General Maths Questions Watch

Kolya
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#121
Report 11 years ago
#121
Sine and cosine rules:

a^2 = b^2+c^2-2bc \ cosA \\\ \frac{sinA}{a} \ = \ \frac{sinB}{b} \ = \ \frac{sinC}{c}

The first can also be written as:
 cosA = \frac{b^2+c^2-a^2}{2bc}
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ImperceptibleNinja
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#122
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#122
You use the cosine rule when you either
-know the lengths of all sides [to find any angle],
-know one angle, and the lengths of the sides meeting at that angle [so you can find the length of the side opposite the angle]

You can use the sine rule if you know an angle and the length of a side opposite, and either one other side or angle:
Say you know A, B, and b
then a = bsinA/sinB
Say you know a, B and b
sinA = asinB/b

Basically, use them whenever you know some stuff about a triangle, and need to find another side length, or another angle.
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Ourkid
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#123
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#123
Could I ask someone to correctly expand Rcos(x+a) for me please?

I have the cos(a+b) = cosAcosB - sinAsinB

which would mean that Rcos(x+a) = Rcos(x)cos(a) - Rsin(x)sin(a)

However, in my notes it has Rcos(x)cos(a) + Rsin(x)sin(a)

If someone could clear up this minor query it would help a lot!

Thanks
w00tt
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jpowell
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#124
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#124
Your notes are wrong. It has a minus sign.
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Ourkid
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#125
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#125
(Original post by jpowell)
Your notes are wrong. It has a minus sign.
Cheers nice one

w00tt
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Ourkid
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#126
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#126
Sorry I have another small query.

If differentiating for example 4sin(2y+6) should I treat (2y+6) as it is? Or should it also be differentiated?

Does the answer become cos(2y+6) or 2cos(2y+6)?
Thanks so much
w00tt
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El Matematico
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#127
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#127
You need the chain rule to differentiate 4sin(2y+6) (with respect to y)

So, let R = 2y + 6
and z = 4sinR

=>dR/dy = 2
dz/dR = 4cosR

=>dz/dy = 2*4cos(2y+6)
=8cos(2y+6)

So yes, you do need to differentiate the 2y+6 bit as part of the chain rule
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Ourkid
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#128
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#128
Excellent thank you very much

One more thing. Does anybody know the PV equations for calculating consequent values for the trigs?

Methinks it is for sin and cos = PV + 360n, and for tan = PV + 180n

Correct?
Thanks
w00tt
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Speleo
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#129
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#129
Assuming PV means previous values, you have (in degrees):
sin: pv + 360n, 180 - pv
cos: pv + 360n, -pv
tan: pv + 180n
There are probably more.
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ad absurdum
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#130
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#130
(Original post by darth_vader05)
how would you integrate (x^3)(e^(1-x)) limits from x=0 to x=1.

can you rewrite that as eINT x^3(e^-x) ?

(i've done it by parts 3 times :eek: and got something like e-16 (cant remember properly )
Yeah, you can do it by parts three times (I got 6e-16). You could take an e outside the integral if you want, but it won't really save you much work when it comes to doing the actual integration.
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Ourkid
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#131
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#131
Would somebody be willing to quickly explain arcsin/arccos/arctan to me? I know that they are sin^-1, cos^-1 tan^-1, but what do I do with them?

Using the derivative of sin x and cos x show that d(tanx)/dx= sec^2x?

So using sin and cos identities to get the RHS then differentiating once you get to tan??

Any help will be appreciated.
w00tt
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Humoured
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#132
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#132
Could someone show the working out on how to answer this question.

An alloy is formed between 3 metals (A, B and C) and weighs a total of 550g.
There is twice as much of A than B and one and a half times as much of B than C.
How much of the metal M is in the alloy?
Express your final answer as a % of the total weight of the alloy.
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jpowell
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#133
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#133
What is M?
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nota bene
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#134
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#134
As jpowell I do not really get the question...
but possibly I could interpret it as 1.5*2A+1.5B+1C=550
And assuming the metals A, B and C have the same relative molecular mass (which they won't have in reality) you can easily solve it. (or well, this depends on what you mean by 'much' if it refers to by mass or by number of particles, if mass there will be no problem of assuming anything...the question is not clear...)

But, IF the above is correct I guess it is just 6/11 of A 3/11 of B and 2/11 of C
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Humoured
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#135
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#135
I mean A lol!

Thanks Nota, your correct
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yapster
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#136
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#136
Can anyone help me with the following question? I keep ending up with an answer that's different from the answer in the back of the book, so I want to check I'm actually doing something wrong.

The question's:

Find the gradient for this curve {xlny^3=6} at (2,e):

I keep ending up with the answer: {-e^3}/4 however the answer in the back is -e/2

Can anyone help me please?!
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jpowell
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#137
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#137
lny = 2/x

y = e^(2/x)

dy/dx = -(2e^(2/x))/x^2

dy/dx at x = 2 is -e/2
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yapster
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#138
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#138
Thanks a lot for the answer. Rep is on it's way!
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bred
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#139
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#139
Q^1/4 = 1796
what's Q?
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Speleo
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#140
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#140
Hint: [x^(1/4)]^4 = x^(1/4 * 4) = x^1 = x
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