Question

Use an appropriate local linear approximation to estimate the value of the given quantity.

(3.02)^4 how do i do this

Answer 1

We could use the function f(x)=(x+3)^4 to help an approximating (3.02)^4
or even f(x)=x^4

Answer 2

So we know the tangent line should look like \[y-f(a)=f'(a)(x-a) \]
So say we used \[f(x)=(x+3)^4 \]
And say we wanted to find the tangent line at a=0

Answer 3

how do you get that

Answer 4

So we would have
\[y=f'(a)(x-a)+f(a) \\ y=f'(0)(x-0)+f(0) \\ f(x) =(x+3)^4 \approx f'(0)(x-0)+f(0) \text{ for values near x=0 } \]

Answer 5

How do you get what?

Answer 6

what happens to .02

Answer 7

and why the zero?

Answer 8

We will approximate (0.2+3)^4 in a sec

Answer 9

I chose 0 because 0.2 was close to the integer 0

Answer 10

oops I mean 0.02 every where I had 0.2

Answer 11

Let me do one example all the way though.
Say we wanted to do a linear approximation for sqrt(10.1)
There are different functions you could choose to help you do this.
But I definitely want my function to involve a sqrt
Say I want to find the tangent line to the curve
\[f(x)=\sqrt{x+10} \] at x=0...
\[y=f(0)+f'(0)(x-0) \\ y= \sqrt{10}+\frac{1}{2 \sqrt{x+10}} |_{x=0} (x-0) \\ y=\sqrt{10} +\frac{1}{2 \sqrt{10}} (x-0) \\ y=\sqrt{10}+\frac{x}{2 \sqrt{10}} \\ \sqrt{x+10} \approx \sqrt{10} +\frac{x}{2 \sqrt{10}} \text{ for values of } x \text{ near x=0 } \]

So we wanted to approximate sqrt(10.1)
so we need x=.1 (this is a value near x=0 so it should be a sorta good approximation)
\[\sqrt{.1+10} \approx \sqrt{10} +\frac{.1}{2 \sqrt{10}} =\sqrt{10}+\frac{1}{20\sqrt{10}}=\frac{20 (10)+1}{20 \sqrt{10}}=\frac{201}{20\sqrt{10}}\]

Answer 12

http://www.wolframalpha.com/input/?i=sqrt%2810.1%29
sqrt(10.1) is approx 3.178 according to a calculator
and our approximation 201/[20*sqrt(10)] is http://www.wolframalpha.com/input/?i=201%2F%2820*sqrt%2810%29%29

Answer 13

As you see 201/[20*sqrt(10)] was a pretty darn good approximation in this case

Answer 14

I feel your example defeats the purpose of the technique. \(\sqrt{10.1}\) is hard to calculate without a calculator and so is \(\sqrt{10}\) . \(\sqrt{9}\) though is easy to calculate without a calculator.

Answer 15

lol.
It is still close numbers.

Answer 16

yes it will give a good approximation

Answer 17

But yeah I know what you mean.

Answer 18

We could have chose f(x)=sqrt(x+9)

Answer 19

9 is a perfect square

Answer 20

I prefer \(f(x)=\sqrt{x}\)

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

x=10.1 and a=9
\[\sqrt{10.1}\approx\sqrt{9}+\frac{1}{2\sqrt{9}}(10.1-9)=3.18\overline{3}\]

Answer 21

\[g(x)=\sqrt{x+9} \\ \sqrt{x+9} \approx f(0)+f'(0)(x-0) \\ \sqrt{x+9} \approx \sqrt{9} +\frac{1}{2 \sqrt{x+9}} |_{x=0} (x-0) \\ \sqrt{x+9} \approx 3 +\frac{1}{2 \sqrt{9}}(x-0) \\ \sqrt{x+9} \approx 3+\frac{1}{2(3)}x \\ \sqrt{1.1+9} \approx 3 +\frac{1}{6} (1.1) \\ \sqrt{10.1} \approx 3+\frac{1.1}{6} \\ \sqrt{10.1} \approx 3+\frac{11}{60}=\frac{180+11}{60}=\frac{191}{60}\]

Answer 22

For some reason I love choosing the function so that I can find the tangent line at x=0

Answer 23

nothing wrong with that

Answer 24

ah i get it thank you your the best