Question

Help.. on Differentials, see below problem.

Use differentials to approximate the quantity.

Sq root of 8.7

Answer 1

\[\sqrt{x} + 1/2\sqrt{x}*Change\]

Answer 2

So it is... \[\sqrt{9} + 1/(2\sqrt{9})*-.3\]

Answer 3

y = sqrt(x);

y' = 1/2sqrt(x)

sqrt(8.7) is close to sqrt(9)

y' = 1/2sqrt(9) = 1/6

linear approximation would be:

a = 1/6 x -9/6 +3

a = (x+9)/6 ; when x = 8.7 we get

a = (8.7+9)/6 = 17.7/6 = 2.95

2.95^2 = 8.7025

Answer 4

hmmm, when I enter 8.7025 into my online hw it says its incorrect but doesn't tell me what is the correct answer. The answer must be 4 decimal places. This is what I did:

f'(x) = (1/2) x^(-1/2) all mult by 0.3(which is the difference of 9 and 8.7) = 0.5

Answer 5

It should be 2.95... Look above at my equation. It is f(x)+ Derivative of f(x) * the small amount

Answer 6

my idea, which prolly aint the right "method" is:

find the equation of the tangent line near 8.7; say at 9
\[y=\frac{x+9}{6}\to\ \frac{17.7}{6}=2.95\]

\[(2.95)^2=8.7025\]\[h=8.7025-8.7000=.0025\]
\[\Delta y=hf'(9)=\frac{.0025}{6}\]
drop the tangent equation by \(\Delta y\)

\[y_n=\frac{x+9-.0025}{6}\]
\[y_n=\frac{8.7+9-.0025}{6}\to\ \frac{17.6975}{6}=2.94958\bar 3\]
\[(2.94958\bar 3)^2=8.7000418402\bar 7\]