Question

Use a linear approximation (or differentials) to estimate the given number.
[4throot](622) and sin61°

Linear Approximation formula: L(x) = f(a)+f '(a)(x-a)
Differentials formula: Δy=f(x+Δx)-f(x)

Answer 1

LaTeX:

\(\Large{\sqrt[4]{622}}\) and \(\large{\sin{(61°)}}\)

Answer 2

\(5^4 = 625\) so for the first you could try...

\( (622)^{\frac{1}{4}} = (625 - x)^{\frac{1}{4}} \), \(x = 3\)

\(= 5 ( 1 - \frac{x}{625})^{\frac{1}{4}} \)

and then Taylor ... or Binomial, easier..... expansion

\(= 5 ( 1 - {\frac{1}{4}} \frac{x}{625} + \mathcal{O}(\frac{x}{625})^2) \)


and maybe use the same kinda idea for the second one....

\(\sin (x + \delta) = \sin x + \delta \cos x + \mathcal{O}(\delta^2)\) with \(x = \frac{\pi}{3}\), and \(\delta = \frac{\pi}{180}\)

Answer 3

I'm not sure what you meant by "expansion" but I've applied the parameter \(\Large{y=\sqrt[4]{x}}\)...
So that would make \(dx=3\) I guess?

Answer 4

I mean... \(dx=-3\)

Answer 5

Well, I started using linear approximation instead of differentials, since I understand that process better

Answer 6

yes, kinda

so you kick off with....
\(y=\sqrt[4]{x - dx}\)

but the expansions work best in form \((1+h)^{\alpha}\) where \(0 < |h| <<1\) so that powers of \(h\) can be ignored as they just get smaller and smaller....

so you might be better looking at:

\[ y=x^{1/4} \sqrt[4]{1 - \frac{dx}{x}} \]

Answer 7

I have no idea what expansions do, haven't covered that :\
...er, I already started with L(x), linear approximation; should I continue?

Answer 8

yes! sorry if i am confusing you. there is a general idea that allows you to expand about a point using calculus. your formula is one representation of that.

so go with your formula.

and put it on here if you want some input.

Answer 9

actually, i think i am confusing you A LOT :-(

do what you were going to do and share

Answer 10

LOL, it's okay...
...well, I'm not quite sure what to apply for \(a\)... I was doing something like \(\Large{\sqrt[4]{x-3}}\)

Answer 11

\(a\) in the linear approximation formula \(\large{L(x)=f(a)+f'(a)(x-a)}\)

Answer 12

yes!

Answer 13

eh?

Answer 14

but what do I put for \(a\)? lol

Answer 15

Intuitively:

\[\large{f(622)=f(625)+f'(625)(-3)}\]

Answer 16

Exactly!
@IrishBoy123
Liinear approximation is what @kittiwitti1 is using, which is taking only the linear term of the expansion. It is used without learning expansions!

Answer 17

I think expansions is covered in Calc II, actually... I'm only in Single-Variable (I) haha

Answer 18

can you **see** what is happening with the approximation. you know the tangent line being used to approximate a non-linear function

sorry if that's patronising but if you can see it you don't really need formulae

Answer 19

Eh... I don't know if I'm good enough to see it...
and don't worry ! I'm not offended at all :-)

Answer 20

if i could draw! let me trawl the net for a good illustration.....

Answer 21

is it possible to explain using the linear approx formula?

Answer 22

or in your formula

\(L(622) = f(625) + f'(625)(622-625)\)

Answer 23

so far I applied this \[\sqrt[4]{x-3}\]but I don't know what to put as \(a\) so I get\[L(x)=f(a)+f'(a)(x-a)\rightarrow\sqrt[4]{a-3}+\frac{1}{4\sqrt[4]{(a-3)^{3}}}(x-a)\]

Answer 24

I guess \(a=625\)?

Answer 25

looks good

x = 622, a = 625

Answer 26

\(f(x + \delta) \approx f(x) + \delta f'(x)\)

\(f(622) \approx f(625) + (622-625) f'(625) \)

\(f(625) = 5\)

\(f(x) = x^{1/4}\)

\(f'(x) = \frac{1}{4}x^{-3/4}\)

\(f'(625) = \frac{1}{4}625^{-3/4}\)

\( = \frac{1}{4}5^{-3} = 1/500\)

Answer 27

that's the intuition. we can pattern match that into your equation if it makes sense and you agree

Answer 28

\[\large{L(622)=\sqrt[4]{625}+\left(\frac{1}{4\sqrt[4]{(625)^{3}}}\right)(622-625)}\]

Answer 29

hmm, okay


(SORRY the server booted me off -_-)

Answer 30

yes, i get bandwidth about 10Mbps and there is no-one online and I am lagging sooo badly too :-(

let me go back to your starting eqn

\(L(x) = f(a) + f'(a) (x - a)\)

\(L(622) = f(625) + f'(625) (622 - 625)\)

for function:
\(f(x) = x^{1/4}\)


the use of the letter L in that formulation is descriptive but unhelpful



[that's why i draw stuff, gets you round the formalism.]

Answer 31

ok well I'm trying to get around the fourth-roots idk

Answer 32

Well,

\[L(\color{blue}{622}) = f(625) + f'(625) (622 - 625)\]

is the same as

\[f(622) = f(625) + f'(625) (622 - 625)\]


or to be more precise

is the same as

\[f(622) \color{green}{\approx} f(625) + f'(625) (622 - 625)\]

Everything on the RHS is do-able on pen and paper w/out a calculator, i guess

if you want to use x = 625 throughout, you get this


\(L(\color{red}{625}) \color{red}{=} f(625) + f'(625) (622 - 625)\), not true

Answer 33

oh

Answer 34

how is it doable without calc? I mean... there are fourth roots

Answer 35

\(\Large \left(\frac{1}{4\sqrt[4]{(625)^{3}}}\right)(622-625) \)

\(\Large = \left(\frac{1}{4 \times (5)^{3}}\right)(622-625) \)

\(\Large = \left(\frac{1}{500}\right)(-3) \)

Answer 36

how did this\[\sqrt[4]{625^{3}}\]go to this?\[(5)^{3}\]isn't the cube under the fourth root o_o

Answer 37

\(625^{\frac{3}{4}}\)

\(= (625^3)^{\frac{1}{4}}\)

\( ( 625^{\frac{1}{4}} ) ^3\)

Answer 38

oh. good point.

Answer 39

wait... I thought it was fourth root of 622 and not 625? because it's this\[\large{\sqrt[4]{x-3},a=625}\]

Answer 40

or are we doing this?\[\large{\sqrt[4]{x}}\]

Answer 41

dunno if this will help, but from above



pattern match the numbers and letters, see if that makes sense

the idea is that 622 is an awkward number so we are trying to use 625 where poss.

Answer 42

what I meant is "what is being used for the theoretical \(f(x)\)?"

Answer 43

is this the process used?\[\huge{L(x)=\sqrt[4]{a}-\left(\frac{1}{4\sqrt[4]{a^{3}}}\right)(x-625)}\]

Answer 44

this is it with the numbers



and

\[\huge{L(x)=\sqrt[4]{a}-\left(\frac{1}{4\sqrt[4]{a^{3}}}\right)(x-\color{red}{a})}\]

Answer 45

er, yes. that.

Answer 46

sorry bout the typo... again lol

Answer 47

(why is there a Clinton poser on this site... this place has become troll bait lol)

Answer 48

and i'm sorry about confusing you :-)

i must learn to read questions properly

Clinton.....well it could be soooo much worse :-)

Answer 49

At least the Trump ones crack (dumb) jokes haha

Answer 50

It's ok, I think I got it now @IrishBoy123 :-)
Thanks for the help!