Differentiation

Derrickkhoo

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Nov 28, 2019
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How to do this?
Given a trigonometric function, f(θ) = cos(θ) + sin(θ).
  • Evaluate the first and second derivative for f(θ).
  • Produce a table for f(θ), f’’(θ), and f’’(θ) in the range of 0 to 360 degrees. Then, use 15 degrees interval for each dataset in the set of {0, 15, 30, …, 345, 360}.
  • Draw three graphs to represent f(θ), f’’(θ), and f’’(θ) using the datasets in Question 2(b).
 
How to do this?
Given a trigonometric function, f(θ) = cos(θ) + sin(θ).
  • Evaluate the first and second derivative for f(θ).
  • Produce a table for f(θ), f’’(θ), and f’’(θ) in the range of 0 to 360 degrees. Then, use 15 degrees interval for each dataset in the set of {0, 15, 30, …, 345, 360}.
  • Draw three graphs to represent f(θ), f’’(θ), and f’’(θ) using the datasets in Question 2(b).
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Please share your work/thoughts about this assignment

Hint:

ddθ[sin(θ)] =cos(θ)\displaystyle \frac{d}{d\theta}\left[sin(\theta)\right] \ = cos(\theta)
 
I presume you are just beginning Calculus. The rules you need are

If f(θ)=u(θ)+v(θ)\displaystyle f(\theta)= u(\theta)+ v(\theta) then f(θ)=u(θ)+v(θ)\displaystyle f'(\theta)= u'(\theta)+ v'(\theta), f(θ)=u(θ)+v(θ)\displaystyle f''(\theta)= u''(\theta)+ v''(\theta) and f(θ)=u(θ)+v(θ)\displaystyle f'''(\theta)= u'''(\theta)+ v'''(\theta)

as well as (sin(θ))=cos(θ)\displaystyle (sin(\theta))'= cos(\theta) and (cos(θ))=sin(θ)\displaystyle (cos(\theta))'= -sin(\theta). Of course the "second derivative" is just the derivative of the first derivative so repeat.
 
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