AP Calc AB (Serrano) Chapter 7 Notes

A function y(t) satisfies the differential equation dy/dt = y⁴ - 6y³ + 5y² a) What are the constant solutions of the equation? b) For what values is y increasing? c) For what values is y decreasing?

***remember that it is ALREADY a derivative! - factor; 0, 5, 1 - where y is increasing (form intervals using solutions) - where y is decreasing

For what values of r does y = e^rx satisfy y" + 5y' - 6y = 0?

1) find derivatives DE: y" + 5y' - 6y = 0 y = e^rx y' = r*e^rx y" = r²*e^rx 2) plug into the equation r²*e^rx + 5(r*e^rx) - 6(e^rx) = 0 take out GCF; e^rx(r² + 5r - 6) = 0; e^rx(x+6)(x-1) x = -6, 1 *cannot take anything out of e^rx because it will never = 0

Is y = x - x⁻¹ a solution of the differential equation xy' + y = 2x?

1) find the derivative y' = 1 + x⁻² 2) plug in the differential equation x(1 + x⁻²) + (x - x⁻¹) = 2x x + x⁻¹ + x - x⁻¹ = 2x 3) check to see if the left = the right; if it does it is a solution 2x = 2x so yes it's a solution

Find the solution of the differential equation that satisfies the given initial condition: du/dt = 2t + sec²t/2u u(0) = -5

2 du = 2t sec² t dt ∫2 du = ∫2t sec² t dt u² = t² + tan t + C u = ±√t² + tan t + 25 bc -5, u = -√t² + tan t + 25

What should you know about Newton's Law of Cooling?

dT/dt = k(T-T(E)) T = temp of object T(E) = environmental temp k = constant T = T(E) + (T₀ - T(E))e^kt

Should you assume that differential equations are separable at first glance?

no, you may have to rewrite the function first them separate

Should you multiply constants when rearranging the equation?

no; they are constants

What should you always do when drawing a slope field?

pay attention to whether the slope is positive or negative

What are differential equations?

rates of change; also derivatives with equal signs that equal something ex. y'x + x²y = y'

What should you know?

the difference between average value and average rate

For what values of k does the function y = cos kt satisfy the differential equation 4y" = -25y?

y' = -sin kt *l y" = - cos kt * k² -4(cos kt)k² + 25 (cos kt) = 0 cos kt(25-4k²) k = +- 5/2 **cos kt can not be used to find solutions because of the k

Solve dy/dx = 1/(3y²-1)

(3y²-1)dy = dx ∫(3y² - 1)dy = ∫1 dx y³ - y + c1 = d + c2 subtract both by c1 = one c value y³ - y = x + c

What two lines do the family of solutions approach?

P = K and P = 0 (gets closer to the initial amount)

What happens if the population is ever zero or at carrying capacity?

P stays that way

What will the initial population of P(t) = ce^kt always be?

P(0) = C

What is true for P(t) if we rule out a population of 0?

P(t) > 0, or is positive, for all t because P can't be negative

If the temperature of a hot cup of coffee is above 140°F, it burns the tongue. If it is below 105°F, it turns into undrinkable sludge. Coffee cools at a rate proportional to the difference between the temperature of the coffee and the temperature of the environment. 1) What differential equation describes this situation? Why is it separable? 2) Use the technique of separation of variables to find the general solution 3) Your general solution should have three parameters (constants). One is the outside temperature, another depends on the initial temperature of the coffee. The third, k, depends on the container holding the coffee. For a typical styrofoam cup, k = -0.05, if t is measured in minutes. Why is this constant negative?

T = temp of the coffee, E = temp of the environment (constant) 1) dT/dt = K(T-E) 2) create a T side and a t side dT/(T-E) = k dt; ∫1/(T-E) dT = ∫k dt ln |T-E| = kt + C e^(ln |T-E|) = e^(kt + C) T - E = e^kt + C or e^kt *e^c T = E + Ae^kt 3) the temperature is decreasing (exponential decay)

A company's revenue is earned at a continuous annual rate of 5% of its net worth. At the same time, the company's payroll obligations amount to $200 million a year, paid out continuously. If the current net worth is $3 billion, in how many years will the company go bankrupt?

W(t) = net worth in millions at t years; dW/dt = revenue rate - payroll (coming in - leaving) dW/dt = 0.05W - 200 dW/dt= 0.05(W-4000) ∫1/W-4000 = ∫0.05 dt ln |W-4000| = 0.05 dt; e^ln|W-4000| = e^0.05 dt W - 40000 = e^0.05 *e^C W = 4000 + Ae^0.05t

What is the proper name for the antiderivative of a differential equation?

a family of solutions

separable equation

a first order DE in which the expression for dy/dx can be factored as a function of x times a function y

What are slope fields (also direction fields or vector fields)?

a tool to graphically obtain the solutions to a first order differential equation. In a slope field, a bunch of little line segments show the slope of the curve at different points

a) Show that every member of the family of functions y - ln x + C/x is a solution of the differential equation x²y' + xy = 1 c) Find a solution of the differential equation that satisfies the initial condition y(1) = 2 d) Find a solution of the differential equation that satisfies the initial condition y(2) = 1

a) **use quotient rule y' = x(1/x) - (ln x + C)/x² y' = 1-ln x - C/x² x²(1-ln x - C/x²) + x(ln x + C/x) 1 - ln x - C + ln x + C 1 = 1 c) ln 1 + C = 2; c = 2 d) ln 2 + C = 2 C = 2 - ln 2

What differential equation models population growth (differential because it contains an unknown function P and its derivative)?

dP/dt = kP where P = the number of individuals in a population, t = time, and k = the proportionality constant proportionality = when one side increases so does the other

A hard boiled egg at 98°C is placed in a sink of 18°C water. After 5 min, the egg's temp is 38°C. Assuming the water is not warmed significantly, how much longer will it take the egg to reach 20°C?

dT/dt = K(T-T(E)) T = (T(0)-T(E))e^kt + T(E) A = 98-18 = 80 38 = 80e^k(5) + 18; k = ln (.25)/5 20 = 80^e^(ln .25/5)t + 18 1/40 = e^(ln .25/5)t ln 1/40 = ln .25/5 t t = 13.305 min

What are rate in/rate out problems?

differential equations: dy/dt = rate in - rate out

What do slope fields do?

draw the slopes at various coordinates for differing values of C

What equation exists if y(t) is the value of a quantity y at time t and if the rate of change of y wrt t is proportional to the size y(t) at any time?

dy/dk = ky

In Exercise 28 in Section 7.2 we discussed a differential equation that models the temperature of a 95°C cup of coffee in a 20°C room. Solve the differential equation to find an expression for the temperature of the coffee at time t.

dy/dt = -1/50(y - 20) ∫1/y-20 dy = ∫-1/50 dt ln |y-20| = -1/50 t + C e^(ln |y-20| = e^-1/50t*e^C y = Ae^11/50t + 20 95 = Ae⁰ + 20; A = 75 y = 75e^-t/50 + 20

What is the equation for Newton's Law of Cooling?

dy/dt = k(Y-R) R = room temp Y = temp of coffee y(0) = upper temperature; approaches model because Y → R and dy/dt → 0

At time t = 0 a tank contains 4 lb of salt dissolved in 100 gal of water. Brine containing 2 lbs of salt per gallon enters the tank at a rate of 5 gal/min and the mixed solution is drained from the tank at the same rate. Find the amount of salt in the tank after 10 minutes. Let y(t) = amount of salt after t minutes in lbs.

dy/dt = rate in - rate out rate in (salt enters): 2 lbs/1 gal * 5 gal/1 min = 10 lbs/min rate out (salt leaves): y lbs/100 gal * 5 gal/1 min = y lbs/20 min dy/dt = 10 lbs/min - y lbs/20 min Rewrite DE as a product via GCF: dy/dt = 1/20 (200-y) **divide using chain ∫1/200-y dy = ∫1/20 dt - ln |200 - y| = 20 t + C e^ ln |200 - y| = e^20t * e^C 200 - y = e^(1/20 t)A y = -200 - Ae^(-t/20) Find A; 200 - 4 = A = 196 y = -200 - 196e^-10/20 y = 81.12 lbs

A function y(x) satisfies the DE (dy/dx) = y² - ey - 6. What are the constant solutions (aka equilibrium solutions) to the equation?

dy/dt, or the derivative, must equal 0 0 = y² - 4y - 6 0 = (y-6)(y+1) y = 6, -1

Solve the differential equation dy/dx = y/x

dy/y = dx/x ∫1/y dy = ∫1/x dx ln |y| = ln |x| + C e^(ln |y|) = e^(ln|x|+C), or e^|ln x| * e^c e^c = a constant = A y = Ax

Solve dy/dx = -4xy², y(0) = 1

dy/y² = -4x dx ∫ 1/y² dy = ∫ -4x dx -1/y = -2x² + c OR y = -1/(-2x² - 1)

Solve the equation e^-y y' + cos x = 0 and graph several members of the family of solutions. How does the solution curve change as the constant C varies?

e^-y dy/dx = - cos x ∫e^-y dy = ∫ - cos x dx e^-y = sin x + c ln (e^-y) = ln (sin x + c) - y = ln (sin x + c) y = - ln (sin x + c)

If the maximum value of f(x) is -3 and if f'(x) = 4-x, then find an equation for f(x).

f'(x) = 0 when x = 4 so the max is at (4,-3). f'(x) = 4 - x; antiderive so f(x) = 4x - x²/2 + C -3 = 4(4) - (4)²/2 + C C = -11 f(x) = 4x - x²/2 - 11

What could be a solution to the differential equation dP/dt = kP, or what could P(t) be?

find a function whose derivative is a constant multiple of itself (exponential)

A tank contains 1000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much salt is in the tank a) after t minutes and b) after 20 minutes?

let y(t) be the amount of salt in kg after t minutes y(0) = 15; salt leaves only dy/dt = -y kg/1000 L (10 L/min) = -y/100 1/y dy = -1/100 dt ln y = 1/100 dt + C if y(0) = 15 then C = ln 15 ln y = -t/100 + ln 15 OR y = e^(-1/100 t)*e^(ln 15) a) y = 15e^(-1/100 t) b) after 20 min y = 15e^(-1/5) = 12.3 kg

What should you remember as an important formula when calculating half life?

ln 2 = -kt or ln (1/2) = kt

What rule should you remember when separating ln x?

ln xy = ln x + ln y ln x/y = ln x - ln y

Should you draw a graph without knowing the initial value?

no

What part of P(t) = ce^kt is important?

only the t > 0 part since time is positive

What is a reasonable assumption about the growth of a population under ideal conditions?

that a model for the growth of a population would be proportional to the size of the population

Explain why the functions with the given graphs can't be solutions of the differential equation dy/dt = e^t(y-1)² - graph 1 has a horizontal asymptote at (1,1) and goes down after an upward trend - graph 2 goes steadily upwards but has a vertical asymptote at (1,1)

the graph should only be increasing since it should be positive for all t. dy/dt = 0 when t = 1, so it should have a horizontal tangent there - graph 1 should never be decreasing - there is no horizontal asymptote; it is vertical

Consider dy/dt = t²(y-y³)⁴ Why can't the following graphs be solutions of the differential equations: - a graph in a W shape with a y-int at y = 1 and two x-ints at x = 2,-2 - a graph going at a slant that crosses the y-axis at y = 4 - a graph with an x-int at x = -1, a y-int at y = 1, and a vertical rest between the two

general: dy/dt = 0 when t = 0 and when y = 1, 0, and -1 dy/dt is never negative because of exponents seems to have dy/dt = DNE at t = -1 for the last graph assuming you plug in (-1,0) as a coordinate pair - increasing/decreasing intervals = suggests negative parts of the graph - there should be a horizontal tangent at t = 0 - there should be additional horizontal tangents at t = 1 and -1, not just t = 0

When is there growth and when is there deacy for dy/dk = ky?

growth: k > 0 decay: k < 0

Carbon-14 has a half life of 5730 years. If there is a sample of 50 grams, how much will be left in 1000 years?

half life of 5730 years so (5730 years, 25 g) and (0 yrs, 50 g) m(t) = 50e^kt 25 = 50e^kt = ln (1/2) = kt OR ln 2 = -kt ln 2 = -k(5730) so k = -ln 2/5730 m(t) =. 50e^(-ln 2/5730)t or m(t) = 5(2)^-t/5730 plug in 1000 years; m(t) = 44.3 g

How can certain types of differential equations be solved?

explicitly

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L/min. How much salt is in the tank a) after t minutes and b) after one hour?

y(0) = 0. Let y(t) be the amount of salt in kg after t minutes. dy/dt = (0.05 kg/L)(5L/min) + (0.04 kg/L)(10 L/min) - (y kg/1000 L)(15 L/min) 0.25 + 0.4 -0.15 y = 0.65 - 0.15 y dy/dt = 130-3y/200 kg/min ∫1/130-3y dy = ∫1/200 dt -1/3 ln |130-3y| = 1/200 t + C -1/3 ln |130-0| = 1/200(0) + C C = -1/3 ln 130 e^(ln |130-3y| = e^(1/200 t) * e^(-1/3 ln 130) [130 - 3y = 130e^(-3/200 t)]/-3 y = 130/3(1 - e^(-3/200 t)) b) after 1 hr (plug in 60) y = 25.7 kg

What happens with the equation y = Ae^kt if the initial value is known?

y(0) = Ae^k(0) = A So, the solution of dy/dt = ky, y(0) = y₀ is: y(t) = y₀e^kt

How can you find a solution to a DE?

you can only find a solution of a differential equation when given an initial value

Solve dy/dx = √x/e^y

∫e^y dy = ∫x¹/² dx ln e^y = ln (3/2 x³/² + C) dx y = ln (2/3 x³/² + C) *no absolute value bars because an exponential function is always positive

Find the solution of the initial-value problem dy/dt = 2t/(y² + t²y²), y(0) = 3

∫y² dy = ∫2t/(1 + t²) dt 1/3 y³= ln |1 + t²| dt + C (3)³/3 = ln |1 + 0| + C C = 9 1/3 y³= ln |1 + t²| dt + 9

Psychologists interested in learning theory study learning curves. A learning curve is the graph of a function P(t), the performance of someone learning a skill as a function of the training time t. The derivative dP/dt represents the rate at which performance improves. a) When do you think P increases most rapidly? What happens to dP/dt as t increases? b) If M is the maximum level of performance of which the learner is capable, explain why the differential equation dP/dt = k(M-P) where k is a positive constant is a reasonable model for learning c) make a rough sketch of a possible solution of this differential equation

**Assume learning for a language such as Russian a) Initially P increases rapidly because the student knows very little at first; there is more room to grow at the beginning b) Assume M = carrying capacity. When P is much less than M then M-P is large; when P is close to M, then M-P is really small and approaches 0. *max- where you're at) c) The graph is not a typical exponential curve. Picture a function that intersects the y-axis just above 0, rises rapidly in an almost linear line, then curves towards P = K and levels out so it is essentially an asymptote. The graph does not exist outside of QI.

Verify that y = sin x cos x - cos x is the solution of the initial-value problem y' + tan(x)y = cos² x, y(0) = -1 on the interval -π/2 < x < π/2

**Use multiplication rule y' = sin(-sin x) + cos x cos x - (-sin x) = cos²x - sin²x + sin x plug in: cos²x - sin²x + sin x + sin²x - sin x cos x **tan = sin x/cos x

Make your own slope field for dy/dx = 2x + y

**plug in x AND y coordinates when x= -2, from y = -2 to y = 2: (-6, -5, -4, -3, -2) when x = -1, from y = -2 to y = 2: (-4, -3, -2, -1, 0) when x = 0, from y = -2 to y = 2: (-2, -1, 0, 1, 2) when x = 1, from y = -2 to y = 2: (0, 1, 2, 3, 4) when x = 2, from y = -2 to y = 2: (2, 3, 4, 5, 6) **Graph each slope for each coordinate on the coordinate

Make your own slope field for dy/dx = 1-x

**plug in x AND y coordinates when x= -2, from y = -2 to y = 2: all 3 when x = -1, from y = -2 to y = 2: all 2 when x = 0, from y = -2 to y = 2: all 1 when x = 1, from y = -2 to y = 2: all 0 when x = 2, from y = -2 to y = 2: all -1 **Graph each slope for each coordinate on the coordinate

Make your own slope field for dy/dx = x/y

**plug in x AND y coordinates when x= -2, from y = 2 to y = -2: (-1, -2, undefined, 2, 1) when x = -1, from y = 2 to y = -2: (-1/2, -1, undefined, 2, 1/2) when x = 0, from y = 2 to y = -2: all 0 when x = 1, from y = 2 to y = -2: (2/3, 2, undefined, -1, -1/2) when x = 2, from y = 2 to y = -2: (1, 2, undefinted, -2, -1) **Graph each slope for each coordinate on the coordinate

Make your own slope field for dy/dx = 1-y

**plug in x AND y coordinates when y= -2, from x = -2 to x = 2: all 3 when y = -1, from x = -2 to x = 2: all 2 when y = 0, from x = -2 to x = 2: all 1 when y = 1, from x = -2 to x = 2: all 0 when y = 2, from x = -2 to x = 2: all -1 **Graph each slope for each coordinate on the coordinate

Let f be a function with f(1) = 3 such that for all points (x,y) on the graph of f, the slope of the tangent is (8x³ + 2x)/y². Find f(x).

*dy/dx is another name for a tan line dy/dx = (8x³ + 2x)/y² ∫y² dy = ∫(8x³ + 2x)dx y³/3 = 2x⁴ + x² + c plug in values; 9 = 2 + 1 + c so c = 6 **do not multiply c; it is a constant y³ = 3(2x⁴ + x²) + c y = ³√(6x⁴ + 3x² + 6)

What should you do for the following types of u-substitution? - definite - indefinite

- calculate new bounds by plugging the current bounds into u (ex. u(x)) and then plugging the new bounds into the unsimplified equation (Ex. u^5/5) - plug the u equation back into the u variable and adding a + C (ex. u^5/5 becomes cos^5 x/5 + C)

Separate the following separable equations: - dy/dx = x/√y - dy/dx = ln x +x/ln y + y - dy/dx = ye^(sin x + cos x)

- cross multiply; x dx = √y dy - cross multiply; ln x + x dx = ln y + y dy - dy/dx = ye^sin x * e^ cos x y/ye^cos y = e^sin x dx

Separate the following DE if possible: - dy/dx = x²/y - dy/dx = 2xy + y

- cross multiply; x²dx = y dy - dy/dx = y(2x + 1) dy/y = (2x+1)dx

What real-world situations can we use rates of change/derivatives to model?

- drug amount and absorption rate - populations and growth rate

What are examples of inseparable equations?

- dy/dx = x + y - dy/dx = ln (xy) - dy/dx = sin (x^y)

Separate the following separable equations: - dy/dx = ln (x^y) - dy/dx = y sin x + xy - dy/dx = (xy+y)/(2x-3xy) - dy/dx = xy - 2x + y - 2

- dy/dx = y ln x dy/y = ln x dx - dy/dx = y(sin x + x) dy/y = (sin x + x)dx - dy/dx = y(x+1)/x(2-3y) dy(2-3y)/y = (x+1)dx/x - group together; dy/dx = x(y-2) + 1(y-2) dy/dx = (x+1)(y-2) dy/(y-2) = (x+1)/dx

Separate the following separable equations: - dy/dx = g(x)fy - dy/dx = g(x)/f(y)

- dy/f(y) = g(x)dx - cross multiply; f(y)dy = g(x)dx

What are the five steps for constructing a slope field?

1. select a coordinate (x,y) and plug it into the differential equation 2. evaluate at that point (x,y) and you will get the slope at (x,y) 3. draw a short line segment, called isoclines, at point (x,y) on the coordinate plane with that slope 4. do this until you have enough isoclines to draw a general solution curve through all the isoclines 5. if given an initial point, then draw the graph with the initial point on the graph

What are the steps for solving a DE once separated?

1. separate 2. integrate BOTH SIDES 3. leave it implicit (no y by itself)

If the temperature of a hot cup of coffee is above 140°F, it burns the tongue. If it is below 105°F, it turns into undrinkable sludge. Coffee cools at a rate proportional to the difference between the temperature of the coffee and the temperature of the environment. 4) A thermos has a constant of k = -0.03. Which does a better of keeping the coffee warm, the styrofoam cup or the thermos? How does knowing the value of k allow you to figure out the answer? 5) The initial temperature of the coffee at 8:30 AM (t = 0) is 16 degrees Fahrenheit. The outside temperature is 42°F. What is the solution for the initial value problem? 6) How long must she wait before she is able to drink the coffee? 7) At what time will the coffee fall below 105°F and become undrinkable? 8) How much time does she have to drink her coffee?

4) The thermos does a better job of keeping the coffee warm because the temperature difference decreases more slowly (k is less negative) 5) T = E + Ae^kt; initial condition k(0) = 160 160 = 42 + Ae^(-0.03*0) A = 118 6) 140 = 42 + 118e^-0.03t t = 6.19 minutes (multiply both sides by ln) 7) 105 = 42 + 118^-0.03t; 20.92 minutes 8) 20.92 min - 6.19 min = 14.73 min

If you use the Trapezoidal Rule with n = 6, how many terms should you multiply by b-a/n/2?

7 terms, from f(0) to f(final)

Find a solution of dy/dt = sin (t) that satisfies y(π/2) = 2

DE = dy/dt = sin (t) general solution: y = -cos (t) + C 2 = - cos (π/2) + C 2 = C solution: y = - cos (t) + 2

What is the general solution to a DE? State the general solution if y' = x³

If y' = x³ then the general solution is given by y = x⁴/4 + C **use the format "if given [__ DE], then the general solution is given by ___"

The temperature, in degrees Celsius (°C), of the water in a pond is a differentiable function W of time t. The table above shows the water temperature as recorded every 3 days over a 15-day period. (t days, W(t)): (0,20), (3,31), (6, 28), (9,24), (12,22), (15, 21) a) Use the data from the table to find an approximation for W'(12). Show the computations that lead to your answer. Indicate units of measure. b) Approximate the average temperature, in degrees Celsius, of the water over the time interval 0 ≤ t ≤ 15 days by using a trapezoidal approximation with subintervals of length ∆t = 3 days.

a) W'(12) = W(15)-W(9)/15-9 = (21-24)/6 = -0.5 °C/day b) 1/15 ∫(15)(0) W(t)dt = (1/15)[(3*1/2)(20 + 2(31) + 2(28) + 2(24) + 2(22) + 21)]² 1/10 (251) = 25.1°C

Let R be the region in the first quadrant enclosed by the graphs of y = e^-x², y = 1 - cos x, and the y-axis. a) Find the area of the region R b) Find the volume of the solid generated when the region R is revolved about the x-axis c) The region R is the base of a solid. For this solid, each cross-section perpendicular to the x-axis is a square. Find the volume of this solid.

a) area (A) = ∫(A)(0) e^-x² - (1 - cos x) dx = 0.591 b) V = π ∫(A)(0) (e^-x²)² - ((1 - cos x))² dx = 1.747 wrt x and washer c) V = ∫(A)(0) (e^-x²- (1 - cos x))² dx = 0.461

Consider the differential equation xy' - 3y = 0. a) determine if y = Cx³ is a solution b) Find the particular solution that satisfies the initial condition y(-3) = 2

a) y' = 3Cx² x(3Cx²) - 3(Cx²y) 3Cx³ - 3Cx³ = 0 yes b) C(-3) 2-27C C = -27/2; y = -2/27x³

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420 a) Find an expression for the number of bacteria after t hours b) Find the number of bacteria after 3 hours c) Find the rate of growth after 3 hours d) When will the population reach 10,000?

a) y(t) = Ae^kt y(t) = 100 e^kt, 420 = 100 e^k(1) so k = ln 4.2 y(t) = 100e^(ln 4.2) or y(t) = 100(4.2)^t b) y(0) = 100, y(1) = 420 y(t) = 100(4.2)³ = 7408.8 bacteria c) rate = derivative; dy/dt = ky so (ln 4.2)(100(4.2)³) = 110,632.254 bacteria d) 10,000 = 100(4.2)^t 100 = 4.2^t ln 100 = t ln 4.2 ln 100/ln 4.2 = t = 3.209 hours

differential equation

an equation that contains an unknown function and some of its derivatives

What happens to dP/dt as P(t) increases?

as P(t) increases, dP/dt becomes larger; in other words, the growth rate increases as the population increases

Prove that P'(t) = kP

assume P(t) = an exponential function with base e SO P(t) = ce^kt P'(t) = ce^kt * k, where ce^kt is the differential equation or P(t) therefore, P'(t) = kP

The temperature, in degrees Celsius (°C), of the water in a pond is a differentiable function W of time t. The table above shows the water temperature as recorded every 3 days over a 15-day period. (t days, W(t)): (0,20), (3,31), (6, 28), (9,24), (12,22), (15, 21) c) A student proposes the function P, given by P(t) = 20 + 10te^(-t/3), as a model for the temperature of the water in the pond at time t, where t is measured in days and P(t) is measured in degrees Celsius. Find P'(12). Using appropriate units, explain the meaning of your answer in terms of water temperature. d) Use the function P defined in part c to find the average value, in degrees Celsius, of P(t) over the time interval 0 ≤ t ≤ 15 days

c) P'(12) = -0.549 °C/day (calculator with integral on P(t)?) On day 12 the temperature of water is decreasing at a rate of -0.549°C per day. d) 1/15 ∫20 + 10te^(-t/3) = 25.757°C

What should you do if y' is given instead of dy/dx?

change it to dy/dx in order to separate it

A population grows according to dP/dt = kP where k is a constant and t is measured in years. If the population doubles every 10 years then what is k?

dP/dt = kP P = Poe^kt 2P(0) = P(0)e^kt P(0) cancels; 2 = e^kt ln 2 = 10 k k = ln 2/10 **basically A = Pe^rt

What do you notice about the slope fields whose differential equation had only y?

the isoclines have the same slopes horizontally (same y coordinate = same slope)

What do you notice about the slope fields whose differential equation had only x?

the isoclines have the same slopes vertically (same x coordinate = same slope

Describe the slope field for dy/dx = x

there are several different parabolas (that resemble the antiderivative, x^2/2) that can be sketched in the slope field with varying values of C; slope must be between isoclines ex dy/dx = 1 when x = 1, dy/dx = 2 when x = 2

What is the significance of differential equations?

they are among the most powerful tools we have for analyzing the world mathematically and are used to formulate the fundamental laws of nature and to model the most diverse physical phenomena

What happens to many populations?

they start out growing exponentially, but will level off when they approach the carrying capacity K

Evaluate the integral ∫ 2/(3x-5) dx

u = 3x-5 so du = 3 dx ∫ 2u⁻¹ dx becomes 2∫ u⁻¹ dx * 1/3 du 2/3 ∫ u⁻¹ dx becomes 2/3 ln |3x + 5| + C

What are real life examples of exponential growth and decay?

usually bacterial colonies and biology, etc.

What can you do sometimes when determining the solution isn't that easy?

verify if a function is a solution to a differential equation

What can we conclude about the population if k > 0?

we can conclude that the population is always increasing

Where are equilibrium solutions?

where the rate of change = 0

What should you draw when the x = 0? When y = 0/the slope is undefined?

x = 0: horizontal line y = 0: vertical line

Solve for y with the equation dy/dk = ky **very important, memorize this equation. it's the solution every time

y = Ae^kt