Calc BC Exam / Expert Strategies, Review of Key
Quizzes, and Practice Questions for Guaranteed
Success’
Terms in this set (45)
If f(x)=cos^2(3x-5) -6sin(3x-5)cos(3x-5)
f'(x)=
∫1/t√t dt= -2t^(-1/2) + c
If f(x)= (5-x)/(x^3+2) (2x^3-15x^2-2)/(x^3+2)^2
f'(x)=
The position of a particle {12,4y"(1)}
moving in the xy-plane is
given by the vector
{4t^3,y(2t)}, where y is a
twice-differeniable function
of t.
At time t=1/2, what is the
acceleration vector of the
particle?
To what number does the π/(π+e)
series
E (-e/pi)^k converge?
, lim as x->a of f(x) ≠ f(a)
Look at graph on #6
Which of the following is
true?
If ∫4 -10 g(x)=-3 8
and ∫4 6 g(x)=5
then ∫-10 6 g(x)=
The length of the curve ∫0 π/6 √1+9cos^2(3x)
y=sin(3x) from x=0 to x=π/6
is given by
The slope of the line tangent 2ln2 + 2
to the graph of y=xe^x
at x=ln2 is
Let y=f(x) be the solution to 15/4
the differential equation
dy/dx= x-y with initial
condition f(2)=8. What is the
approximation for f(3)
obtained by using Euyler's
method with two steps of
equal length, starting at x=2?
If x^2+xy-3y=3, then at the 5
point (2,1)
dy/dx=
~(3x+1)/(x^2-4x+3)= 5ln[x-3] - 2ln[x-1] + c
Quizzes, and Practice Questions for Guaranteed
Success’
Terms in this set (45)
If f(x)=cos^2(3x-5) -6sin(3x-5)cos(3x-5)
f'(x)=
∫1/t√t dt= -2t^(-1/2) + c
If f(x)= (5-x)/(x^3+2) (2x^3-15x^2-2)/(x^3+2)^2
f'(x)=
The position of a particle {12,4y"(1)}
moving in the xy-plane is
given by the vector
{4t^3,y(2t)}, where y is a
twice-differeniable function
of t.
At time t=1/2, what is the
acceleration vector of the
particle?
To what number does the π/(π+e)
series
E (-e/pi)^k converge?
, lim as x->a of f(x) ≠ f(a)
Look at graph on #6
Which of the following is
true?
If ∫4 -10 g(x)=-3 8
and ∫4 6 g(x)=5
then ∫-10 6 g(x)=
The length of the curve ∫0 π/6 √1+9cos^2(3x)
y=sin(3x) from x=0 to x=π/6
is given by
The slope of the line tangent 2ln2 + 2
to the graph of y=xe^x
at x=ln2 is
Let y=f(x) be the solution to 15/4
the differential equation
dy/dx= x-y with initial
condition f(2)=8. What is the
approximation for f(3)
obtained by using Euyler's
method with two steps of
equal length, starting at x=2?
If x^2+xy-3y=3, then at the 5
point (2,1)
dy/dx=
~(3x+1)/(x^2-4x+3)= 5ln[x-3] - 2ln[x-1] + c
