a) Use Differentials to estimate $\sqrt {3.99}$
b)Find a point on the Graph of $y=x^3-3x$ where the tangent line $y=9x+8$, and write the equation of the tangent line at the point.
The derivative of $\sqrt {3.99}$ is $1/2(3.99)^{-1/2}$
How do I use $L(x)= f(a)+f'(a)(x-a)$ with this problem?
Best Answer
Here's how to find the linear approximation:
Let $y=\sqrt{x}$. $\sqrt{3.99} \approx \sqrt{4}=2$. So the difference between $\sqrt{3.99}$ and $2$ should approximately be equal to the differential:
$$dy=y'(x)dx = \left.(\sqrt{x})'\right|_{x=4}(3.99-4)$$
Thus $$\color{red}{\sqrt{3.99} \approx \sqrt{4} + \left.(\sqrt{x})'\right|_{x=4}(3.99-4)}$$
Now let's check our answer. WolframAlpha gives the following values: $$\sqrt{3.99} = 1.99749843554381789\dots \\ \sqrt{4} + \left.(\sqrt{x})'\right|_{x=4}(3.99-4) = 2+\frac {1}{4}(3.99-4) = 1.9975$$
So you can see that if we get $5$ digits of accuracy (round to the $5$th digit). That's pretty good. Just for fun, though, let's see what the quadratic approximation yields:
The quadratic approximation is $$\color{blue}{\sqrt{3.99} \approx \sqrt{4} + \left.(\sqrt{x})'\right|_{x=4}(3.99-4) +\frac 1{2!}\left.(\sqrt{x})''\right|_{x=4}(3.99-4)^2}$$
Using WolframAlpha again gives:
$$2+ \frac 14(3.99-4) -\frac 1{2!}\frac 1{32}(3.99-4)^2 = 1.9974984375$$
Now we've $9$ digits of accuracy. I wonder how many terms of the Taylor series I need for $100$ digits... :)