Need some Calculus Help on a few problems.

[Justinx93]

I'm not entirely bright at math hehe.
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Use Differentials (or equivalently, a linear approximation) to approximate sin(25°) as follows: Let f(x) = sin(x) and find the equation of the tangent line to f(x) at a 'nice' point near 25°. Then use this to approximate sin (25°)

Approximation = _________

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Use differentials (or equivalent, a linear approximation) to approximate ln(0.97) as follows: Let f(x) = ln(x) and find the equation of the tangent line to f(x) at a 'nice' point near 0.97. Then use this to approximate ln(0.97)

Approximation = __________

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The circumference of a sphere was measured to be 90.000 cm with a possible error of 0.50000 cm. Use linear approximation to estimate the maximum error in the calculated surface area. _______

Estimate the relative error in the calculated surface area. _________

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Thanks in advance. it would be great if I can get some explanations to go with it.

 

[pka]

Use Differentials (or equivalently, a linear approximation) to approximate sin(25°) as follows: Let f(x) = sin(x) and find the equation of the tangent line to f(x) at a 'nice' point near 25°. Then use this to approximate sin (25°)
Approximation = _________

From the limit definition for derivative we get this approximation:
\(\displaystyle f(x_0+\Delta x)\approx \Delta x\cdot f'(x_0)+f(x_0)\).

\(\displaystyle x_0=\dfrac{\pi}{6}~\&~\Delta=\dfrac{-5\pi}{180}\)

 

[soroban]

Hello, Justinx93!

I'll do the last one . . . It's the trickiest.

The circumference of a sphere was measured to be 90.000 cm with a possible error of 0.50000 cm.

(a) Use linear approximation to estimate the maximum error in the calculated surface area.

(b) Estimate the relative error in the calculated surface area.

(a) We know: .\(\displaystyle C \,=\,2\pi r \quad\Rightarrow\quad r \,=\,\dfrac{C}{2\pi}\)
. . We are given: .\(\displaystyle C = 90.\) . Hence: .\(\displaystyle r \,=\,\dfrac{90}{2\pi} \,=\,\dfrac{45}{\pi}\) .[1]

We have: .\(\displaystyle dC \,=\,2\pi\,dr\)
. . We are given: .\(\displaystyle dC = \frac{1}{2}.\) . Hence: .\(\displaystyle \frac{1}{2} \,=\,2\pi\,dr \quad\Rightarrow\quad dr \,=\,\dfrac{1}{4\pi}\) .[2]

We know: .\(\displaystyle A \,=\,4\pi r^2 \quad\Rightarrow\quad dA \,=\,8\pi r\,dr\)

Substitute [1] and [2]: .\(\displaystyle dA \:=\:8\pi\left(\dfrac{45}{\pi}\right)\left(\dfrac{1}{4\pi}\right) \:=\:\dfrac{90}{\pi}\)


(b) Relative error \(\displaystyle = \:\dfrac{dA}{A}\)

We have: .\(\displaystyle A \:=\:4\pi r^2 \:=\:4\pi\left(\dfrac{45}{\pi}\right)^2 \:=\:\dfrac{8100}{\pi}\)

Therefore: .\(\displaystyle \dfrac{dA}{A} \:=\:\dfrac{\frac{90}{\pi}}{\frac{8100}{\pi}} \:=\: \dfrac{1}{90} \;=\;1.1111\%\,\text{ to 5 significant figures.}\)