Question

Using differentials to approximate (16.1)^(3/2). Compare the approximation tomthe result given by the calculator. Have no idea what to do here!

Answer 1

i think its refering to derivatives .... and recurrsive inputs

Answer 2

let $$y = x^{(3/2)}\\
dy/dx = (3/2)\sqrt{x}\\
\Delta y=(3/2)\sqrt{x}\Delta x$$
x = 16, dx = 0.1
dy = 1.5*4*0.1 = 0.6
y+dy = 16*sqrt(16)+0.6 = 16*4 + 0.6 = 64.6

Answer 3

We have not learn anything about recurrsive inputs yet.

Answer 4

i sat there the other night and developed an approximator for square roots\[\sqrt{p}\\ \\\frac{p-g^2}{2g}+p=new ~g\]

Answer 5

So how do I know that x=16 and dx=0.1? Is the decimal that indicator?

Answer 6

since 3/2 = 1 1/2 .... this reduces to finding the sqrt of 16.1

that +p is a +g :) so my guess is 4
\[\frac{16.1-4^2}{2(4)+4}=new~g\]
\[\frac{16.1-(new~g)^2}{2(new~g)+new~g}=another~g\]

3 repetitions would get me to a very close approx

Answer 7

look at the value of "x" = 16.1
16 is ok and you can play with it easily.. a small error (causing the big problem) is "0.1" part

Answer 8

electros posts are prolly more attuned to the nature of the problem tho :)

Answer 9

@amistre64 hold it pal. you are going by an asymptote

Answer 10

Oh ok.