Day 11, Mat128: Calculus A

Please read section 2.1 to prepare, and do the preview exercise prior to our Monday meeting.

In a way this is review. That's the good news!

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:

      

  • Let's have a look at a couple of the exercises at the end of this section:
    2. Exercise 2: Local linearization of a graph
    3. Exercise 3: Estimating with the local linearization
    4. Exercise 4: Predicting behavior from the local linearization

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