7.3 Sketching Slope Fields

Question: 7.3: Sketching Slope Fields
Answer: A plot of short line segments with slopes f(x,y) at lattice points (x,y) in the plane.
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Question: A slope field
Answer: a differential equation on an xy-plane.
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Question: A slope field represents
Answer: the "slope" of all the particular solutions to the differential equation
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Question: What does a slope field show
Answer: plug (x,y) coordinates into differential equation

draw short segments representing slope at each point
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Question: Slope field
Answer: plug in the coords into the differential equation for the slope of the tangent line

1 - (-1)(-2) = 1 - 2
= -1 (slope)

now plug everything into point slope formula
y-y1=m(x-x1)

y + 2 = -1 (x + 1)
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Question: The figure below shows the slope for the differential equation dy/dx = 1 - xy

Let f be the function that satisfies the given differential equation. Write an equation for the tangent line to the curve y = f(x) through the point (-1. -2)
Answer: y-y1=m(x-x1)
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Question: point slope formula
Answer: check where the equation will have a slope of 0 or unefined and check the graph

cross out ones that dont have it there

then pick a point and find the general slope to compare the equations to

keep doing this till u get all the answers
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Question: know how to match a slope field equation to a slope field picture
Answer: a.)
= 1/2

b.)
y - 1 = 1/2 (x - 1)

y - 1 = 1/2 (1.2 - 1)

y = 1.1
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Question: Consider the differential equation given by dy/dx = xy/2 and its slope field shown below.

a.) Calculate dy/dx at the point (1, 1) and verify that the result agrees with the figure.

b.) Let f be the function that satisfies the given differential equation. Write an equation for the tangent line to the curve y = f(x) through the point (1, 1). Then use your tangent line equation to estimate the value of f(1.2).
Answer: DO IT!
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Question: do hw questions for practice since i cant put pics in here
Answer: The process of finding the derivative of a dependent variable in an implicit function by differentiating each term separately, by expressing the derivative of the dependent variable as a symbol, and by solving the resulting expression for the symbol.

implicit is when it does not = y

x^2 + y^2 = 9
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Question: 7.6: Separation of Variables (General Solutions)
Answer: 2x + 2y(dy/dx) = 0
dy/dx = (-x/y)
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Question: Implicit Differentiation
Answer: implicit differentiation backwards
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Question: differentiate x^2 + y^2 = 9
Answer: Step 1: Multiply by dx and move the y to the other side to get like terms on each side
y * dy = x^2 dx

Step 2: Integrate & move C to right side
āˆ«y dy = āˆ«x^2 dx

y^2 / 2 + c= x^3 / 3 + c
move + c to the right

Step 3: Get y alone on the left side

multiply by 2 then square root (remember to do + or - for square roots)

y = +- āˆš((2/3)x^3 + c)

^^ Now it's explicit!
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Answer: the integral with + c
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Question: Find the general solution of each differential equation.

dy/dx = x^2 / y
Answer: y = (1 / cosx + c)
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Question: General Solution
Answer: y = e^((x^2)/2) * c
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Question: dy/dx = (sinx) y^2
Answer: y = ce^(x^2) - 2
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Question: dy/dx = xy
Answer: y = āˆš(2x^3 + c)
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Question: dy/dx = 2xy + 4x
Answer: y = e^(8/3)x^3 + c

aka

y = ce^(8/3)x^3
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Question: dy/dx = 3x^2 / y
Answer: y = 1 / (-e^x + c)
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Question: dy/dx = 8x^2 y
Answer: y = c/e^x^2 + 3
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Question: dy/dx = e^x y^2
Answer: y = ce^sinx
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Question: dy/dx = -2x (y - 3)
Answer: y = ce^((x^2)/2) + 2x) - 5
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Question: dy/dx = ycosx
Answer: y = ln(e^x + c)
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Question: dy/dx = (y + 5)(x + 2)
Answer: it becomes an exponent to the number (or equation) behind ln
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Question: dy/dx = e^(x-y)
Answer: y = ln (āˆš(2x^2 + c)
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Answer: find the general solution first like how you did in the previous lesson
-1/y = (x^3 / 3) + c

then take the numbers you were given and plug-in for c
c= -4/3

then, plug the c into the general solution to get the particular solution
-1/y = (x^3 / 3) - (4 / 3)
-1/y = (x^3 - 4)/3

finally, make sure y is alone and positive and not a fraction
y = -(3 / x^3 - 4)
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Question: dy/dx = 2x / e^2y
Answer: check notes for help

y = +āˆš(2x^2 - 7)

only pos bc its graphed in a positive position

use the coordinate point to determine which half of the curve you should graph for the square root ones
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Question: Lesson 7.7: Separation of Variables (Particular Solutions)
Answer: 1.) āˆ«1 / y+2 * dy = āˆ«e^x * dx

2.) ln|y+2| = e^x + c

3.) ln|-1+2| = e^0 + c
-1 = c

4.) ln|y+2| = e^x - 1
e^(ln|y+2|) = e^(e^x - 1)

5.) y + 2 = +- e^(e^x - 1)

6.) y = +- e^(e^x - 1) - 2

then find if the graph is positive or negative

7.) Plug in x-point again to both positive and negative versions to see if u get the correct y-value

+e^(e^0 - 1) - 2 = e^0 - 2 = -1 [CORRECT ONE SO IT'S POS]

-e^(e^0 - 1) - 2 = e^0 - 2 = -3 [WRONG SO NOT NEG]

8.) FINAL VERSION:
y = +e^((e^x) -1) - 2
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Question: Find the solution to the differential equation dy/dx = (xy)^2 with the initial condition y(1) = 1.
Answer: y = 4e^sinx
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Question: Find the solution y = f(x) to the differential equation dy/dx = 2x/y with initial condition f(2) = 1. (for square roots)
Answer: y = 2e^-cosx
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Question: Find the solution to the differential equation dy/dx = (y + 2)e^x with initial condition y(0) = -1. (for abs values)
Answer: y = -āˆš(4x^3 + 4)

it's neg cuz the y-val of the coords is neg
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Question: dy/dx = ycosx and y = 4 when x = 0
Answer: y = -2 e^(-1/5x) + 8
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Question: dy/dx = ysinx if y(Ļ€/2) = 2
Answer: y = 6e^(1/2 x^2 + 2x) - 5
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Question: dy/dx = 6x^2 / y if y(0) = -2
Answer: y = -āˆš(2e^x + 14)

make negative cuz y is negative
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Question: dy/dx = 1/5 (8 - y) and y = 6 when x = 0.
Answer: dy/dx = e^x / e^y

āˆ« e^y = āˆ« e^x dx

e^y = e^x + c

e^2 = e^0 + c

e^2 - 1 = c

y = ln(e^x + e^2 -1)
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Question: dy/dx = (y + 5)(x + 2) when f(0) = 1.
Answer: y = (-2 / x^2) + 1

AKA
y = (1 / (1 - (x^2)/2))
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Question: dy/dx = e^x / y if y(0) = -4.
Answer: y = a(b)^t

a is the initial value

b is the growth/decay factor

when b > 1, growth
when b < 1, decay
(when raised to a power)


dy/dt = a b^t * ln(b)
dy/dt = y * ln(b)

when it's dy/dt = k * y!!
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Question: Find the particular solution to dy/dx = e^(x-y) when f(0) = 2. Sketch the graph of this particular solution on the slope field provided.
Answer: The rate of change of a quantity is proportional to the size of the quantity.
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Question: Find the particular solution to dy/dx = xy^2 if y = 1 when x = 0. Sketch the graph of this particular solution on the slope field provided.
Answer: dw/dt = k * w
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Question: 7-8: Exponential Models with Differential Equations
Answer: dp/dt = kp

k will be negative cuz of decay but we dont need to write it bc we could just make k negative
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Question: Exponential Growth and Decay
Answer: y = ce^kt
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Question: What represents an exponential model?
Answer: exponention model of a differential equation

c is the initial value

e is the growth/decay factor

t is the x

k is the constant you have to find

kt will be the part of the right of the original equation (dy/dt = ky)

SO

kt is whatever the dy/dx equals just make the x into a t
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Question: The weight of an animal is increasing at a rate proportional to its weight.
Answer: y = 5e^6t
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Question: A bacteria population is shrinking at a rate proportional to its population size.
Answer: y = 4e^-3t
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Question: Solve dy/dt = ky using separation of variables
Answer: 1.) dw/dt = kw

2.) w = 3e^kt

3.) plug one of the coords ur given into the equation to get k
4 = 3e^3k
4/3 = e^3k
ln (4/3) = 3k
k = 0.0989 (STORE THE WHOLE THING IN UR CALCULATOR SO THERE IS NO ERROR. YOU CAN ONLY ROUND AT THE END.)

4.) plug the 5 back into x aka the t value and get the answer
w(5) = 3^e(the stored thing of 0.0989)(5)
w(5) = 4.845 pounds (you can round the final answer)
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Question: y = ce^kt
Answer: do not round any answers until the very end.

make sure to store it or just type the whole thing in the next equation
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Question: The solution to dy/dt = ky is y = Ce^kt, where C represents the initial value of the model.
Answer: positive

y = 5e^2t
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Question: Find the particular solution for each differential equation.

dy/dt = 6y and y = 5 when t = 0
Answer: negative

y = 7e^-3t

= y = 7(1/e)^3t
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Question: dy/dx = -3y and y = 4 when x = 0
Answer: yā‚€ stands for c which is the inital amount

y = ce^kt

2yā‚€ = yā‚€ e^k(7)
put 2 on the left bc its doubling, if it was halfing, put 1/2

2 = 1e^7k
the yā‚€ aka c cancels out on both sides

ln(2) = 7k
k = 0.099
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Question: WILL BE ONE ON THE TEST LIKE THIS:

An animal weighs 3 pounds at birth and 4 pounds just three months later. The weight of the animal is increasing at a rate proportional to it weight. Set up a differential equation for this scenario. How much will the animal weigh when its 5 months old?
Answer: 1280 ppl
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Question: reminder:
Answer: 0.063
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Question: A growth model will have a ______ exponent.
Answer: y = -2e^8t
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Question: A decay model will have a ______ exponent.
Answer: k = 0.01

w(t) = 30e^0.01t
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Question: doubling time/half life problem

A population y grows according to the equation dy/dt = ky, where k is a constant and t is measured in years. If the population doubles every 7 years, what is the value of k?
Answer: Pā‚€ = 295 flies
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Question: 5.) During a certain epidemic, the number of people that are infected at any time increases at a rate proportional to the number of people that are infected at that time. If 500 people when the epidemic is first discovered, and 800 people are infected 5 days later, how many people are infected 10 days after the epidemic is first discovered?
Answer: k = 0.0277
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Question: A population y grows according to the equation dy/dt = ky, where k is a constant and t is measured in years. If the population doubles every 11 years, then what is the value of k?
Answer: t = 3.506 hrs
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Question: dy/dt = 8y and y = -2 when t = 0
Answer: k = -0.0023
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Question: A certain animal weighs 30 grams at birth. During the first 4 weeks after the animal's birth, its weight in grams is given by the function W that satisfies the differential equation dW/dt = 0.01W, where t is measured in days. What is an expression for W(t)?
Answer: A(t) = 100e^-0.5t
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