Show that for small values of h, sqroot(4+h) is approx equal to 2 + (h/4)

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- Nov 26th 2006, 07:38 PMmightymaxWeird problem
Show that for small values of h, sqroot(4+h) is approx equal to 2 + (h/4)

- Nov 26th 2006, 07:51 PMThePerfectHacker
This is a

*linearzation problem*.

Consider the function,

$\displaystyle \sqrt{4+h}$

And its tangent line at point $\displaystyle h=0$.

We know that a small change in the actual function $\displaystyle \Delta h$ is approximately equal to a small change in the tangent line $\displaystyle dh$.

That is,

$\displaystyle \Delta h\approx dh$

But the derivative is,

(Hint: $\displaystyle \sqrt{4+h}=(4+h)^{1/2}$)

$\displaystyle \frac{4+h}{2}=2+(h/2)$

(The reason why we look at the derivative is because the derivative**is**the tangent line. So a small change in the derivative is a small change in the function). - Nov 26th 2006, 08:02 PMmightymax
Thanks for the help

but I dont really understand how to answer the question. And do I look at the derivative of the sqroot(4+h) function? and if so what do i do next? - Nov 26th 2006, 11:06 PMearboth
Hello,

I'll try to explain what TPH meant.

You've got a function $\displaystyle v(h)=\sqrt{4+h}$ which gives for any h the value of the square-root. You know that v(0) = 2.

Now consider the tangent line at this point. For small amounts of h the tangent line and the curve of the function v are practivally the same. (see attachment. I've enlarged a small surrounding of (0, v(0)))

To calculate the tangent line you need the slope of the line which is the same as the gradient of v for h = 0:

$\displaystyle v'(h)=\frac{1}{2}\cdot \frac{1}{\sqrt{4+h}}$. Thus $\displaystyle v'(0)=\frac{1}{4}$

Therefore the tangent line has the equation:

$\displaystyle t(h)=\frac{1}{4}\cdot h +2$

Because t(h) is the same as v(h) for small values of h, you can write (or calculate): $\displaystyle \sqrt{4+h}\approx \frac{1}{4}\cdot h +2$

EB