3.3.04 AP Calculus Exam 2nd derivative

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Discussion Overview

The discussion revolves around the calculation of the second derivative from a given first derivative equation in the context of an AP Calculus exam problem. Participants explore the steps involved in deriving the second derivative and evaluate it at specific points, while also referencing answer choices from the exam.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant suggests starting the derivation by dividing both sides of the equation $$\dfrac{dy}{dx}=\dfrac{2x-y}{x+2y}$$.
  • Another participant provides the expression for the second derivative, $$\dfrac{d^2y}{dx^2} = \dfrac{(x+2y) \left(2 - \frac{dy}{dx} \right) - (2x-y)\left(1+2\frac{dy}{dx}\right)}{(x+2y)^2}$$, noting a specific value for $$\dfrac{dy}{dx}$$ at the point (3,0).
  • Some participants discuss substituting specific values of $$x$$ and $$y$$ into the second derivative expression, leading to calculations that yield $$-\frac{10}{3}$$.
  • There is a correction regarding the answer choice, indicating that the initial assumption of the answer being D is incorrect, prompting a reevaluation of the answer choices.
  • A later reply presents an alternative derivation approach, leading to the same conclusion of $$\frac{d^2y}{dx^2}=-\frac{10}{3}$$.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct answer choice, as there are conflicting interpretations of the results and the answer options provided in the exam.

Contextual Notes

There are unresolved aspects regarding the derivation steps and the implications of the answer choices, which may depend on the definitions and interpretations of the problem presented.

karush
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Ok not sure if I fully understand the steps but presume the first step would be divide both sides deriving$$\dfrac{dy}{dx}=\dfrac{2x-y}{x+2y}$$offhand don't know the correct answer
$\tiny{from College Board}$
 

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Re: 297 AP Calculus Exam 2nd direvative

$\dfrac{d}{dx} \left(\dfrac{dy}{dx} = \dfrac{2x-y}{x+2y}\right)$

$\dfrac{d^2y}{dx^2} = \dfrac{(x+2y) \left(2 - \frac{dy}{dx} \right) - (2x-y)\left(1+2\frac{dy}{dx}\right)}{(x+2y)^2}$note that $\dfrac{dy}{dx} \bigg|_{(3,0)} = 2$

finish the evaluation ...
 
Re: 297 AP Calculus Exam 2nd direvative

are you suggesting a plug in of

$$\dfrac{(x+2y) \left(2 - (2) \right) - (2x-y)\left(1+2(2)\right)}{(x+2y)^2}$$
 
Re: 297 AP Calculus Exam 2nd direvative

karush said:
are you suggesting a plug in of

$$\dfrac{(x+2y) \left(2 - (2) \right) - (2x-y)\left(1+2(2)\right)}{(x+2y)^2}$$
... And we know what x and y are...

-Dan
 
Re: 297 AP Calculus Exam 2nd direvative

topsquark said:
... And we know what x and y are...

-Dan
so with $x=3$ and $y= 0$
$$\dfrac{(x+2y) \left(2 - (2) \right) - (2x-y)\left(1+2(2)\right)}{(x+2y)^2}
=\dfrac{(3+2(0)) \left(2 - (2) \right) - (2(3)-(0))\left(1+2(2)\right)}{((3)+2(0))^2}=-\frac{10}{3}$$

which is D of the selection hopefully
 
Re: 297 AP Calculus Exam 2nd direvative

karush said:
so with $x=3$ and $y= 0$
$$\dfrac{(x+2y) \left(2 - (2) \right) - (2x-y)\left(1+2(2)\right)}{(x+2y)^2}
=\dfrac{(3+2(0)) \left(2 - (2) \right) - (2(3)-(0))\left(1+2(2)\right)}{((3)+2(0))^2}=-\frac{10}{3}$$

which is D of the selection hopefully

not D ... take another look at the answer choices
 
$$(x+2y)\frac{dy}{dx}=2x-y$$

$$x\frac{dy}{dx}+2y\frac{dy}{dx}=2x-y$$

$$\frac{dy}{dx}+x\frac{d^2y}{dx^2}+2\left(\frac{dy}{dx}\right)^2+2y\frac{d^2y}{dx^2}=2-\frac{dy}{dx}$$

$$2+3\frac{d^2y}{dx^2}+8=0$$

$$\frac{d^2y}{dx^2}=-\frac{10}{3}$$
 
Re: 297 AP Calculus Exam 2nd direvative

skeeter said:
not D ... take another look at the answer choices
then A

https://dl.orangedox.com/6rStfn4eMFHuHvAKuX
 
Last edited:

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