MHB 9.2.2 AP Calculus Exam Slope Fields

[karush]

View attachment 9322
I'm just going to post this image now since my tablet won't render the latex. This is a free response question..
But my experience is that the methods of solving are more focused here at mhb saving many error prone steps..

Mahalo ahead...

 

[HallsofIvy]

What is your purpose in posting this? Do you just want someone to do your homework for you?

If you understand what a "slope field" is then (a) should be straight forward.

The tangent line to y= f(x) at (a, b) is y= f'(a)(x- a)+ b. In (b) you are told that a= 0 and b= 1. What is f'(0) when you are also told that dy/dx= (3- y)cos(x)? Using that equation, what is y when x= 1?

The equation, \frac{dy}{dx}= (3- y)cos(x) is "separable" as \frac{dy}{3- y}= cos(x) dx. Integrate!

 

[karush]

It's not a homework assignment

 

[skeeter]

View attachment 9323
(b) $f(0)=1$

$\dfrac{dy}{dx}\bigg|_{(0,1)} = (3-1)\cos(0) = 2$

tangent line at $(0,1)$ is $y-1 = 2(x-0) \implies y = 2x+1$

$f(0.2) \approx y = 2(0.2)+1 = 1.4$

(c) $\dfrac{dy}{3-y} = \cos{x} \, dx$

$\dfrac{dy}{y-3} = -\cos{x} \, dx$

$\ln|y-3| = -\sin{x} + C$

$y-3 = e^{-\sin{x} + C} = e^C \cdot e^{-\sin{x}} = Ae^{-\sin{x}}$

$y = 3+Ae^{-\sin{x}}$

initial condition is $(0,1)$ ...

$1 = 3 + Ae^0 \implies A = -2$

$f(x) = 3-2e^{-\sin{x}}$

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[karush]

(b) $f(0)=1$

$\dfrac{dy}{dx}\bigg|_{(0,1)} = (3-1)\cos(0) = 2$

tangent line at $(0,1)$ is $y-1 = 2(x-0) \implies y = 2x+1$

$f(0.2) \approx y = 2(0.2)+1 = 1.4$

(c) $\dfrac{dy}{3-y} = \cos{x} \, dx$

$\dfrac{dy}{y-3} = -\cos{x} \, dx$

$\ln|y-3| = -\sin{x} + C$

$y-3 = e^{-\sin{x} + C} = e^C \cdot e^{-\sin{x}} = Ae^{-\sin{x}}$

$y = 3+Ae^{-\sin{x}}$

initial condition is $(0,1)$ ...

$1 = 3 + Ae^0 \implies A = -2$

$f(x) = 3-2e^{-\sin{x}}$

Ok I really appreciate the help
I've always had difficulty in understanding slope Fields
I'll do some more and see if I can go thru it all the way.

I am basically reviewing this it never showed up when I took the class