Estimate relative error using differentials

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SUMMARY

The discussion focuses on estimating the relative error of the area of a right triangle using differentials, specifically when the hypotenuse H is 4 cm and the angle theta is 30 degrees with a measurement error of 15 minutes of arc. The formula for the area A is given as A=(1/4)H^2sin(2theta). Participants emphasize the need to find the differential dA and clarify that one minute of arc is precisely 1/60 of a degree, which is crucial for accurate calculations.

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  • Familiarity with trigonometric functions and their derivatives
  • Knowledge of angle measurement, specifically degrees and minutes of arc
  • Ability to apply formulas for area in geometry
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  • Learn about the application of trigonometric identities in calculus
  • Research the concept of relative error and its significance in measurements
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DevoMci
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Area of right triangle with hypotenuse H is

A=(1/4)H^2sin(2theta)

where theta is one of the acute angles.

Use differentials to estimate the relative errors of the area A if H=4cm and theta is measured to be 30 degrees with an error of measurement of 15 minutes of arc.

note: a minute of arc, 1' is approximately equal to (1/60) of a degree.

I'm not quite sure what a measurement of arc is and whether I'm supposed to find the derivative or what? I am so lost, thanks in advance.
 
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DevoMci said:
Area of right triangle with hypotenuse H is

A=(1/4)H^2sin(2theta)

where theta is one of the acute angles.

Use differentials to estimate the relative errors of the area A if H=4cm and theta is measured to be 30 degrees with an error of measurement of 15 minutes of arc.

note: a minute of arc, 1' is approximately equal to (1/60) of a degree.

I'm not quite sure what a measurement of arc is and whether I'm supposed to find the derivative or what? I am so lost, thanks in advance.
You need to find the differential of A, dA. Your book should have some examples of how to find the differential.

Also, unless I'm missing something, one minute of arc, 1', is exactly equal to 1/60 of a degree.
 

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