Hey everyone...
I am about to start university next week, where I will have a maths test just testing our mathematical ability. However, since I have been on a gap year, my maths skills have gone somewhat rusty, and this is me being polite to myself!
So, I was wandering if you guys could help me with a few maths problems? Please only feel free to answer which ever ones you can, and I am terribly sorry I am asking such a huge favour without contributing anything first. I have got the answers to the questions, but just don't know how to arrive to them!
1) Rearrange the following formula to give d in terms of D and f:
(2 / D - d) + (2 / D + d) = (1/f)
It does look simpler if you write it out!
The answer is d = (D^2 - 4Df)^(1/2)
2) Given that y = loge (Ae^(-Bx)), find x in terms of y, A and B.
The answer is 1/B(loge A - y)
Differentiation Questions
3) Differentiate y = ln(3x^(2) +1) with respect to x.
The answer is 6x/(3x^(2) +1)
4) Differentiate y = (x^(2) + x)^(-2) with respect to x
The answer is -(4x+2)(x^(2)+x)^(-3)
5) By differentiating the relation x^2 + y^2 = e^k (where k is a constant) with respect to x, find dy/dx.
The answer is -x/y
I would really appreciate it you guys could help me out here, as I genuinely require some help.
Thank you everyone in advance
1) Rearrange the following formula to give d in terms of D and f:
(2 / D - d) + (2 / D + d) = (1/f)
It does look simpler if you write it out!
The answer is d = (D^2 - 4Df)^(1/2)
The idea is to do anything so that d will be isolated by itselt in the Lefthand side (LHS).
I think that should be
[2/(D-d)] +[2/(D+d)] = 1/f
Combine the two fractions in the LHS.
[2(D+d) +2(D-d)] /[(D-d)(D+d)] = 1/f
[2D +2d +2D -2d] /(D^2 -d^2) = 1/f
[4D] /(D^2 -d^2) = 1/f
Cross multiply,
(4D)(f) = (D^2 -d^2)(1)
4Df = D^2 -d^2
d^2 = D^2 -4Df
Take the square roots of both sides,
d = sqrt[D^2 -4Df]
Or,
d = [D^2 -4Df]^(1/2)
2) Given that y = loge (Ae^(-Bx)), find x in terms of y, A and B.
The answer is 1/B(loge A - y)
y = ln[A*e^(-Bx)]
y = ln[A] +ln[e^(-Bx)]
y = ln[A] +(-Bx)ln[e]
y = ln[A] -(Bx)(1)
y = ln[A] -Bx
Bx = ln[A] -y
x = (1/B)(ln[A] -y)
We used:
---Log(ab) = Log(a) +Log(b)
---Log(a^b) = b*Log(a)
---Log(to the base a)[a] = 1
Differentiation Questions
3) Differentiate y = ln(3x^(2) +1) with respect to x.
The answer is 6x/(3x^(2) +1)
y = ln[3x^2 +1]
dy/dx = (1/[3x^2 +1])*(6x)
dy/dx = (6x) /(3x^2 +1)
4) Differentiate y = (x^(2) + x)^(-2) with respect to x
The answer is -(4x+2)(x^(2)+x)^(-3)
y = (x^2 +x)^(-2)
dy/dx = {(-2)[x^2 +x]^(-3)}*(2x +1)
dy/dx = (-4x -2)[x^2 +x]^(-3)
dy/dx = -(4x +2)[x^2 +x]^(-3)
5) By differentiating the relation x^2 + y^2 = e^k (where k is a constant) with respect to x, find dy/dx.
The answer is -x/y
x^2 +y^2 = e^k
Differentiate both sides,
(2x)dx +(2y)dy = 0
(x)dx +(y)dy = 0
Divide both sides by dx,
x +y(dy/dx) = 0
y(dy/dx) = -x
dy/dx = -x/y