Infinite Intervals:
Limit
Replace inf with b (on top) or a (on bottom)
Find the antiderivative (u-sub, arctrig, etc.)
Subtract with bounds
If the answer is undefined (inf), DIVERGES
If the answer is a value, CONVERGES TO value
Infinite-Infinite Intervals:
Limit addition
Integral (c/-inf) + integral (inf/c)
Replace -inf with a, inf with b
Antiderivative (u-sub, arctrig, etc.)
Subtract with bounds each of the two limits
If one diverges, it DIVERGES
If all answers are values, CONVERGES to the sum of values
Infinite Discontinuities:
Identify the discontinuity (L or R)
Write the limit
Find the antiderivative
Subtract with bounds
Converges/Diverges
Infinite Discontinuities Internally
Identify the value that makes the integrand discontinuous
Limit to discontinuous value for both sides of c
Find antiderivative of each
If one diverges, it DIVERGES
If all answers are values, CONVERGES to the sum of values
Direct variation
Linear functions: y = kx
y varies directly as x
y is directly proportional to x
y is “proportional” to x
Inverse variation
Function: y = k/x
y varies inversely as x
y is inversely proportional to x
Double check:
At a rate?
dX/dt (assumed it’s over time)
K = constant of proportionality
Differences, sums, products, quotients
Verifying solutions
Find the derivatives (1st, 2nd, 3rd…)
Plug in derivatives or y’s to differential equation
Check if it is a solution (= #) by setting them equal to each other
Double check:
When subtracting a group, multiply by negative 1 to each value
Solve for k
Differential equations
Write an equation of the tangent line to y = f(x)
y - y1 = m(x - x1)
Use a tangent line to approximate a value of f(x)
f(d) ≈ m(d - x1) + y1
Find d2y/dx2
fig - fgi/g2 (quotient rule)
fig + fgi (product rule)
Plug in for x and y terms
Find or check for critical values and relative extrema
dy/dx = 0, derivative tests
Derivatives
Find out a squared value (always positive)
Find out how dy/dx can become negative (or positive)
State how dy/dx can be less (or greater) than 0
Sketching solution curves
Passing through a coordinate (x,y)
Stop at x-axis, BOTH sides
Range of solution curve y = f(x)
[y1, y2], ex: [4,6] (in terms of y)
Matching slope fields
If slopes are the same vertically, only x
If slopes are the same horizontally, only y
If slopes have a combination, both x and y
Double check:
Plug in multiple coordinate points (x,y) to each equation
Process of elimination
Euler’s method
Separate variables (dy/dx)
Start at given point
y(0) = 1
Plug in x and y into dy and multiply by dx (change in x, ∆x)
dy = (x-y)dx = (0-1)(0.5) = -0.5
Add dy to original y for new y
1 + (-0.5) = 0.5
Insert new y into new x (change in x, dx, ∆x)
0 → 0.5 (∆x = 0.5), y = 0.5
y(1.5) ≈ 0.75; under approximation because d2y/dx2 > 0 (plug in new y and new x)
d2y/dx2 = 1 - dy/dx = 1-(x-y) = 1-x+y = 1-1.5 + 0.75 > 0
Double check:
If f’’ > 0 → concave up, approxication is under (U)
If f’’ < 0 → concave down, approximation is over (n)
Approximation f(x) ≈
Separating variables
Multiply dy/dx with dx
Divide y-variable
y-values in left side, x-values in right side
Integrate (antiderivative)
Find y (square rooting, arctrig, e, etc.)
Find particular solution
Multiply both sides of the equation by dx
Separate the variables (multiplying, dividing)
y on left, x on right
Integrate both sides (u-sub, etc.)
Use initial condition to find C
Plug in (x,y) to find C
Solve for y + C
Determine positive or negative when square rooting
f(1) = -4
Exponential Growth and Decay
y = Cekt
dy/dt = ky
Growth: k > 0
Decay: k < 0
Half-lives of radioactive isotopes
Ex: Carbon (14C) = 5730 years
Decay: Applications of half-life
Question: Suppose that 15 grams of caesium isotope Cs-137 that was released during the 1986 Chernobyl nuclear accident and absorbed into the groundwater. How long will it take for the 15 grams to decay to 1 gram and become safer to drink?
Plug in given (x,y)
y = Cekt
(0, 15) → 15 = Cek(0)
Find C
15 = C
Plug in half-life (x,y)
Caesium (137Cs) = 30 years
(30, 7.5) → 7.5 = 15ek(30)
Find k (natural log)
7.5 / 15 = 15ek(30) /15
ln(1/2) = ln(e30k) → ln(1/2) /30 = 30k /30
k = ln(1/2)/30
Plug in given to find t (log rules)
y = 15eln(1/2)/30t → 1 = 15eln(1/2)*t/30
1 /15 = 15(1/2)t/30 /15
ln(1/15) = ln(1/2)t/30 → ln(1/15) /ln(1/2) = t/30*ln(1/2) /ln(1/2)
ln(1/15)/ln(1/2) /30 = 30t /30
t = 30*ln(1/15)/ln(1/2)* (don’t have to simplify if not given a calculator)
Growth: Application of population
Question: An absent-minded graduate student at a university is studying a population of fruit flies in a biology life. She works under the premise that this experimental population of fruit flies increases according to the law of exponential growth. She counts 100 flies after the second day of the experiment and 300 flies after the fourth day, however, she forgot to record the number of fruit-flies she initially had at the beginning of the experiment. Approximately how many flies were in the original population? Round to the nearest fly.
Plug in both given (x,y)
y = Cekt
100 = Cek(2) /e2k, 300 = Cek(4) /e4k
Find C by equalling both equations
C = 100/e2k, C = 300/e4k
100/e2k = 300/e4k → e4k/e2k = 300/100
e2k *ln = 3 *ln
Find k (natural log)
ln(e2k) = ln(3) → 2k /2 = ln(3) /2
Plug in found to find y (0,__)
y = (100/e2(ln(3)/2)eln(3)/2(0) → 100/3e0 → 100/3 flies ≈ 33 flies
Growth: More applications
Question: Fish are being introduced into a man-made lake. The rate of fish, F, with respect to time, t, is directly proportional to 900-F, where t is measured in years. When t=0, there are 400 fish in the lake and 3 years later, there are 600 fish in the lake.
Solving the differential equation, write the equation
dF/dt *dt /900-F = k(900-F) *dt /900-F→ dF/900-F = kt
Integrate
-ln|900-F| = kt +C, u=900-F du=-dF
Solve for C by plugging in a point (x,y)
(0,400) → -ln|900-400| = k(0) +C → -ln|500| = C
Solve for k by plugging in C and plugging in another point (x,y)
(3, 600) → -ln|900-600| +ln|500| = k(3) - ln|500| +ln|500|
ln|500/300| /3 = k(3) /3 → ln|5/3|/3 = k
Isolate F by plugging in known values
-ln|900-F| *-1 *e = 1/3ln(5/3)t - ln|500| *-1 *e
F = 900-e-1/3ln(5/3)t + ln(500)* (do not have to simplify)
Find the fish population in another 3 years
t = 6, F(6) = 900-500(3/5)6/3 = 720 fish
Find lim (t→inf)F(t) and explain what the answer means
lim(t→inf) (900-500(3/5)t/3) = 900-0=900
The maximum number of fish that could inhabit the lake would be 900 fish.
Logistic differential equation form
dy/dt = ky(1-y/L)
Logistic growth equation form
y = L/1+be-kt
Double Check:
L = carrying capacity
k = growth constant
k is negative in growth form