I need help :S I will give 50 000 credits if you help with all questions :3

[25thhour]

Questions I Couldn't Do:

Newton's Universal Law of Gravitation asserts that two point masses m1 and m2 separated by a distance s, attract one another with a force
F = Km1m2/(s^2) , K being a universal constant.
If s = 8 meters, use differentials to estimate the change in s that will decrease the force by 11%

What I got so far
d/ds = ?
d/dF= .110

d/dF = (-2Km1m2)/s) dF/ds

Im kinda stuck from there on out? I can't remember how to do these It will probably come back to me like right away once someone shows me how to do it.

Number 2

Question:
Use differentials to determine approximately what percentage does the perimeter of a square table increase if its diagonal increased from 26cm to 26.351cm. Express answer as a percentage.

p = perimeter = 4x
r = diagonal = sqrt(2x^2)
d/dp = ?
d/dr = .351cm

dp/dx = 4
dr/dx = 1/(2x)

Not sure what to do? I am pretty sure Im screwing something up with my d/d(x)'s... :S

Number 3
The volume of a cube is increased from 512 cubic centimeters to 546.612 cubic cm. Use differentials to determine by approximately how many square cm does the surface area of the cube increase.

v = x^2
SA = 6x^2
d/dv = 34.612cm

I have gotten 2 different answers one was 17.11 by doing simple math and the other was 40.75998 by doing stuff I think is right...

Question 4 I have finished just need someone to check my work
Find slopes of all lines tangent to the graph of the relation:

y^3 + (8x^12)y^2 - 20y = 120 - 120x^2

at the point on the graph where x = 1.

There are 3 answers:

What I did:

Implicitly took the derivative to get:

(240 - 96y^2) / (-3y^2 - 16y + 20) = y'(x)

I then plugged in x = 1 to the original equation to find the y values

Original equation at x(1) = y^3 + 8y^2 - 20
solved: y(y+10)(y-2)
y = 0, -10, and 2.

I then plugged them into the derivative and I know that y'(x) = slope so whatever I got would be the slope.

My 3 slopes were 6, 78, and 12

Question 5

Find the y-coordinate of all points on the curve:

2x + (y + 2)^2 = 0

where the normal line to the curve passes through the point (-63, -122) (not on the curve) (There are 3 y-coords)

What I did
What I do know is that the slope of the normal line to the curve is = -1/m1

but that was basically all I got...

Question 6
I answered just need a check they gave me the answer in the question but I did figure it out using math

Let g be a function such that
g'(x) = -11x^10(sin(kx^7) - 7kx^17(cos(kx^7) , g(1) = 13/7. Where k is the real number given by tan(k) = 1/sqrt48 , 0 < k < pi/2... What is g(1)

HINT: what is d/dx{x^11(sin(kx^7)

g(1) = 13/7...

Question 7 Need a check

Find the curve y = F(x) that passes through the point (-1,0) and satisfies dy/dx = 6x^2 + 6x...

I basically took the integral and found what c was by plugging in the given points... easy.

y = 2x^3 + 3x^2 - 1

Question 8
Let f(x) = 8x^3 - 15x^2 - 3 , x >2
Assume f has an inverse g, find the value of g'(622) given that f(5) = 622

What I did:
used the relation g'(x) = 1/(f'(g(x))

I know the f(5) = 622 also means that g(622) = 5

evaluate for g'(622) = 1/f'(5) = 1/450

Answer = 1/450

Question 9 Need a check

Let f(x) = (1/3)x^3 +9x - 1 and let y = f(inverse)(x) be the inverse function for f. Determine the x-cords of the two points on the graph of the inverse function where the tangent line is perpendicular to the straight line y = -18x + 21.

I had trouble with this guy but I think I figured it out eventually

I know the slope of the straight line is -18. So the slopes of the perpendicular line has to be 1/18.

I then used that if point (a,b) on the inverse function was used on the regular function it would actually be (b,a) on f(x).

I then used the formula m = 1/f'(f(inverse)(a)) which the bolded function is actually equal to b.

1/f'(b) --- I then subbed in b to the d/dx on the original function:

skip a few steps I got:

1/18 = 1/(b^2 + 9)

cross multiply = b^2 + 9 = 18

b = +-3

I then solved for the x-cords to get 89 and 79

[L.B.D.D]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

chegg.com is your friend

[rushyrulz]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

wolframalpha.com is also your friend

[dag12]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

ok, I will help. I will edit this post as I go along.

1. F = Km1m2/(s^2)
dF/ds = -2Km1m2/(s^3)
ds = dF / [-2Km1m2/(s^3)]

In the limit of small changes, we approximate the differentials dF and ds as ΔF and Δs, respectively:
Δs = ΔF / [-2Km1m2/(s^3)]

Now simply substitute the value of s, as well as ΔF. ΔF is given by -0.11F.
You will need to evaluate F for s = 8m using the formula given.

2.A simple use of the Pythagorean theorem yields:
P = 4*l / sqrt(2), where P is the perimeter and l is the length of the diagonal.
dP = 4/sqrt(2) * dl
Using the same method above,
ΔP = 4/sqrt(2) * Δl
Since we want the percent change of the P, we want ΔP/P, which is given by:
ΔP/P = (4/sqrt(2) * Δl) / (4*l / sqrt(2))
ΔP/P = Δl /l

Since the diagonal changes from l = 26cm to 26.351, Δl = 0.351.
So the answer is simply:
ΔP/P = 0.351 / 26
ΔP/P = 0.0135
Expressed as a percent, this would be 1.35%.

Note that this is consistent with the real answer, which can be found by evaluating:
((4*26.351/sqrt(2))-(4*26/sqrt(2)))/(4*26/sqrt(2))

3. Letting the length of the side of the cube be x, we have
V = x^3
A = 6*x^2
dV/dx = 3x^2
dA/dx = 12x
Dividing one by the other (which mathematicians will tell you to be wary of, but the details of that are unnecessary here) we have
dV/dA = x/4
Hence under the small error approximation,
ΔA = ΔV*4 / x
Note that this expression is dimensionally consistent, as V has units of length^3, and dividing by x [length] will give units of length^2.
ΔV = (546.612-512) cm^3
and x = cube root[512]

Substitution yields
ΔA = 17.31 cm^2

This is approximately equal to the real answer, which can be found by evaluating:
6*[(546.612)^(1/3)]^2-6*[(512)^(1/3)]^2 = 17.12 cm^2

[dag12]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

4. Your work is largely correct.
However,
(240 - 96y^2) / (-3y^2 - 16y + 20) = y'(x) is not correct; what you've attempted to find is y'(1).

Furthermore, 240 should be -240; hence the correct expression is:
y'(1) = (-240 - 96y^2) / (-3y^2 - 16y + 20)
or more simply
y'(1) = (240 + 96y^2) / (20 - 3y^2 -16y)
Re-evaluation at y = 0, 2, and -10 should yield
y'(1) = 12, -26, and -82, respectively.

To demonstrate that your slopes are wrong, see the graph here:

http://www.wolframalpha.com/input/?i=Plot+y%5E3+%2B+%288x%5E12%29y%5E2+-+20y+%3D+120+-+120x%5E2+from+x+%3D+0.95+to+1.05

Note that at (1,2), the slope there is a negative value.

5. Evaluating the differential of the equation 2x + (y + 2)^2 = 0 implicitly, we have
2 + 2(y+2) dy/dx = 0, hence
dy/dx = -2/(2(y+2))
dy/dx = -1/(y+2)

Since the slope at any given point along the curve is -1/(y+2), a normal line through that point a given point would have the slope (y+2).

Next we can evaluate the slope of the line that goes through the point (-63, -122) and any given point along the curve.
The points along the curve all satisfy the equation 2x + (y + 2)^2 = 0, hence they have the coordinates:
(-(y+2)^2/2,y), where y is the free variable.
The slope between this point and (-63,-122) is given by:
y+122 / (-(y+2)^2/2+63).
This slope must be equal to the normal slope at the corresponding point. Hence
y+122 / (-(y+2)^2/2+63) = (y+2)
Solving this equation yields
y = -14, 0, 8

6. Could you check the expression? There are missing parenthesis, perhaps missing k variables, and maybe wrong exponents. Don't really want to try unless I'm sure the expression is correct...

7, 8. These are correct. Good job.

9. Almost correct, except for the last line.
Quite simply, as you note, the slope of the perpendicular line has to be 1/18.
Thus dy/dx for the inverse function must be equal to 1/18.
Since slope is given by (y2-y1)/(x2-x1), and since we interchange x and y for the inverse function, this value corresponds to
(x2-x1)/(y2-y1).
Hence it's not hard to see (in a non-rigorous way, admittedly) that dy/dx in the inverse function corresponds to dx/dy in the original function.
So we simply find
dx/dy = 1/18 in the original function, or
dy/dx = 18.

This yields
x = -3, 3 (as you obtained)
with corresponding coordinates (-3,-37), (3,35).
In the inverse function then, these points would be (-37,-3) and (35,3).

[25thhour]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

(-11x^10)(sin(kx^7)) - (7kx^17)(cos(kx^7))

is the correct equation just missed a few brackets

[dag12]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

I still don't really understand the question.

Integration of the expression yields
g(x) = -x^11*sin(kx^7) + C
You can solve for C by substitution of x = 1, but then they want you to find the value of g(1) when they've given it to you already?

[25thhour]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

I still don't really understand the question.

Integration of the expression yields
g(x) = -x^11*sin(kx^7) + C
You can solve for C by substitution of x = 1, but then they want you to find the value of g(1) when they've given it to you already?

Yes, I was totally confused with the question. Lol. Not sure if they are trying to trick me or not.

haha

Sory for the late reply went to sleep early.

[dag12]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

It's ok. Hope I answered all your questions. If you have further questions, feel free to ask.

[21992]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

Physics C? man and I thought B was hard...

[MarioNintendo]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

dag is the nicest math guy in the universe.

[25thhour]

Re: I need help :S I will give 50 000 credits if you help with all questions :3

Physics C? man and I thought B was hard...

First Year Calculus... Really easy stuff, just couldn't grasp it well.

edit;
thanks a lot Dag, I have been away for a few days, I will send the credits ASAP and I do have a few questions for you :P But I will PM them