anonymous
  • anonymous
Help.. on Differentials, see below problem. Use differentials to approximate the quantity. Sq root of 8.7
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sqrt{x} + 1/2\sqrt{x}*Change\]
anonymous
  • anonymous
So it is... \[\sqrt{9} + 1/(2\sqrt{9})*-.3\]
amistre64
  • amistre64
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

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anonymous
  • anonymous
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
anonymous
  • anonymous
It should be 2.95... Look above at my equation. It is f(x)+ Derivative of f(x) * the small amount
amistre64
  • amistre64
|dw:1333398406504:dw| 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 \(\Delta y\) \[y_n=\frac{x+9-.0025}{6}\] \[y_n=\frac{8.7+9-.0025}{6}\to\ \frac{17.6975}{6}=2.94958\bar 3\] \[(2.94958\bar 3)^2=8.7000418402\bar 7\]
anonymous
  • anonymous
I am certain the answer is 2.5
anonymous
  • anonymous
**2.95
amistre64
  • amistre64
:) im not too concerned about the answer perse since im not really up to par with the method
phi
  • phi
try 2.9500
phi
  • phi
This is the general idea \[ f(x_0 + \Delta x )\approx f(x_0) + f'(x_0)\Delta x\] In this case \( f(x)= x^{\frac{1}{2}} \) and I hope you can find the derivative of f(x) \(f'(x)= \frac{1}{2\sqrt{x}} \) Now let's use the differentials to find (approximately) the square root of 8.7= 9-0.3 We pick 9 because we know 3= sqrt(9). We see \( \Delta x= -0.3\) because we want \(x_0=9 \) and \(x_0+\Delta x = 8.7 \) To use \[ f(x_0 + \Delta x )\approx f(x_0) + f'(x_0)\Delta x\] we need \( f(x_0)= \sqrt{9}= 3 \) we need \( f'(x_0)= \frac{1}{2\sqrt{9}}= \frac{1}{6} \) we already know \( \Delta x= -0.3 \) Putting it all together: \[ \sqrt{8.7}= \sqrt{9-0.3}\approx 3-\frac{1}{6}\cdot\frac{3}{10}= 3-\frac{1}{20} =2.95\]
anonymous
  • anonymous
awesome, thank you everyone. When I put in 2.9500 it accepted it. Thank you for the workout of the problem, that will help me understand how to get the answer! :)

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