Thursday, November 09, 2006

Chapter 3 Test Question 12

a) Use linear approximation techniques to estimate the fourth root of 256.8. Leave your answer as a number plus or minus a fraction. Show your work.
b) Determine the calculator-generated value of the fourth root of 256.8
c) To how many decimal places is your estimate similar to the calculator-generated value?

Ok. I'm gonna concentrate on how to do part A, just because part B is putting something into your calculator and part C is subracting your answer from that number.

The trick to the problem is visualization. There's an equation for linear approximation, but it just makes a pretty simple concept really complicated.
NOTE: THIS IS NOT A DIAGRAM OF THE 4TH ROOT OF X. IT IS AN EXAMPLE (because the graph of y=the 4th root of x looks really straight)
This is like the diagram that explains linear approximation in the book. It makes very little sense, so I'll try to explain what's going on. We want to find f(256.8). You may have noticed that f(256) is equal to 4. We can imagine that f(256.8) would be really close to 4, but it should be a tiny bit larger. To find that tiny bit, we're going to get the slope (derivative) at 256 and pretend that it hardly changes between 256 and 256.8. If we imagine a tangent line with that slope that only touches at 256, if we go to 256.8 on that line, the Y value there will be about the same as f(256.8). If the black line in the diagram is f(x) and we set A equal to 256, we can calculate the tangent line (the purple one) at that point. Now, to approximate f(256.8), we're just going to trace that tangent line out .8 units farther and we'll add that change to the original f(256). In the graph above, .8 is represented by the variable X.

To put it mathematically...
Basic Equation
Y=X^.25 (same as fourth root)

Y(256) =4

Derivative Equation
Y'=.25X^-.75 (by the power rule)
Y'(256) = .25 (256^-.75)
Simplifies to...
Y'(256) = 1/256

Distance down the Tangent Line
Since we know f(256) and we want f(256.8), the distance we need to go down the tangent line is .8. Since we know the slope at 256, the change in Y will be the slope times .8, or:
Change in Y = (.8)(.25 (256^-.75))

The Final Answer
Now we just take our original number (f(256)=4) and add the change in Y when we move .8 units to the right (Change in Y = (.8)(1/256)), and we get the answer

256.8^.25 = (approximately) 4 + (1/320)

Alternate method
An (arguably) easier way to do this question would be with the equation:
f(A+X) = f(A) + X*f'(A)
Where A and X are the same as described above.


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