Differentials Solution 12

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SOLUTION 12: Begin with the function

$f(x)= \arcsin x$
and choose

$x-values: 0.5 \rightarrow 0.45$
so that

$\Delta x = 0.45-0.5 = -0.05$
The derivative of

$\ y=f(x) \$ is

$f'(x)= \displaystyle{ 1 \over \sqrt{1-x^2} }$
The exact change of

$y-$ values is

$\Delta y = f(0.45) - f(0.5)$

$= \arcsin(0.45) - \arcsin(0.5)$

$= \arcsin(0.45) - \displaystyle{ \pi \over 6 }$

$\approx \arcsin(0.45) - 0.5236$
The Differential is

$dy = f'(0.5) \ \Delta x$

$= \displaystyle{ 1 \over \sqrt{1-(0.5)^2} } \cdot (-0.05)$

$= \displaystyle{ 1 \over \sqrt{1-(1/2)^2} } \cdot (-0.05)$

$= \displaystyle{ 1 \over \sqrt{1-(1/4)} } \cdot (-0.05)$

$= \displaystyle{ 1 \over \sqrt{3/4} } \cdot (-0.05)$

$= \displaystyle{ 1 \over \sqrt{3}/2 } \cdot (-0.05)$

$= \displaystyle{ 2 \over \sqrt{3} } \cdot (-0.05)$

$= \displaystyle{ -0.1 \over \sqrt{3} }$

$\approx \displaystyle{ -0.1 \over 1.7321 }$

$\approx -0.0577$
We will assume that

$\Delta y \approx dy \ \ \ \ \longrightarrow$

$\arcsin(0.45) - 0.5236 \approx -0.0577 \ \ \ \ \longrightarrow$

$\arcsin(0.45) \approx 0.5236 -0.0577 \ \ \ \ \longrightarrow$

$\arcsin(0.45) \approx 0.4659$

NOTE: The number 0.5 was chosen for its proximity to 0.45 and for it's convenient inverse tangent. Check the accuracy of the final estimate using a CALCULATOR:

$\arcsin(0.45) \approx 0.4668$

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