AP Calculus BC Exam Review Questions and Answers
Average Rate of Change - Answer-Slope of secant line between two points, use to
estimate instantanous rate of change at a point.
/.Instantenous Rate of Change - Answer-Slope of tangent line at a point, value of
derivative at a point
/.Definition of Derivative - Answer-limit as h approaches 0 of [f(a+h)-f(a)]/h or limit as x
approaches a of [f(x)-f(a)]/(x-a)
/.When f '(x) is positive, f(x) is - Answer-increasing
/.When f '(x) is negative, f(x) is - Answer-decreasing
/.When f '(x) changes from negative to positive, f(x) has a - Answer-relative minimum
/.When f '(x) changes fro positive to negative, f(x) has a - Answer-relative maximum
/.When f '(x) is increasing, f(x) is - Answer-concave up
/.When f '(x) is decreasing, f(x) is - Answer-concave down
/.When f '(x) changes from increasing to decreasing or decreasing to increasing, f(x) has
a - Answer-point of inflection
/.When is a function not differentiable - Answer-corner, cusp, vertical tangent,
discontinuity
/.Product Rule - Answer-uv' + vu'
/.Quotient Rule - Answer-(uv'-vu')/v²
/.Chain Rule - Answer-f '(g(x)) g'(x)
/.Particle is moving to the right/up - Answer-velocity is positive
/.Particle is moving to the left/down - Answer-velocity is negative
/.absolute value of velocity - Answer-speed
/.y = sin(x), y' = - Answer-y' = cos(x)
,/.y = cos(x), y' = - Answer-y' = -sin(x)
/.y = tan(x), y' = - Answer-y' = sec²(x)
/.y = csc(x), y' = - Answer-y' = -csc(x)cot(x)
/.y = sec(x), y' = - Answer-y' = sec(x)tan(x)
/.y = cot(x), y' = - Answer-y' = -csc²(x)
/.y = sin⁻¹(x), y' = - Answer-y' = 1/√(1 - x²)
/.y = cos⁻¹(x), y' = - Answer-y' = -1/√(1 - x²)
/.y = tan⁻¹(x), y' = - Answer-y' = 1/(1 + x²)
/.y = cot⁻¹(x), y' = - Answer-y' = -1/(1 + x²)
/.y = e^x, y' = - Answer-y' = e^x
/.y = a^x, y' = - Answer-y' = a^x ln(a)
/.y = ln(x), y' = - Answer-y' = 1/x
/.y = log (base a) x, y' = - Answer-y' = 1/(x lna)
/.To find absolute maximum on closed interval [a, b], you must consider... - Answer-
critical points and endpoints
/.Linearization - Answer-use tangent line to approximate values of the function
/.left riemann sum - Answer-use rectangles with left-endpoints to evaluate integral
(estimate area)
/.right riemann sum - Answer-use rectangles with right-endpoints to evaluate integrals
(estimate area)
/.Trapezoidal rRle - Answer-use trapezoids to evaluate integrals (estimate area)
/.[(h1 - h2)/2]*base - Answer-area of trapezoid
/.definite integral - Answer-has limits a & b, find antiderivative, F(b) - F(a)
/.indefinite integral - Answer-no limits, find antiderivative + C, use inital value to find C
, /.area under a curve - Answer-∫ f(x) dx integrate over interval a to b
/.area above x-axis is - Answer-positive
/.area below x-axis is - Answer-negative
/.average value of f(x) - Answer-= 1/(b-a) ∫ f(x) dx on interval a to b
/.To find particular solution to differential equation, dy/dx = x/y - Answer-separate
variables, integrate + C, use initial condition to find C, solve for y
/.To draw a slope field, - Answer-plug (x,y) coordinates into differential equation, draw
short segments representing slope at each point
/.slope of horizontal line - Answer-zero
/.slope of vertical line - Answer-undefined
/.methods of integration - Answer-substitution, parts, partial fractions
/.∫ u dv = - Answer-uv - ∫ v du
/.dP/dt = kP(M - P) - Answer-logistic differential equation, M = carrying capacity
/.P = M / (1 + Ae^(-Mkt)) - Answer-logistic growth equation
/.given rate equation, R(t) and inital condition when
t = a, R(t) = y₁ find final value when t = b - Answer-y₁ + Δy = y
Δy = ∫ R(t) over interval a to b
/.given v(t) and initial position t = a, find final position when t = b - Answer-s₁+ Δs = s
Δs = ∫ v(t) over interval a to b
/.given v(t) find displacement - Answer-∫ v(t) over interval a to b
/.given v(t) find total distance travelled - Answer-∫ abs[v(t)] over interval a to b
/.area between two curves - Answer-∫ f(x) - g(x) over interval a to b, where f(x) is top
function and g(x) is bottom function
/.volume of solid with base in the plane and given cross-section - Answer-∫ A(x) dx over
interval a to b, where A(x) is the area of the given cross-section in terms of x
/.volume of solid of revolution - no washer - Answer-π ∫ r² dx over interval a to b, where r
= distance from curve to axis of revolution
Average Rate of Change - Answer-Slope of secant line between two points, use to
estimate instantanous rate of change at a point.
/.Instantenous Rate of Change - Answer-Slope of tangent line at a point, value of
derivative at a point
/.Definition of Derivative - Answer-limit as h approaches 0 of [f(a+h)-f(a)]/h or limit as x
approaches a of [f(x)-f(a)]/(x-a)
/.When f '(x) is positive, f(x) is - Answer-increasing
/.When f '(x) is negative, f(x) is - Answer-decreasing
/.When f '(x) changes from negative to positive, f(x) has a - Answer-relative minimum
/.When f '(x) changes fro positive to negative, f(x) has a - Answer-relative maximum
/.When f '(x) is increasing, f(x) is - Answer-concave up
/.When f '(x) is decreasing, f(x) is - Answer-concave down
/.When f '(x) changes from increasing to decreasing or decreasing to increasing, f(x) has
a - Answer-point of inflection
/.When is a function not differentiable - Answer-corner, cusp, vertical tangent,
discontinuity
/.Product Rule - Answer-uv' + vu'
/.Quotient Rule - Answer-(uv'-vu')/v²
/.Chain Rule - Answer-f '(g(x)) g'(x)
/.Particle is moving to the right/up - Answer-velocity is positive
/.Particle is moving to the left/down - Answer-velocity is negative
/.absolute value of velocity - Answer-speed
/.y = sin(x), y' = - Answer-y' = cos(x)
,/.y = cos(x), y' = - Answer-y' = -sin(x)
/.y = tan(x), y' = - Answer-y' = sec²(x)
/.y = csc(x), y' = - Answer-y' = -csc(x)cot(x)
/.y = sec(x), y' = - Answer-y' = sec(x)tan(x)
/.y = cot(x), y' = - Answer-y' = -csc²(x)
/.y = sin⁻¹(x), y' = - Answer-y' = 1/√(1 - x²)
/.y = cos⁻¹(x), y' = - Answer-y' = -1/√(1 - x²)
/.y = tan⁻¹(x), y' = - Answer-y' = 1/(1 + x²)
/.y = cot⁻¹(x), y' = - Answer-y' = -1/(1 + x²)
/.y = e^x, y' = - Answer-y' = e^x
/.y = a^x, y' = - Answer-y' = a^x ln(a)
/.y = ln(x), y' = - Answer-y' = 1/x
/.y = log (base a) x, y' = - Answer-y' = 1/(x lna)
/.To find absolute maximum on closed interval [a, b], you must consider... - Answer-
critical points and endpoints
/.Linearization - Answer-use tangent line to approximate values of the function
/.left riemann sum - Answer-use rectangles with left-endpoints to evaluate integral
(estimate area)
/.right riemann sum - Answer-use rectangles with right-endpoints to evaluate integrals
(estimate area)
/.Trapezoidal rRle - Answer-use trapezoids to evaluate integrals (estimate area)
/.[(h1 - h2)/2]*base - Answer-area of trapezoid
/.definite integral - Answer-has limits a & b, find antiderivative, F(b) - F(a)
/.indefinite integral - Answer-no limits, find antiderivative + C, use inital value to find C
, /.area under a curve - Answer-∫ f(x) dx integrate over interval a to b
/.area above x-axis is - Answer-positive
/.area below x-axis is - Answer-negative
/.average value of f(x) - Answer-= 1/(b-a) ∫ f(x) dx on interval a to b
/.To find particular solution to differential equation, dy/dx = x/y - Answer-separate
variables, integrate + C, use initial condition to find C, solve for y
/.To draw a slope field, - Answer-plug (x,y) coordinates into differential equation, draw
short segments representing slope at each point
/.slope of horizontal line - Answer-zero
/.slope of vertical line - Answer-undefined
/.methods of integration - Answer-substitution, parts, partial fractions
/.∫ u dv = - Answer-uv - ∫ v du
/.dP/dt = kP(M - P) - Answer-logistic differential equation, M = carrying capacity
/.P = M / (1 + Ae^(-Mkt)) - Answer-logistic growth equation
/.given rate equation, R(t) and inital condition when
t = a, R(t) = y₁ find final value when t = b - Answer-y₁ + Δy = y
Δy = ∫ R(t) over interval a to b
/.given v(t) and initial position t = a, find final position when t = b - Answer-s₁+ Δs = s
Δs = ∫ v(t) over interval a to b
/.given v(t) find displacement - Answer-∫ v(t) over interval a to b
/.given v(t) find total distance travelled - Answer-∫ abs[v(t)] over interval a to b
/.area between two curves - Answer-∫ f(x) - g(x) over interval a to b, where f(x) is top
function and g(x) is bottom function
/.volume of solid with base in the plane and given cross-section - Answer-∫ A(x) dx over
interval a to b, where A(x) is the area of the given cross-section in terms of x
/.volume of solid of revolution - no washer - Answer-π ∫ r² dx over interval a to b, where r
= distance from curve to axis of revolution
