Tangent Approximation

masterpizza

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Jun 4, 2010
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Can someone please explain to me how to do this/what this accomplishes/shows us so I can have a working understanding of it. My instructor went over it but I am a bit confused.

EX ( ?101
EX) [sup:1kk8yvht]3[/sup:1kk8yvht] ?7.9
 
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Hello, masterpizza!

These are problems in differential approximations.

This is the basic idea . . .

We have a function: .f(x)=y\displaystyle f(x)\:=\:y

If we change x\displaystyle x by some small amount, Δx\displaystyle \Delta x
. . the value of y\displaystyle y will also change:. . f(x+Δx)=y+Δy\displaystyle f(x + \Delta x) \:=\:y + \Delta y

For small values of Δx\displaystyle \Delta x, the differential dy\displaystyle dy is approximately Δy.\displaystyle \Delta y.
. . That is:. . f(x+Δx)    y+dy\displaystyle f(x+\Delta x) \;\approx\;y + dy


(1)  101\displaystyle (1)\;\sqrt{101}

The function is the square-root function: .f(x)=x=x12\displaystyle f(x) \:=\:\sqrt{x} \:=\:x^{\frac{1}{2}}

Note that 101 is "close" to 100, and we know that: .100=10\displaystyle \sqrt{100} = 10

So, 100+1\displaystyle \sqrt{100+1} is equal to 10 "plus a little more".


We have: .x=100,  dx=1, and y=x12\displaystyle x = 100,\;dx = 1,\:\text{ and }\,y \:=\:x^{\frac{1}{2}}

The differential is:   dy=12x12dx=dx2x\displaystyle \text{The differential is: }\;dy \:=\:\tfrac{1}{2}x^{-\frac{1}{2}}dx \:=\:\frac{dx}{2\sqrt{x}}

Substitute: dy=12100=120=0.05(the "little more")\displaystyle \text{Substitute: }\:dy \:=\:\frac{1}{2\sqrt{100}} \:=\:\frac{1}{20} \:=\:0.05\quad\text{(the "little more")}


Therefore: 101    10+0.05=10.05\displaystyle \text{Therefore: }\:\sqrt{101} \;\approx\;10 + 0.05 \:=\:\boxed{10.05}


Check

The actual value of 101\displaystyle \sqrt{101} is: .10.04987562...\displaystyle 10.04987562...

So our approximation is very good!




(2)  7.93\displaystyle (2)\;\sqrt[3]{7.9}

We have: f(x)=x3=x13\displaystyle \text{We have: }\:f(x) \:=\:\sqrt[3]{x} \:=\:x^{\frac{1}{3}}

We note that 7.9 is close to 8, and we know that: .83=2\displaystyle \sqrt[3]{8} = 2

So that 80.13\displaystyle \sqrt[3]{8 - 0.1} is equal to 2 "minus a little less".


We have: x=8,  dx=0.1,  y=x13\displaystyle \text{We have: }\:x = 8,\;dx = -0.1,\;y \,=\,x^{\frac{1}{3}}

The differential is: dy=13x23dx=dx3x23\displaystyle \text{The differential is: }\:dy \:=\:\tfrac{1}{3}x^{-\frac{2}{3}}dx \:=\:\frac{dx}{3\sqrt[3]{x^2}}

Substitute: dy  =  0.13823=0.112(the "little less")\displaystyle \text{Substitute: }\:dy \;=\;\frac{-0.1}{3\sqrt[3]{8^2}} \:=\:-\frac{0.1}{12}\quad\text{(the "little less")}


Therefore: 7.93    20.112  =  1.9916666...\displaystyle \text{Therefore: }\:\sqrt[3]{7.9} \;\approx\;2 - \frac{0.1}{12} \;=\;\boxed{1.9916666...}


Check

7.93  =  1.991631701...\displaystyle \sqrt[3]{7.9} \;=\;1.991631701...


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Question: Why do we have to know this?

Suppose you are on Survivor (without a calculator).

And your very lives depend on knowing 101\displaystyle \sqrt{101}.

You can work out 10.05, writing in the dirt with a stick . . .


Trust me on this, guys.
Chicks really dig this stuff!


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Before you start composing your hate mail . . .

Yes, I know that "chicks" is a sexist term.

But saying "chicks" is not a felony
. . . . . although it is a Ms-demeaner.

 
Baby chickens always like short-cuts - what is this noise about felony and what-not????
 
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