3.3.04 AP Calculus Exam 2nd derivative

In summary, the first step would be to divide both sides to get $$\dfrac{dy}{dx}=\dfrac{2x-y}{x+2y}$$
  • #1
karush
Gold Member
MHB
3,269
5
View attachment 9319

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
$\tiny{from College Board}$
 

Attachments

  • 297.PNG
    297.PNG
    4.5 KB · Views: 106
Physics news on Phys.org
  • #2
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 ...
 
  • #3
Re: 297 AP Calculus Exam 2nd direvative

are you suggesting a plug in of

$$\dfrac{(x+2y) \left(2 - (2) \right) - (2x-y)\left(1+2(2)\right)}{(x+2y)^2}$$
 
  • #4
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
 
  • #5
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
 
  • #6
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
 
  • #7
\(\displaystyle (x+2y)\frac{dy}{dx}=2x-y\)

\(\displaystyle x\frac{dy}{dx}+2y\frac{dy}{dx}=2x-y\)

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

\(\displaystyle 2+3\frac{d^2y}{dx^2}+8=0\)

\(\displaystyle \frac{d^2y}{dx^2}=-\frac{10}{3}\)
 
  • #8
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:

FAQ: 3.3.04 AP Calculus Exam 2nd derivative

What is the purpose of the 2nd derivative in calculus?

The 2nd derivative is used to describe the rate of change of the rate of change of a function. It helps to determine the concavity and inflection points of a function, and can also be used to find the maximum and minimum values of a function.

How is the 2nd derivative calculated?

The 2nd derivative is calculated by taking the derivative of the first derivative of a function. In other words, it is the derivative of the slope of the original function.

What does a positive 2nd derivative indicate?

A positive 2nd derivative indicates that the slope of the original function is increasing. This means that the function is concave up and has a minimum point at that particular value.

What does a negative 2nd derivative indicate?

A negative 2nd derivative indicates that the slope of the original function is decreasing. This means that the function is concave down and has a maximum point at that particular value.

How is the 2nd derivative used in real-world applications?

The 2nd derivative is used in various fields such as physics, engineering, and economics to model and analyze real-world phenomena. For example, it can be used to determine the acceleration of an object or the optimal production level for a company.

Similar threads

Replies
6
Views
2K
Replies
4
Views
2K
Replies
6
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
4
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
Back
Top