Question

In problem set 3, question 2A 14* (b), I have simplified the exact thickness of the bubble gum into a cubic equation: r^3 + 30r^2 + 300r = 3(pi). Is there any way to find the exact answer from this? I know Newton's method would give an approximate answer, but is there a way to find it exactly?

Answer 1

without knowing what the details of the "problem set" are; it would be difficult to give a cogent response.

I assume this forum has some sort of course material in it that I do not possess :)

Answer 2

Sorry about the confusion, the question reads as follows:

(From 18.01 Exercises, Unit 2, 2A Approximation)

2A-14* Suppose that a piece of bubble gum has volume 4 cubic centimeters.
a) Use a linear approximation to calculate the thickness of a bubble of
inner radius 10 centimeters.
(Hint: Start with the relation between the volume V of a sphere and the radius
r, and derive the approximate relation between ΔV and Δr.)
b) Find the exact answer.

The answer is not listed within the solutions to the exercise.

Answer 3

hmmm.... if i read it correctly; the volume of gum never changes; just the shape, and the shape is assumed to be a sphere with an inner radius of 10

V{gum} = Volume of outer sphere - Volume of inner sphere; this should give us our radiuses

4 = (4/3) pi R^3 - (4/3) pi 10^3

4 = (4/3) pi (R^3 - 10^3)

4
------- = R^3 - 10^3
(4/3) pi


4(3)
----- + 10^3 = R^3
4 pi


3
--- + 1000 = R^3
pi

cbrt[(3 + 1000pi)/pi] = R

R-r should then equal the thickness of the gum.

cbrt[(3+1000pi)/pi] -10 = thickness of gum

\[\sqrt[3]{\frac{3+1000\pi}{\pi}}-10\]
perhaps?

Answer 4

4 = (4/3) pi R^3 - (4/3) pi r^3

4 = (4/3) pi (R^3 - r^3)

3 = pi (R^3 - r^3)

3/pi = R^3 - r^3

(3/pi) + r^3 = R^3

cbrt[(3/pi) + r^3] = R ; therefore
...................................
R-r = thickness

cbrt[(3/pi) + r^3] - r = thickness
\[\text{thickness =}\left(\sqrt[3]{\frac{3}{pi}+r^3}\right)-r^3\]
i cant tell if im doing it according to the instructions tho :)

Answer 5

The cubic equation
a₀x³+3a₁x²+3a₂x+a₃=0
is reduced by the transformation x=((y-a₁)/(a₀)) to the form:
y³+3Hy+G=0
where
H=a₀a₂-a₁² , G=a₀²a₃-3a₀a₁a₂+2a₁³
Cardan's solution can be used in all cases, but it is simpler to use the Cardan solution when G²+4H³>0 and use the trigonometric solution when G²+4H³<0, in this case, the roots are real if the coefficients are real. If G²+4H³=0 there will be a repeated root of the equation in y , whose value will be √(-H) or -√(-H). You have to prove which of these solutions satisfies the equation.

Cardan solution (because Tartaglia)
If p is any one of the values (-(G/2)+(1/2)√(G²+4H³))^{(1/3)}
and if q=-(H/p). Then the roots of the equation in y are:
p+q and pw+qw² and pw²+qw
where w is a complex cube root of unity

If G²+4H³>0, then you can choose to p positive real value, and the roots will be

p+q and -(1/2)(p+q)+((√3)/2)(p-q)i and -(1/2)(p+q)-((√3)/2)(p-q)i

Trigonometric solution for the case where G²+4H³<0

We take cosφ=-(1/2)G(-H)^{-(3/2)} where we take the positive value of (-H)^{(3/2)}
and the roots will be

2(+√(-H))cos((1/3)φ) and 2(+√(-H))cos((1/3)φ+(2/3)π) and 2(+√(-H))cos((1/3)φ+(4/3)π)

where +√(-H) indicates the positive root

Answer 6

I have use Scientific workplace for place the equation

Answer 7

just to rectify small errors
r^3 + 30r^2 + 300r = 3/(pi); hokie4ever
\[\text{thickness =}\left(\sqrt[3]{\frac{3}{\pi}+r^3}\right)-r\];amistre64