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AP Calculus AB

An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

Tuesday, November 29, 2005

Tangent Line Approximation

Stan the Clause - another one of the usual related rates problem Mr. K always gives every morning. As to what I did, I never preassumed that a wide set of concepts are to be considered in answering these kinds of problems. Claiming to that perception, I twisted my mind a little bit to come up with not "The Answer" but something just "close" to it. And to my dismay, I got way far from it. (laughs!)

A lot of mathematical ideas come into use when these kinds of questions are given:
- Find the rate at which the angle of elevation from the naughty boy's house to the flying can is changing in radians per second when the angle of elevation is pi over 12 -

To make life easier, visualize the whole problem. Try to illustrate. Know all the given facts or numbers. Then, give it your best shot! ;) It is just like solving an everyday math question, only it is more complex. Students might have to apply all the stuff they have learned throughout their mathlife. Regarding this problem, I was not able to get this one completely because I never thought this would include trigonometric concepts and all. I was stuck when I came to that point because I have slightly forgotten some stuff about it. So now I think I might have to work on refreshing myself with trigo.

In the remaining minutes of the period, we learned a new lesson that is the Tangent Line Approximation. This is basically applied in (as to what Mr. K used as an example) finding the value of a square root function with a non-perfect square number.

Given f(x) = √37 find the value

First, formulate its derivative equation - f ' (x) = 1 / (2 √x)
Then, obtain a perfect square number closest to the given number, in this example the closest is 36.
Substitute the x variables in the derivative equation by 36 and by doing that, the slope at that point will result. - f ' (x) = 1 / (2√36) = 1 / 12
Use this slope to form a derivative function for the given number 37 using the point-slope form. Y - Y1 = m (X - X1)

-> f(X) - 6 = 1 / 12 (37 - 36)

So, f (37) = 1 / 12 + 6 = 73 / 12

And using the calculator, the √37 is 6.08333 that is, almost as close to 73 / 12

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