AP Calculus BC Midterm Review

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Last updated 6:26 PM on 12/13/22
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28 Terms

1
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1st derivative of a parametric
**y’=(dy/dt)/(dx/dt) “derivative of y(t) over derivative of x(t)”**
**y’=(dy/dt)/(dx/dt) “derivative of y(t) over derivative of x(t)”**
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2nd derivative of a parametric
 **y’’=[d/dt(dy/dx)]/(dx/dt) “derivative of dy/dx over derivative of x(t)”**
 **y’’=[d/dt(dy/dx)]/(dx/dt) “derivative of dy/dx over derivative of x(t)”**
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Horizontal tangent line (parametrics)
dy/dt=0 “slope is 0, top is 0”
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Vertical tangent line (parametrics)
dx/dt=0 “slope is undefined, bottom is 0”
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Speed (aka “magnitude of the velocity vector”)
“the square root of dx/dt squared + dy/dt squared”
“the square root of dx/dt squared + dy/dt squared”
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Arc length (aka “distance travelled”) parametrics
“the integral of the square root of dx/dt squared + dy/dt squared”
“the integral of the square root of dx/dt squared + dy/dt squared”
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Arc length (aka “distance travelled”) cartesian/rectangular
“the integral of the square root of 1 + f’(x) squared”
“the integral of the square root of 1 +  f’(x) squared”
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Vector-Valued functions (position, velocity, acceleration)
Given as <x(t), y(t)>, keep separate, 1st derivative is velocity 2nd derivative is acceleration
Given as <x(t), y(t)>, keep separate, 1st derivative is velocity 2nd derivative is acceleration
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Taylor Polynomials
knowt flashcard image
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MacLaurin Polynomials
Taylor polynomial centered at x=0 (a=0 in equation)
Taylor polynomial centered at x=0 (a=0 in equation)
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Lagrange Error Bound
M is the max value of the (n+1)st derivative
M is the max value of the (n+1)st derivative
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FTC 1
top minus bottom, remember indefinite integrals have a +C term
top minus bottom, remember indefinite integrals have a +C term
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Integration- if the derivative of the bottom is on the top
General solution: “ln of the absolute value of x + C”
General solution: “ln of the absolute value of x + C”
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Other techniques of integration (there are 8)
is the derivative of the bottom on the top, is the derivative of the inside next to it, can you split it into fractions (single term in denominator), can you expand it, can you use synthetic/long division, can you use the rules of logs to simplify, can you use trig substitution, u-substitution
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Integration by parts
“uv minus the integral of vdu (think voodoo)”, remember LIATE for assigning u values
“uv minus the integral of vdu (think voodoo)”, remember LIATE for assigning u values
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LIATE
Logs

Inverse Trig

Algebraic

Trig functions

Exponential
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Integration of powers of sin and cos
Remember: sin^x + cos^x = 1

\- If the power of one is odd and positive, save one of those terms and convert the rest to the other term

\
Remember: sin^2(x) = \[1-cos(2x)\]/2,

cos^2(x) = \[1+cos(2x)\]/2

\-Use these identities if both terms exponents are even and positive
Remember: sin^x + cos^x = 1

\- If the power of one is odd and positive, save one of those terms and convert the rest to the other term

\
Remember: sin^2(x) = \[1-cos(2x)\]/2,

cos^2(x) = \[1+cos(2x)\]/2

\-Use these identities if both terms exponents are even and positive
18
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Integration of powers of tan and sec
Remember info in pic!

A lot of trial and error
Remember info in pic!

A lot of trial and error
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Method of partial fractions
factor denominator, split fraction into two parts, cross multiply, solve for A and B, then plug back into the fractions and integrate
20
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Exponential Growth and Decay (differential equations)
“the rate of change of Y is proportional to Y”, dy/dt=ky, y=Ce^(kt) where C is the starting value
“the rate of change of Y is proportional to Y”, dy/dt=ky, y=Ce^(kt) where C is the starting value
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“Other” Exponential Growth and Decay (differential equations)
“the rate of change of Y is proportional to BLANK”, dy/dt = (get from word problem) y = use separation of variables
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Newton’s Law of Cooling (differential equations)
“the rate of cooling of an object is proportional to the difference between the object’s temp and room temp”, dT/dt = k(T0 - Ts) where Ts is a constant representing room temp and T0 is initial temp, T = (T0-Ts)e^kt + Ts
“the rate of cooling of an object is proportional to the difference between the object’s temp and room temp”, dT/dt = k(T0 - Ts) where Ts is a constant representing room temp and T0 is initial temp, T = (T0-Ts)e^kt + Ts
23
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Logistics Growth (differential equations)
“rate of change of y is proportional to y and the difference between the upper limit, L, and y”, dy/dt=ky(L-y), y=L/\[1+Ce^(-kLt)\]
“rate of change of y is proportional to y and the difference between the upper limit, L, and y”, dy/dt=ky(L-y), y=L/\[1+Ce^(-kLt)\]
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Taylor series to memorize
sin: starts pos to neg, uses odd powers

cos: starts pos to neg, uses even powers
sin: starts pos to neg, uses odd powers 

cos: starts pos to neg, uses even powers
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d/dx(Arccos(x)) =
\-1/sqrt(1-x^2)
\-1/sqrt(1-x^2)
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d/dx(Arcsin(x)) =
1/sqrt(1-x^2)
1/sqrt(1-x^2)
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d/dx(Arctan(x)) =
1/(1+x^2)
1/(1+x^2)
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sin(2x) =
2sin(x)cos(x)