IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: Chingoo on December 27, 2010, 02:19:21 pm
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Given that y = (1-sinx)/cosx, show that dy/dx = 1/(1+sinx).
It seems simple but idk why I'm not getting the answer =/
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Let S be sinx & C be cosx.
Use the y=u/v => dy/dx = [Udv/dx - Vdu/dx]/V2
Therefore
dy/dx = [(1-S)(-S) - C(0-C)]/C2
[-S + S2 + C2]/C2
Using identity: C2=1-S2
[-S + S2 + 1 - S2]/(1-S2)
(-S+1)/(1-S2))
Use identity: a^2 - b^2 = (a+b) (a-b)
(1-S)/(1+S)(1-S)
Finally dy/dx = 1/(1+sinx)
(:
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Let S be sinx & C be cosx.
Use the y=u/v => dy/dx = [Udv/dx - Vdu/dx]/V2
Therefore
dy/dx = [(1-S)(-S) - C(0-C)]/C2
[-S + S2 + C2]/C2
Using identity: C2=1-S2
[-S + S2 + 1 - S2]/(1-S2)
(-S+1)/(1-S2))
Use identity: a^2 - b^2 = (a+b) (a-b)
(1-S)/(1+S)(1-S)
Finally dy/dx = 1/(1+sinx)
(:
+ rep. ;)
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Thanks. (:
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Thanks, but now I'm confused. Isn't the quotient rule for differentiation
y=u/v => dy/dx = [v.du/dx - u.dv/dx]/v^2
Granted you managed to get the answer, but I'm pretty sure that's the quotient rule.
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Sorry, I haven't used the rule for months now - rusty. :S
My bad. So you probably got the answer after using the correct rule right?
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&& you get the correct answer using the product rule
y=(1-S)(C-1)
dy/dx = (1-S)(S/C2)+(C-1)(0-C)
=S(1-S))/(1-S2) + 1/C (-C)
=S(1-S)/(1+S)(1-S) - 1
=(S + 1 - S)/(1+S)
=1/(1+S)
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Thanks a lot! =D +rep
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No problemm. (:
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i kind of find maths AS very hard....especially differentiation.....i have a doubt here:
find the coordinates of the point on the curve y=2x^2 -x-1, where the tangent is parallel to the line; y=3x-2?
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Practice and you'll get used to it eventually. (:
gradient of line y=3x-2 is 3 b/c in y=mx+c form m is the gradient.
So now we have to find the point on the curve where gradient is 3.
dy/dx= 4x-1
tangent to curve parallel where 4x-1=3
4x=4
x=1
at x=1, y=2-1-1=0
therefore coordinates should be (1,0)
(:
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Differentiation is easy and Integration is easier.
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Practice and you'll get used to it eventually. (:
gradient of line y=3x-2 is 3 b/c in y=mx+c form m is the gradient.
So now we have to find the point on the curve where gradient is 3.
dy/dx= 4x-1
tangent to curve parallel where 4x-1=3
4x=4
x=1
at x=1, y=2-1-1=0
therefore coordinates should be (1,0)
(:
thank you so much wispher
i got it now...........;)
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i have a doubt here;
The production cost per kilogram C(in thousand pounds) when x kilogram of a chemical are made, is given by;
C= 3x+100/x , x>0. find the value for x for which the cost is minimum and the minimum cost??
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Differentiate the equation. Set dy/dx = 0 and solve for x.
Determine if the stationary point is a minimum or maximum. Select the minimum point and plug it into the original equation.
This is the minimum cost.
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Differentiate the equation. Set dy/dx = 0 and solve for x.
Determine if the stationary point is a minimum or maximum. Select the minimum point and plug it into the original equation.
This is the minimum cost.
sorry ari if ur getting fedup of me asking so much questions........but i got the answer.......thanks alot!! ;D
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thank you so much wispher
i got it now........... ;)
Glad to help. (:
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sorry ari if ur getting fedup of me asking so much questions........but i got the answer.......thanks alot!! ;D
No, I'm not fedup.
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No, I'm not fedup.
cool
8)
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ok......so i hv yet another doubt;
a rectangle box with a lid is made from thin metal. its length is 2x cm and its width is x cm. if the box volume is 72 cm^3
a. show that the area of the metal used is equal to 4x^2 +216/x.
b. find the value of x so that the area A is minimum?
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l = 2x cm
w = x cm
V = 72 cm3
V = l * b * h
71 = 2x * x * h
h = 72/ 2x2
Area = 2 ( lb + lh + bh)
= 2 ( 2x * 72/ 2x2 + x * 72/ 2x2 + 2x * x )
= 2 ( 72/x + 36/x + 2x2 )
= 2 ( 108/x + 2x3 )
= 216/x + 4x2
or
4x2 + 216/x
Hence, proved.
Identify the domain.
4x2 >= 0
x>= 0
however, 1/x > 0 and not = 0
so domain is x belongs to positive integers.
dA/dx = 8x - 216/x2
substitute this as 0
you get x = 3
check whether it is the point of minimum.
Yes, it is, so x = 3 is the value where A is minimum.
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l = 2x cm
w = x cm
V = 72 cm3
V = l * b * h
71 = 2x * x * h
h = 72/ 2x2
Area = 2 ( lb + lh + bh)
= 2 ( 2x * 72/ 2x2 + x * 72/ 2x2 + 2x * x )
= 2 ( 72/x + 36/x + 2x2 )
= 2 ( 108/x + 2x3 )
= 216/x + 4x2
or
4x2 + 216/x
Hence, proved.
Identify the domain.
4x2 >= 0
x>= 0
however, 1/x > 0 and not = 0
so domain is x belongs to positive integers.
dA/dx = 8x - 216/x2
substitute this as 0
you get x = 3
check whether it is the point of minimum.
Yes, it is, so x = 3 is the value where A is minimum.
thanks vin.... :D
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l = 2x cm
w = x cm
V = 72 cm3
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Yes, it is, so x = 3 is the value where A is minimum.
+Rep (:
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ok......so i hv yet another doubt;
a rectangle box with a lid is made from thin metal. its length is 2x cm and its width is x cm. if the box volume is 72 cm^3
a. show that the area of the metal used is equal to 4x^2 +216/x.
b. find the value of x so that the area A is minimum?
Essentially all of these questions follow the same general idea :
Differentiate equation
Set equal to zero
Solve to find max and min points
Use max and min points in original equation depending on what the question requires.
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Essentially all of these questions follow the same general idea :
Differentiate equation
Set equal to zero
Solve to find max and min points
Use max and min points in original equation depending on what the question requires.
thanks ari
+rep
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a new make of personel organizer costs $100 each to manufacture and market. market research indicates that the number, N thousand, of organizers that will be sold in the first year is given by; N=-1S+150, where $S is the selling price. the numbers which wll be sold at two particular selling prices are as given below
S N
120 30
140 10
a. express the profit per organizer in terms of S and hence show that the total profit in the first year is $T thousands, where T=-15000+250 S-S^2
b. find the selling price, which will produce the maximum possible total profit in the first year and find this total profit( you should show that this profit is maximum and not minimum)?
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here