1. Approximation

Hi, I have some problems with approximation.I'll be glad if you could help.We just learnt basics.I can't solve this kind of questions.

App.(2.9)³

2x^4 + 5x³ - 7x ^2 +8x -3 X= 1.03

How accurately must the edge of a cube be measured so that the volume will be correct to within 3% ?

Thanks in advance. I have an exam on Saturday ! Your help will be much appreciated.

2. Originally Posted by truevein
Hi, I have some problems with approximation.I'll be glad if you could help.We just learnt basics.I can't solve this kind of questions.

App.(2.9)³

2x^4 + 5x³ - 7x ^2 +8x -3 X= 1.03

How accurately must the edge of a cube be measured so that the volume will be correct to within 3% ?

Thanks in advance. I have an exam on Saturday ! Your help will be much appreciated.
Hopefully, the "basics" included using the derivative to find the tangent line approximation to a graph: if y= f(x) then the tangent line at x= a is y= f'(a)(x- a)+ F(a).

For the first problem, find the tangent line to $y= x^3$ at x= 3. The derivative of $x^3$ is $3x^2$ and at x= 3, that is 27. Also $3^3$ is 27. The tangent line is y= 27(x- 3)+ 27. Now let x= 2.9

Do the same with $y= 2x^4 + 5x³ - 7x ^2 +8x -3$. Find the tangent line to the graph at x= 1 and let x= 1.03.

If a side of a cube is "s", its volume is $V= s^3$. The derivative is $\frac{dV}{ds} = 3s^2$ and, in terms of differentials $dV= 3s^2 ds$. You can think of "dV" and "ds" as the errors in V and s. Saying that you want "the volume will be correct to within 3%" means that you want $\frac{dV}{V}= \frac{3s^2}{s^3}ds= \frac{3}{s}ds$ less than .03. "How accurately must the edge of a cube be measured", as a percentage, is $\frac{ds}{s}$.

3. Thanks a lot! I could solve the first two myself with the help of f(x) = f'(xo)(x-xo) +f(xo). I didn't memorise it, btw, it is the slope formula But the third question was really bugging me. Thanks for your time and effort again.