Question

I need help understanding how to do a problem using linear approximations or differentials.

Answer 1

post a problem

Answer 2

\[\sqrt[3]{1001}\]

Answer 3

y= x^1/3
y(1000)=10
\[y(x_1+\Delta x)=y' \Delta x + y(x)\]

\[y'=\frac{1}{3}x^{-2/3}\]

\[y(1001)=\frac{1}{3}1000^{-2/3}*(-1)+10\]

Answer 4

Please disregard the last line

Answer 5

\[y(1001)=\frac{1}{3}*1000^{-2/3}*(1)+10\]

Answer 6

y(1001)=10.003

Answer 7

any question?

Answer 8

im writing it out step by step trying to understand it if i get stuck ill let you know. thanks!

Answer 9

ok.

Steps to these problem

1) find number close by that returns easy answers
in our case it was 1000 since 1000^1/3 was just 10 plus it is close 1001

2) Write your general function ; y=x^1/3 in our case and differentiate

3) write this
\[y(x_1+\Delta x)=y' \Delta x + y(x)\]

\[\Delta x=+1 \]
since 1001 is 1 bigger than 1000

Answer 10

ok got it, is it the same as using y= f(a) + f'(a)(x-a)

Answer 11

oh yeah

Answer 12

oh okay could u explain using those terms please?

Answer 13

remember
y=mx+b

?

Answer 14

yes

Answer 15

this is actually the same

m= f'(a)

x=(x-a)

b= f(a)

Answer 16

ohh wow i see

Answer 17

y= f(a) + f'(a)(x-a)
a=1000
f(a)=10

\[f'(x)=\frac{1}{3}x^{-2/3}\]

plug in 1000
f'(a)=\[\frac{1}{3}1000^{-2/3}\]

x=1001

a=1000

(x-a)=1

Answer 18

so its y=10 + 1/3 1000^-2/3(1)

Answer 19

yes

Answer 20

okay sec im gonna solve.

Answer 21

okay i got 10.0033333333 which is wat u got thanks!

Answer 22

Calc I?

Answer 23

umm i think so.