Midterm - Terms to Know Flashcards
instantaneous velocity
lim(h–>0) of [f(x1 + h) - f(x)]/h
lim(x–>0) of sinx/x
1
lim(x–>inf.) of sinx/x
0
lim(x–>c) of A
A
lim(x–>c) of x
c
lim(x–>c) of [f(x) +/- g(x)]
lim(x–>c) of f(x) +/- lim(x–>c) of g(x)
lim(x–>c) of [kg(x)]
k*lim(x–>c) of g(x)
lim(x–>c) of [f(x)^n]
[lim(x–>c) of f(x)]^n
Sandwich Theorem
if g(x) </= f(x) </= h(x) in some interval about c, and lim(x–>c) of g(x) = lim(x–>c) of h(x), then lim(x–>c) of f(x) = L
a function is continuous if…
the value exists
the limit exists
the limit equals the value
intermediate value theorem
if a function is continuous between points a and b, it will take on every value between f(a) and f(b)
tangent line equation
ytan = m(x-x1) + y1
what is a normal line?
a line perpendicular to the tangent line; has the opposite reciprocal slope
when is a function differentiable?
if it has a derivative at every point in its domain, and is both continuous and smooth
power rule
d/dx(x^n) = nx^(n-1)
constant multiple rule
d/dx(cu) = c*d/dx(u)
sum and difference rule
d/dx(u +/- v) = du/dx +/- dv/dx
product rule
d/dx(uv) = du/dxv + dv/dxu
quotient rule
“low d high minus high d low, square the bottom and away we go”
d/dx(u/v) = (du/dxv - dv/dxu)/v^2
position, velocity, acceleration relationships
velocity is the first derivative of position
acceleration is the first derivative of velocity and second derivative of position
velocity and acceleration
when velocity is + or - but increasing, acceleration is +
when velocity is constant, acceleration is 0
when velocity is decreasing, acceleration is -
d/dx(sinx)
cos(x)
d/dx(cosx)
-sin(x)
d/dx(tanx)
sec^2(x)
d/dx(secx)
secxtanx
d/dx(cscx)
-cscxcotx
d/dx(cotx)
-csc^2x
chain rule
dy/dx = dy/du * du/dx
y’ = f’(g(x)) * g’(x)
inverse functions
derivative of a function’s inverse is the reciprocal of that function’s derivative
d/dx(sin-1x)
1/(rt[1-x^2])
d/dx(cos-1x)
-1/(rt[1-x^2])
d/dx(tan-1x)
1/(1+x^2)
d/dx(cot-1x)
-1/(1+x^2)
d/dx(secx)
1/(|u|*rt[u^2-1])
d/dx(cscx)
-1/(|u|*rt[u^2-1])
d/dx(e^x)
e^x
d/dx(lnx)
1/x
extreme value theorem
if f is continuous over a closed interval, then f also has a maximum and a minimum value over that closed interval, including at its endpoints
–> maximum/minimum are where f’ = 0
first derivative test
extrema exist where f exists and f’ changes signs (+ –> - or - –> +)
neg. to pos.: minimum
pos. to neg.: maximum
mean value theorem
if f(x) is continuous over [a, b] and differentiable over (a, b), then at some point c between a and b, f’(c) = (f(b)-f(a))/(b-a)
increasing/decreasing functions
increasing if: when x1 < x2, f(x1) < f(x2)
decreasing if: when x1 < x2, f(x1) > f(x2)
antipower rule
if f(x) = ax^n, F(x) = (a/([n+1])x^(n+1)
curve sketching
concave up: when f ‘’ (x) > 0
concave down: when f ‘’ (x) < 0
concave up/increasing: f’ > 0, f ‘’ > 0
concave up/decreasing: f’ < 0, f ‘’ > 0
concave down/increasing: f’ > 0, f ‘’ < 0
concave down/decreasing: f’ < 0, f ‘’ < 0
related rates tip
find the rate at which one quantity is changing by relating it to other quantities, whose rates of change are known
1. draw a picture
2. write down what you know and what you’re looking for
3. write an equation relating knowns and unknowns
4. take derivative, usually with respect to time (t)
5. solve (by plugging in what you know) and interpret the answer
rectangle/square formulas
P = 2l + 2w
A = lw
parallelogram formulas
P = 2a + 2b
A = bh (h = height, not side length)
triangle formulas
P = a + b + c
A = (1/2)bh
trapezoid formulas
P = a1 + a2 + b1 + b2
A = (1/2)h(b1 + b2)
circle formulas
P = 2pir
A = pir^2
rectangular prism formulas
V = lwh
SA = 2lw + 2lh + 2wh
cone formulas
V = (1/3)pir^2h
SA = pir^2 + pi*rs (s = length of side)
sphere formulas
V = (4/3)pir^3
SA = 4pir^2
pyramid formulas
V = (1/3)(ab)h
SA = (1/3)(2a + 2b)(s)(ab)
where s = height of triangle face
cylinder formulas
V = pir^2h
SA = 2pir^2 + 2pi*rh
linearization of f at a
L(x) = f(a) + f’(a)(x-a)
Newton’s method
used to find approximate roots of equations using linearizations
1. find function & derivative –> take a guess at a root, then find the linearization of the curve at that point, where L(x) = 0
2. use x-value found from linearization to guess again; x(n+1) = xn - (f[xn]/f’[xn])
3. find where x stops changing
optimization steps
- draw scenario
- write equation for quantity you want to optimize in terms of one variable
- find first derivative of equation and set equal to 0
- check endpoints if necessary
approximation methods for integrals
LRAM: A = h(y0 + y1 + y2 + y3) where y-values are from left corner
RRAM: A = h(y1 + y2 + y3 + y4) where y-values are from right corner
MRAM: A = h(y0.5 + y1.5 + y2.5 + y3.5)
calculating area under the curve - definite integral notation
lim(n–>inf.) of the sum of all f(ck)(delta of x) = intg. from b –> a of (f(x)dx)
integral evaluation theorem
intg. from b–>a of (f(x)*dx) = F(b) - F(a)
rules for integrals
reversing limit changes the sign (+/-)
if upper and lower limits are equal, area = 0
constant multiples can be moved outside
integrals can be added/subtracted
intervals can be added/subtracted
neither integrals nor intervals can be multipled/divided
mean value theorem for intervals
1/(b-a) * intg. from b–>a of (f(x)*dx)
fundamental theorem
d/dx of (intg. from x–>a of f(t)*dt) = f(x)
need parameters:
- derivative of integral
- derivative matches upper limit of integration (d/dx matches x)
- lower limit is a constant (a)
*integral can be rearranged to suit these parameters
trapezoidal rule
T = (h/2)(y0 + 2y1 + 2y2…+2y(n-1) + yn)
Simpson’s Rule
T = (h/3)(y0 + 4y1 + 2y2 + 4y3…+2y(n-2) + 4y(n-1) + yn)
integration by parts formula
intg. of (udv) = uv - intg. (vdu)
u must differentiate to 0, dv must be easy to integrate
*choose u using LIPET
tabular integration works for…
integrals of the form intg. (f(x) * g(x)*dx), where f(x) differentiates to 0 and g(x) integrates forever
slope fields
if x is affecting, change horizontally
if y is affecting, change vertically
