rule its derivative wrt t => {t[ ∂(x)/∂t] - x( ∂t/∂y)}/t2 Subtituting t and the values of the other derivatives in the above formula, yields=> = - x(2y)/t2) = -2yx/(y2 + x2)2 Hence ∂[x/(y2 + x2)]/∂y => -2yx/(y2 + x2)2 Which is also the 2nd derivative of the implicit function => :. ∂2[arcTan(y/x)]/∂y2 => -2yx/(y2 + x2)2 2 Likes |
Re: Nairaland Mathematics Clinic by Richiez(m): 11:42am On Jan 20, 2013 |
I'm impressed |
Re: Nairaland Mathematics Clinic by ositadima1(m): 1:58pm On Jan 20, 2013 |
arctan(y/x)=v so, tan(v)=y/x I am going to manipulate with sin^2(v)+cos^2(v)=1 Now, dividing sin^2(v)+cos^2(y/x)=1 by cos^2(v)... u know sin^2(v)/cos^2(x)=tan^2(v) and 1/cos^2(v)=sec^2(v) 1+tan^2(v)=sec^2(v) So, sec^2(v)=1+(y/x)^2 recalling arctan(y/x)=v sec(arctan(y/x))=sec(v) square both sides sec^2(arctan(y/x))=sec^2(v)=1+(y/x)^2 sec^2(arctan(y/x))=1+(y/x)^2 now, using chain rule and taking partial derivative of both sides with respect to x we have... ∂[sec^2(arctan(y/x))]/∂x=∂[1+(y/x)^2]/∂x ∂[arctan(y/x))]/∂x*2[sec^2(arctan(y/x))]*tan[(arctan(y/x)]*(-(y/x)^2)=-2*y^2/x^3 now, tan[(arctan(y/x)]=y/x sec^2(arctan(y/x)=1+(y/x)^2 substituting back ∂[arctan(y/x))]/∂x*2*[1+(y/x)^2]*(y/x)*(-(y/x)^2)=-2*y^2/x^3 ∂[arctan(y/x))]/∂x*[1+(y/x)^2]*-2*y^3/x^3=-2*y^2/x^3 ∂[arctan(y/x))]/∂x*[1+(y/x)^2]=y ∂[arctan(y/x))]/∂x=y/[1+(y/x)^2] second partial derivative ∂^2[arctan(y/x))]/(∂x)^2=∂[y/[1+(y/x)^2]/∂x using chain rule again... p ∂^2[arctan(y/x))]/(∂x)^2=-y[1+(y/x)^2]^(-2)*-2*y^2/x^3 ∂^2[arctan(y/x))]/(∂x)^2=2xy(x^2+y^2)^2 DoubleDx there is no minus in the final solution, check again |
Re: Nairaland Mathematics Clinic by Nobody: 3:06pm On Jan 20, 2013 |
kasbeats: i salute all ma 5* generals for here lo........abeg its this question i would like sum1 to solve oo.....i knw its gonna b too simple for ppl like Richiez,ositadima,biolabee and d likes........find §^2V/§y^2 wen V=tan`(y/x)........§ means partial differentiation,tan` means arc tan......tanks bros @Osita, I think you didn't get the question properly, it says we should differentiate wrt to y and not x. You did a good job but you solved yours wrt x. That explains why your first derivative differs from mine. There is no need removing your workings. It's perfect if we are to differentiate with respect to x. That could help some folks, just indicate that you are differentiating wrt x then check the one wrt y. 1heart bro. |
Re: Nairaland Mathematics Clinic by SpicyMimi(f): 4:01pm On Jan 20, 2013 |
I NEED YOUR ASSISTANCE GUYS?!! You dont know how happy i am to find this thread, feeling so relaxed now! Please help me solve the following questions on Matrix. The Matrix of units of products produced by Daddy K cream factory in three shifts is given: (2 3 7) ( X-Y 1 3) (x+Y 3 5) You are required to determine: (a) Rhe values of x and y, if Matrix A is Symmetric. (b) The gross revenue vector (AP) for each product, if the price $P per unit of the product is given by the vector P = (45) (30) (55) please i need the solution before tomorrow!!! Pleaseeeeeeee!!! |
Re: Nairaland Mathematics Clinic by SpicyMimi(f): 4:06pm On Jan 20, 2013 |
Seriously! You guys are GURUS!!! |
Re: Nairaland Mathematics Clinic by ositadima1(m): 5:16pm On Jan 20, 2013 |
doubleDx: ha haha, my bad... |
Re: Nairaland Mathematics Clinic by ositadima1(m): 5:24pm On Jan 20, 2013 |
SpicyMimi: I NEED YOUR ASSISTANCE GUYS?!! You dont know how happy i am to find this thread, feeling so relaxed now! Please help me solve the following questions on Matrix. You want us to do your homework for you, while u dey chill out abi? |
Re: Nairaland Mathematics Clinic by Nobody: 5:53pm On Jan 20, 2013 |
SpicyMimi: I NEED YOUR ASSISTANCE GUYS?!! You dont know how happy i am to find this thread, feeling so relaxed now! Please help me solve the following questions on Matrix. Questiob 1 a [ 2 3 7] [(x - y) 1 3] [(x+y) 3 5] A matrix A is said to be symmetric if A equals A transpose. i.e A => AT which means that if we interchange rows and columns of A, the matrix remains unchanged. :. If the above matrix is A, then A transpose will be => [2 x-y x+y] [3 1 3] [7 3 5] Looking at the two matrices, A and AT. The number on the first row second column of AT is given by the expession (x - y) which equals 3 in the matrix A. Hence x - y = 3 ......(1) Also, the number on the first row third column of AT is given by the expression (x + y) which equals 7 in the matrix A. :. x + y = 7 .....(2) Adding equation (1) and (2) yields=> x - y = 3 x + y = 7 2x = 10 x = 5 subtitute x in equation (1) 5 - y = 3 y = 5 - 3 y = 2 :. x = 5 and y = 2. 1 Like |
Re: Nairaland Mathematics Clinic by ositadima1(m): 5:59pm On Jan 20, 2013 |
doubleDx: See babe I wan chike..., u don quickm answer finish, finish it up joo... |
Re: Nairaland Mathematics Clinic by Nobody: 6:04pm On Jan 20, 2013 |
ositadima1: Lol bro, I'm on mobile, typing and walking help her finish the b part. Carry on bruv, I'm with you all the way |
Re: Nairaland Mathematics Clinic by ositadima1(m): 6:09pm On Jan 20, 2013 |
doubleDx: brother me dey walka too, branch some where and end what you started. |
Re: Nairaland Mathematics Clinic by Nobody: 6:16pm On Jan 20, 2013 |
ositadima1: Lol, what I started? When did we start doing that on this thread? That's quite funny! If that's the case then I think only the op should answer questions on here. Na wa oh Well, I'm helping just like you are doing so when I get home I'll help her finish the b part. |
Re: Nairaland Mathematics Clinic by ositadima1(m): 6:18pm On Jan 20, 2013 |
doubleDx: Yes sir! |
Re: Nairaland Mathematics Clinic by Nobody: 6:36pm On Jan 20, 2013 |
ositadima1: Oh, all the questions you've been solving here is so you can gain popularity and get chics online? Well, some of us are married and we do this to help people and not for some e-chics |
Re: Nairaland Mathematics Clinic by ositadima1(m): 6:44pm On Jan 20, 2013 |
doubleDx: lol, what did I do to you this Sunday |
Re: Nairaland Mathematics Clinic by Nobody: 6:47pm On Jan 20, 2013 |
ositadima1:hahahahahahahaha, u na bad guy |
Re: Nairaland Mathematics Clinic by Nobody: 7:02pm On Jan 20, 2013 |
ositadima1: No beef bruv. I've got no beef with you! But the way you addressed my post was uncalled for. It's not compulsory to help folks. That others are not willing to help doesn't mean we are the best on here, I studied Engineering and not BEd Mathematics. I'm just giving the best I can out of my free will. If I decide not to log on to nairaland again, I'll leave and never come back. Nothing matters to me here because nairaland doesn't pay me. If you are quoting my post next time, make sure you write something reasonable. Gosh! We are not kids anymore bruv! |
Re: Nairaland Mathematics Clinic by SpicyMimi(f): 7:07pm On Jan 20, 2013 |
ositadima1:hahhahahhahahahhahahahahhahahahahhah:...haba nah! U r sooooooooooo hilariousssssssssss!!!! Actually i aint o! I got loads of assignments on my table...so have been so busy surfing d internet! Chill ke? Nothn like dat o! God bless You Doubledx! You ve just made ma day, cause right now, am the most happppiiiieeeeeeeesssst girl!!!! Words cant describe how excited and grateful i am right now! God Bless you and your family abundantly in Jesus name.Amen!!! |
Re: Nairaland Mathematics Clinic by ositadima1(m): 7:08pm On Jan 20, 2013 |
Let it go...bro |
Re: Nairaland Mathematics Clinic by SpicyMimi(f): 7:17pm On Jan 20, 2013 |
doubleDx:Thank you!!! Thank You!!! Thank you!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! ThaNk You!!! Thank You!!! And Thank You!!! God am sooooo happpyyyy!!!! And sooooo Grateful!!!!!! 1 Like |
Re: Nairaland Mathematics Clinic by Richiez(m): 7:31pm On Jan 20, 2013 |
Doubledx Ride on, you're a star @Ositadima1 common show dis chic what you got! |
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