# differencials approximation

• Mar 22nd 2007, 11:46 AM
UMStudent
differencials approximation
I have a test tomorrow in my calc class and have a few problems I'm having trouble with after going through the chapter review problems. So any help and explaination would be great, thanks.

1) Find the linearization of f(x) = (1+3x)^(1/3) at a = 0. State the corresponding linear approximation and use it to give an approximate value for (1.03)^(1/3)

2) The volume of a cube is increasing at a rate of 10cm^3/min. How fast is the surface area increasing when the length of an edge is 30cm?

3) An oil tanker runs aground and ruptures. Suppose the leaking oil spreads out in a cicle centered at the tanker and the area of that circle increases at a constate rate at five square miles per hour. Compute the rate of change of the radius of the oil spill at hte moment when the radius is four miles.

Thanks again for any help.
• Mar 22nd 2007, 05:18 PM
topsquark
Quote:

Originally Posted by UMStudent
1) Find the linearization of f(x) = (1+3x)^(1/3) at a = 0. State the corresponding linear approximation and use it to give an approximate value for (1.03)^(1/3)

You can think of this either as a binomial expansion, or a Taylor expansion about x = 0. (This last is probably what your professor wants.) Either way:
f(x) (is approximately) 1 + (1/3)(3x)x + ... = 1 + x + ...

So
f(1.03) = f(1 + 3*0.01) (is approximately) 1 + 0.01 = 1.01.

-Dan
• Mar 22nd 2007, 05:21 PM
topsquark
Quote:

Originally Posted by UMStudent
2) The volume of a cube is increasing at a rate of 10cm^3/min. How fast is the surface area increasing when the length of an edge is 30cm?

Call the side of the cube x, the surface area S, and the volume V.
V = x^3

So
dV/dt = 3x^2*dx/dt = 10 cm^3/min

So when x = 30 cm, dx/dt = 1/810 cm/min.

Now,
S = 6x^2

dS/dt = 12x*dx/dt = 4/9 cm^2/min.

-Dan
• Mar 22nd 2007, 05:23 PM
topsquark
Quote:

Originally Posted by UMStudent
3) An oil tanker runs aground and ruptures. Suppose the leaking oil spreads out in a cicle centered at the tanker and the area of that circle increases at a constate rate at five square miles per hour. Compute the rate of change of the radius of the oil spill at hte moment when the radius is four miles.

This is similar to problem 2).

A = (pi)r^2

dA/dt = 2(pi)r*dr/dt = 5 mi^2/h

So when r = 4 mi, dr/dt = 5/(8*(pi)) mi/h (is approximately) 0.198944 mi/h

-Dan
• Mar 22nd 2007, 06:14 PM
Jhevon
Quote:

Originally Posted by UMStudent
1) Find the linearization of f(x) = (1+3x)^(1/3) at a = 0. State the corresponding linear approximation and use it to give an approximate value for (1.03)^(1/3)

Ok, so topsquark's solution for this was correct, but i don't think you know about taylor expansions as yet. here is the way to do it using the good ol' fashion linearization formula from calc 1.

the linearization formula for approximation is:

f(x) ~= f(a) + f ' (a)(x - a)

where f(x) is the value you want to approximate because you don't know the exact value of it. a is a value very close to x for which you do know the value of f(a) and f ' (a).

notice for the question that you have (1.03)^(1/3) when x = 0.01
your professor says to take a = 0, that is a value relatively close to 0.01 that you do know the answer for. let's see how this works.

f(x) ~= f(a) + f ' (a)(x - a)

when a = 0 we have

f(x) ~= f(0) + f ' (0)(x - 0)

= f(0) + f ' (0)(x)

okay, now we find the pieces, we know f(x) and therefore can find f(0), but what is f ' (x)?

f(x) = (1+3x)^(1/3)
=> f ' (x) = (1/3)(1 + 3x)^(-2/3) * (3) = (1 + 3x)^(-2/3)
ok.
now f(0) = (1+3(0))^(1/3) = 1
f ' (0) = (1 + 3(0))^(-2/3) = 1

now we go back to the formula:

f(x) ~= f(0) + f ' (0)(x) = 1 + (1)x = 1 + x
=> (1.03)^(1/3) = f(0.01) ~= 1 + 0.01 = 1.01
• Mar 22nd 2007, 06:35 PM
Jhevon
Quote:

Originally Posted by topsquark
Call the side of the cube x, the surface area S, and the volume V.
V = x^3

So
dV/dt = 3x^2*dx/dt = 10 cm^3/min

So when x = 30 cm, dx/dt = 1/810 cm/min.

Now,
S = 6x^2

dS/dt = 12x*dx/dt = 4/9 cm^2/min.

-Dan

I believe you made an error somewhere Dan. I don't think dx/dt = 1/810

Let V be the volume of the cube, S be the surface area and x be the length of each side, then

V = x^3 and S = 6x^2

=> dV/dt = 3x^2 dx/dt
dV/dt = 10 and we are concerned with the instant x = 30 cm
=> 10 = 3(30)^2 dx/dt = 2700 dx/dt
=> dx/dt = 1/270 when x = 30

now S = 6x^2
=> dS/dt = 12x dx/dt
when x = 30, dx/dt = 1/270
=> dS/dt = 12(30)(1/270) = 4/3 cm^2/min