Question

Help with College Calculus..

Answer 1

The nearest square to 84 is 81.
So we want to find the tangent line to y=sqrt(x) at x=81.
And then we are going to use that tangent line to approximate sqrt(84)

Answer 2

Can you explain how you got 81? I have absolutely no idea how to solve this.

Answer 3

81 is the nearest square number to 84

Answer 4

And you figured this out by squaring each number below 84?

Answer 5

well I didn't have to do that

Answer 6

81 is 9^2
100 is 10^2
84 is between 81 and 100

Answer 7

but 84 is closer to 81

Answer 8

this is the reason I'm using 81 and not 100

Answer 9

do you know how to find the tangent line to y=sqrt(x) at x=81?

Answer 10

line with slope m and point (a,f(a)) looks like
\[y-f(a)=m(x-a)\]
m is the slope of the tangent line here
and the slope at x=a is f'(a)

so m=f'(a)

y-f(a)=f'(a)(x-a)

adding f(a) on both sides

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

we are going to use this L(x)=f'(a)(x-a)+f(a) as an approximate for values near x=a

Answer 11

do you know how to find f'(x) given f(x)?

Answer 12

Sorry but why are you talking about tangent line?

Answer 13

the equation you have written down in the attachment above says to use the tangent line

Answer 14

Freckles is giving the theory

Answer 15

\[\Delta y =f'(x) \Delta x \\ y-f(a)=f'(a)(x-a)\]

Answer 16

when you have time, see
https://www.khanacademy.org/math/differential-calculus/derivative-applications/local-linearization/v/local-linearization-intro

Answer 17

Okay well I did this:

Square root 84 --> square root 81 (by squaring every number below 84)

triangle x = dx= 3

y= square root x= x^1/2
dy= f`(x)dx= 1/2 x^-1/2 dx

= 1 over 2x^1/2 dx

x= 81 and dx=3
dy= 1 over 2(81)^1/2 (3)= 0.166666 -> 0.167

square root 84 = f(81+3)= f(81)+0.167
=9.167

Answer 18

The video was not helpful. And I still do not see why you were talking about tangent line.

Answer 19

I don't exactly understand what you did... your answer is a bit off from what I have.
Your question says to approximate using the tangent line... this is why I was talking about the tangent line.

Answer 20

No it doesnt. My question does not say tangent line anywhere.

Answer 21

My answer is correct according to Pearson. I have no clue what your doing.

Answer 22

oops i changed 83 to 84

Answer 23

i mean the other way around

Answer 24

yes it does

Answer 25

delta y=f'(a)delta x

Answer 26

this is tangent line at x=a

Answer 27

\[\Delta y=f(x)-f(a) \\ \text{ where as } \Delta x=x-a \\ f(x)-f(a)=f'(a)(x-a)\]
this is the tangent line at x=a

Answer 28

\[f(x)\approx f'(x)(x-a)+f(x)\]

Answer 29

What did you get for the answer using your equation?

Answer 30

\[\text{ the tangent line \to } y=\sqrt{x} \text{ at } x=81 \\ \text{ is } y-f(81)=f'(81)(x-81) \\ \text{ we are going to use this to approximate } \sqrt{x} \text {for values near } x=81 \\ \\ \sqrt{x} \approx f'(81)(x-81)+f(81)\]
\[\sqrt{x} \approx \frac{1}{2 \sqrt{81}} (x-81)+\sqrt{81} \\ \sqrt{x} \approx \frac{1}{18}(x-81)+9 \\ \\ \text{ now just replace } x \text{ with } 84 \\ \sqrt{84} \approx \frac{1}{18}(84-81)+9\]

Answer 31

you just simplify right hand side
I don't use decimals by the way
I just simplify as a fraction

Answer 32

The only time ive used tangent line was like this

https://www.youtube.com/watch?v=b6I6RlyL_jA

Answer 33

as I have been saying the formula they asked you to use in your picture is telling you to use tangent line so you are still using it :p

Answer 34

you do know Delta x means change in x
while Delta y means change in y?

Answer 35

Ok. Well you make no sense. So im going to stick with how Pearson shows to solve it. Its less complicated and you have yet to explain where you got 81 from.

Answer 36

i have told you 81 is the nearest number to 84 that is a square number

Answer 37

a square number is a number that you can take the square root of and get an integer

Answer 38

and @Destinyyyy I'm sure I have helped you in the past.... Your comments to me seem kind of rude. I'm kind of shocked.

Answer 39

But if you don't want me to help you in the future, I'm sorry I won't :(

Answer 40

I will say one more thing...
We had a square root on that number 83...
we know sqrt(83) is not a nice number because 83 is not a perfect square
but 81 is a perfect square
we know sqrt(81)=9
we know know that 81 is the nearest number to 83 that is a perfect square
So this is why I chose to find the tangent line at x=81



just like if wanted to approximate:
\[\sqrt[4]{9}\]
I would find the tangent line at x=16 since 16 is the nearest "perfect 4th power" number to x=9
So I'm going to find the tangent line to
\[f(x)=\sqrt[4]{x} \text{ at } x=16 \\ f(16)=\sqrt[4]{16}=2 \\ f'(x)=(x^\frac{1}{4})'=\frac{1}{4} x^{\frac{-3}{4}}=\frac{1}{4 x^{\frac{3}{4}}} \\ \\ f'(16)=\frac{1}{4 (16)^\frac{3}{4}}=\frac{1}{4(2)^3}=\frac{1}{4(8)}=\frac{1}{32} \\ \text{ so the tangent line is } \\ y-f(16)=f'(16)(x-16) \\ y-2=\frac{1}{32}(x-16) \\ y=\frac{1}{32}(x-16)+2 \text{ okay this is the tangent line } \\ \text{ we are going to use } \\ f(x) \approx \frac{1}{32}(x-16) +2 \text{ for values of } x \text{ near } x=16 \\ \\ \sqrt[4]{x} \approx \frac{1}{32}(x-16)+2 \\ \text{ we wanted to approximate } \sqrt[4]{9} \\ \text{ so we replace } x \text{ with } 9 \\ \sqrt[4]{9} \approx \frac{1}{32}(9-16)+2 \\ \sqrt[4]{9} \approx \frac{1}{32}(-7)+2 \\ \sqrt[4]{9} \approx \frac{-7}{32}+2 \\ \sqrt[4]{9} \approx \frac{-7+2(32)}{32} \\ \sqrt[4]{9} \approx \frac{-7+64}{32} \\ \sqrt[4]{9} \approx \frac{57}{32}\]

Answer 41

http://www.sosmath.com/calculus/diff/der06/der06.html
also here is a site that has said the exact thing I have said
about your equation \[\Delta y=\Delta x f'(x_0)\]
All this is the tangent line at x_0