# Use differentials to approximate?

• Nov 30th 2006, 07:57 AM
math34
Use differentials to approximate?
Use differentials to approximate the value of 7.8^3??

Should be easy, but can´t find the correct answer. How do you do this... help....
• Nov 30th 2006, 08:26 AM
Soroban
Hello, math34!

Quote:

Use differentials to approximate the value of $7.8^3$

Evidently, we have the function: . $f(x) \:=\:x^3$
Then: . $df \:=\:3x^2dx$

We know that: . $f(8) \,= \,8^3 \,= \,512$

Let $x = 8,\;dx = -0.2$

Then: . $dy \:=\:3(8^2)(-0.2)\:=\:-38.4$

Hence: . $7.8^3\:\approx\:512 - 38.4 \:=\:\boxed{473.6}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

How good is this approximation?

The actual value is: . $7.8^3\:=\:474.552$

Our answer is short by only: . $474.552 - 473.6 \:=\:0.952$

As a percent: . $\frac{0.952}{474.552}\:=\:0.002066103 \:\approx\:0.2\%$

We were off by only two-tenths of a percent (2 parts in 1000) . . . not bad!

• Nov 30th 2006, 08:30 AM
CaptainBlack
Quote:

Originally Posted by math34
Use differentials to approximate the value of 7.8^3??

Should be easy, but can&#180;t find the correct answer. How do you do this... help....

F(a+e)~=F(a)+eF'(a)

Put F(x)=x^3, F'(x)=3x^2, and a=8 and e=-0.2, then:

F(7.8)=F(8)-0.2*3*8^2=512 - 0.6*64=473.6,

which compares with an exact value of 474.552.

RonL
• Nov 30th 2006, 09:02 AM
math34
How did you get -0.2 for dx?
Thanks
• Nov 30th 2006, 09:13 AM
math34
How did you get -0.2 for dx? ,,
xdfgsdfg
Quote:

Originally Posted by Soroban
Hello, math34!

Evidently, we have the function: . $f(x) \:=\:x^3$
Then: . $df \:=\:3x^2dx$

We know that: . $f(8) \,= \,8^3 \,= \,512$

Let $x = 8,\;dx = -0.2$

Then: . $dy \:=\:3(8^2)(-0.2)\:=\:-38.4$

Hence: . $7.8^3\:\approx\:512 - 38.4 \:=\:\boxed{473.6}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

How good is this approximation?

The actual value is: . $7.8^3\:=\:474.552$

Our answer is short by only: . $474.552 - 473.6 \:=\:0.952$

As a percent: . $\frac{0.952}{474.552}\:=\:0.002066103 \:\approx\:0.2\%$

We were off by only two-tenths of a percent (2 parts in 1000) . . . not bad!

• Nov 30th 2006, 09:29 AM
ThePerfectHacker
Because that was the decrease in the value. You went from 8 to 7.8 thus the change was -.2