Homework Help Forum: Calculus

Posted by Calvin on Wednesday, October 28, 2009 at 11:40pm.

The measurement of one side of a right triangle is found to be 9.5 inches, and the angle opposite that side is 26°45' with a possible error of 1.5'. Approximate the percent error in computing the area of the triangle.

• Calculus - MathMate, Thursday, October 29, 2009 at 8:55am

  Area of right triangle =(1/2)L*(L)cot(θ) =(1/2)L²cot(θ) L=9.5, θ=26°45' ± 1.5' =0.46688 rad ± 0.000002315 rad. If you have not done calculus yet, calculate the area based on the given θ, then calculate the largest and smallest possible value of θ to give the higher and lower limits. If you have done calculus, set A(θ)=L²cot(θ) and differentiate with respect to θ to get e=A'(θ). Multiply e by the error in θ to give the error in area. I get ±0.000516 sq.in. using both methods.

  
  

• Calculus - MathMate, Thursday, October 29, 2009 at 9:03am

  =0.46688 rad ± 0.000002315 rad. should read 0.46688 rad ±0.000436 rad. The calculated error should therefore read: "I get ±0.097 sq.in. using both methods."

  
  

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