Day 11, Mat128: Calculus A

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I had a couple of questions on it, and we really didn't make enough progress to justify submitting it before class. You can do that after class!

The upshot is that linear functions are good approximations to smooth functions (functions with derivatives), if you zoom in close enough.

• An important first step was using what we call the "linearization", the tangent line: we use the fact that \[ f(a+h) \approx f(a) + hf'(a) \] This comes straight out of the limit definition, where we throw away the limit. That's why we have to write "$\approx"$: \[ f'(a) \approx \frac{f(a+h)-f(a)}{h} \]

• So if we define $x=a+h$, then \[ f(x) \approx f(a) + f'(a)(x-a) \] or \[ f(x) \approx y = f(a) + f'(a)(x-a) \] The tangent line serves as an impressive approximating machine, provided $h=x-a$ is small enough. (How small is "small enough?")

  We can see why this works well if we get "small enough", if we visit the website suggested by our authors.

• That linear function given by the tangent line, $y = f(a) + f'(a)(x-a)$, is called the local linearization of $f$ at the point $(a,f(a))$:
  $L(x)=f'(a)(x-a)+f(a)$
  • For example, suppose you need to compute $\sqrt{15.96}$. Now the easy way is to use your calculator, but the cowboy way would be to use the linearization! Let's have a go at that one.
    • We know the square root of 16, which is close: our answer should be near 4.

    • Draw a picture! The picture will demonstrate two different types of quantities that we want to discuss: differentials $dy$, and increments $\Delta y$.

      Differentials versus increments

      
      The increment is the true change in the function value; the differential approximates the true change: One way to remember the difference:
      • Differentials $dy$ follow the direction suggested by the derivative
      • Increments $\Delta y$ are actual changes in the function value.

      $dy \approx \Delta y$

      We want the increment, but may settle for the easily computed differential.

    • Compute the derivative of $f(x)=\sqrt{x}$ at $x=16$, and we can write down the tangent line equation: \[ L(x) = (x-16)/8 + 4 \]

    • We use the tangent line for our approximation, and, when we're done, we can ask "is our answer an over- or an under-estimate?"

      The answer takes us back to the second derivative:

      

  • For today's worksheet, we'll work some activities and homework problems from section 1.8.