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Calculus/linear approximation

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Expert: - 11/23/2009


Question
Hi! I really appreciate your help. I am having trouble with these problems and hope you can help me. I have to use the differential to approximate each quantity. Then use a calculator to approximate the quantity, and give the absolute value of the difference in the two results to four decimal places. Thank you so much!

1) √.99
2)e^.01
3)ln 1.05
4)sin(.03)

Answer
1) This looks like y = √x, and we take x0 = 1, so x = x0 - 0.01.

The derivative is y' = 1/(2√x).  y'(1) = 1/2, so to approximate √0.99, it is 1 - 0.01/2
= 1 - 0.005 = 0.995, or 0.9950 to four places.  Now 0.005² = 0.990025, so that is fairly close.

2) y=e^x, y'=e^x, and x0 = 0, and y0 = e^0 = 1.
0.01 is 0.01 above 1, so y0 + 0.01(y'(0)) = 1 + 0.01(1) = 1.01.
Now e^(0.01) is really 1.01005...

3) The ln(1.05) can be found by letting x0=1.
Now since y=ln(x), y'=1/x, and y'(1)=1.  Also, at x0, y is 0.
So y0 + hy'(x0) = 0 + 0.05(1) = 0.05.
The ln(1.05) is really 0.04879...

4) Use x0=0, the y(x0) = 0.
It is known that y' = cos(x), and y'(x0) = 1.
Thus, sin(0.03) = 0 + 0.03(1) = 0.03.
Now the sin(0.03) is really 0.029996, which is very close.
In fact, to four digits, it is 0.03000.  

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