MHB 3.3.04 AP Calculus Exam 2nd derivative

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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