Given depth function depth(t) = .02 t² - 12 t + 30, we wish to find the instantaneous rate which depth is changing at clock time t = 40.

We first find the rate of depth change function.

• The depth function is of the form y = depth(t)  = a t² + b t + c, a quadratic, so that the rate of depth change function is the derivative depth ' (t) = .04 t - 12 (obtained from our previous knowledge that the derivative of the quadratic is y' = 2 a t + b).

Given the rate function rate(t) = .03 t - 15, what is the corresponding the depth function, provided that depth at clock time t = 0 is 100?

• The rate is the derivative rat(t) = depth ' (t), so we have depth ' (t) = .03 t - 15.
• This is of the form y' = mt + b which, as we have seen, is the derivative of the function y = .5 m t² + b t + c.

• Since m = .03 and b = -15, we see that        y = .5 m t² + b t + c = .015 t² - 15 t + c.

• If we know that the depth at clock time t = 0 is 100, then depth (0) = .015 (0)² - 15 (0) + c = c.
• We also know that depth (0) = 100, so c = 100.
• Our depth function is therefore depth (t) = .015 t² - 15 t + 100.

video clip 01

http://youtu.be/34QFdK1SvdE

We proceed to find the derivative of the function y = a x³.

• The graph below depicts this function.
• In order to find the derivative of the function at the point x, we follow the usual procedure:

• We evaluate the function at x and at x + `dx, then
• we determine the change `dy in y values corresponding to the change `dx at x.
• We then take the limit of this expression as `dx -> 0.

As shown on the graph the y values are y(x) = a x³ and y(x + `dx) = a (x + `dx)³.

• The rise of the slope triangle is therefore rise = `dy = a(x+`dx)³ - a x³, and the run is of course `dx.
• So to find y'(x) we will take the limit of the expression [a(x + `dx)³ - a x³] / `dx.

In order to compute the desired limit we first expand the expression (x + `dx)³.

• This is done by repeated application of the distributive law.
• We could also have used the binomial theorem, which will use later when we calculate the general derivative of the function y = a xⁿ. 
• The details of the calculation are shown below.

We now calculate the desired limit.

• We note that the a x³ term cancels out in the third step, leaving only terms with `dx, `dx² or `dx<sup>3</sup> in the numerator.
• This is a good thing, because we're dividing by `dx, which approaches 0 so it had better cancel out with everything in the numerator.
• It does, leaving us with the expression 3 a x² + 3 a x `dx + `dx².
• The limit of this expression as `dx approaches 0 is clearly 3 a x².
• This is our desired limit and the derivative of the function y = a x³.  Thus d (a x³) / dx = 3 x².

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http://youtu.be/sPtdq1DH6Rk

We apply the derivative to the following situation:

If our specific model of sandpile amount vs. diameter is y = a x³ = .002 x³, then how much additional sand would we have to add to a pile whose diameter is 15, in order to raise the diameter by .25 units, to 15.25?

• We could of course figure out the answer by substituting 15 and then 15.25 into the volume function y = .002 x³.
• However, it is often expedient to use knowledge of the derivative to estimate such a difference.

• Estimation is sometimes more efficient and the results provide greater insight into the way volume builds.

• The 'green' triangle on the graph depicts the rise corresponding to the 'run' of .25 from 15 to 15.25. The horizontal distance and, consequently, the vertical distance are greatly exaggerated for clarity.
• Observe that if we knew the slope of the green triangle, then since we know the run we could easily find the rise (rise = run * slope).

• We note that the derivative at x = 15 is the limiting value of the slopes of the triangles starting from the x = 15 point on the graph, as the run approaches 0.
• The run of the present graph, .25, is pretty small on the scale of the graph, so we expect that the slope of the green triangle is pretty close to the value of the derivative. So we use this derivative as the approximate slope to find the rise.

• The derivative y' = 3 a x² is, for this function, y' = .006 x².

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http://youtu.be/2jEokcXh-pc

The figure below is a closer look at the green triangle of the former graph.

• The lower left hand corner of the slope triangle is a point (15,6.75), representing the diameter x = 15 and the resulting volume, obtained by substituting the diameter into the formula y = volume = .002 x³.
• Note that as mentioned before we could find the volume change easily enough by substituting 15.25 into the same formula and subtracting the results to get the rise of the triangle.

At x = 15 the derivative of the function is y' = .006 x² = .006 (15²) = 1.35.

• This is the rate at which the function is changing at x = 15.
• To remind ourselves of the meaning of this rate, we note that since x is measured in diameter units and y in tablespoons (tbsp), the rate is 1.35 tbsp / diam unit.
• If volume continues changing at this rate, then as x changes by .25 units of diameter the volume must change by

• volume change = 1.35 tbsp / diam unit * (.25 diam units) = .35 tbsp.

• Of course the volume doesn't continue changing at precisely this rate, but if the rate doesn't vary too much our estimate will still be good.

The figure shows the graphical representation of our reasoning.

• The tangent line to the point (15, 6.75), with its slope y ' (15) = 1.35, is sketched in purple.

Since there is an upward curvature to the graph, with a resulting steady increase in the slope, the actual slope between the x = 15 and the x equal 15.25 points of the graph will be slightly greater than the 1.35 slope of the tangent line.

• The curvature of the graph and the resulting discrepancy between the slope of the tangent line and that of the slope triangle have been deliberately exaggerated in this figure; over the distance .25 between the x values for this function in this region, the graph would be very nearly straight and the difference between the two slopes would be very slight.

Assuming that the slopes are in fact nearly the same, so that the slope of the slope triangle is near 1.35, we can easily calculate the difference in the y values between the two points.

• This difference will be the rise of the triangle.
• If the run is .25, then the rise will be `dy = rise = slope * run = 1.35 * .25 = .35. 
• This rise corresponds to the .35 tbsp increase we found above.

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http://youtu.be/r74Cbq9gMgs

We see that in general at a graph point (x,y) of a function y(x) the slope is y'(x) and that, for small `dy corresponding to the change in x coordinate from x to x + `dx will be very close to  `dy = y ' (x) * `dx.

This is the essence of the concept of the differential. This is a very important concept that often eludes first-year calculus students. Hopefully this common-sense first exposure to the idea will make things clearer when a rigorous study of the differential is undertaken.

video clip 08

http://youtu.be/N1gD3SnzKV8