9.1.317 AP calculus exam multiple choice derivatives of sin wave

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SUMMARY

The discussion focuses on the derivatives of sine and cosine waves as part of the 9.1.317 AP Calculus exam multiple-choice questions. Key observations include a wave with a period of 12 and an amplitude of 3, which the user attempted to replicate using Desmos but was unsuccessful. The conversation also highlights the relationship between the accumulation function and its derivatives, specifically noting the function \( h(x) = \int_0^x f(t) \, dt \) and its derivatives \( h'(x) = f(x) \) and \( h''(x) = f'(x) \). The user raises questions about the behavior of these functions at specific points, particularly \( h(6) < 0 \) and the implications of \( h'(6) = 0 \) and \( h''(6) > 0 \).

PREREQUISITES
  • Understanding of calculus concepts, particularly derivatives and integrals.
  • Familiarity with sine and cosine wave properties, including amplitude and period.
  • Proficiency in using graphing tools like Desmos for visualizing functions.
  • Knowledge of accumulation functions and their derivatives in calculus.
NEXT STEPS
  • Explore the properties of sine and cosine functions in detail.
  • Learn how to effectively use Desmos for graphing calculus functions.
  • Study the Fundamental Theorem of Calculus and its implications for accumulation functions.
  • Investigate the behavior of functions at critical points and inflection points.
USEFUL FOR

Students preparing for the AP Calculus exam, educators teaching calculus concepts, and anyone interested in understanding the derivatives of trigonometric functions.

karush
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ok just posted an image due to macros in the overleaf doc

this of course looks like a sin or cos wave and flips back and forth by taking derivatives
looks like a period of 12 and an amplitude of 3 so...
but to start I was not able to duplicate this on desmos

altho I think by observation alone I think which choices represent the graph.
 

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typo in the accumulation function. should be ...

$$h(x) = \int_0^x f({\color{red}t}) \, dt$$$h'(x) = f(x) \implies h''(x) = f'(x)$$h(6) < 0$ ... why?

$h'(6) = f(6) = 0$

$h''(6) = f'(6) > 0$ ... why?
 

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