What is the limit as x approaches to infinity for sin(x)?
g(x)=(6x2)/(2-x)
Solution: g'(x)=(24x-6x2)/(2-x)2
What is the coefficient of π₯2 in the Taylor series for π ππ(π₯)2 about 0 ?
1
The polar curve C has equation
r = β 2 (cos(ΞΈ)sin(ΞΈ) ), 0 β€ ΞΈ< 2Ο .
Find a Cartesian equation for C.
Answer: (x-1)2+(y+1)2=2
Evaluate β« x(3/2) + 2x + 3 dx
(2/5)x(5/2) + x2 +3x + c
When does the limit not exist?
oscillating behavior
Unbounded behavior
Removable discontinuity
The first and second option
Differentiate h(t)=(t3)-(t2)sin(t)
h'(t)=(3t2)-2tsin(t)-(t2)(cos(t))
2. Determine if the series converges or diverges:
The summation of (-1)n-1 /(7+2n). (n=1 and ontop of sumation is infinity)
So, both of the conditions in the Alternating Series Test are met and so the series is convergent. (limit is 0 and it is decreasing)
The curve C has polar equation: r = 6cos(3ΞΈ) , β Ο< ΞΈβ€ Ο .
Find the exact value of area enclosed by the C , for (-Ο/6) < ΞΈ β€ (Ο/6)
Answer: Area: 3Ο
Evaluate β« x7 - 48x11 - 5x16 dx
(1/8)x8 - 4x12 - (5/17)x17 + c
What is the limit as x approaches 3 for (x^2 -3x +2)/(x-2)?
2
Z(v)= (v+tanv)/(1+cscv)
Z'(v)= ((1+sec2(v))(1+cscv)-(v+tanv)(-csc(v)cot(v))/(1+cscv)2
2. The coefficient of (π₯ β (π /4) )^3 in the Taylor series about π/4 of π(π₯) = πππ π₯ is
1/(6β2)
The curve C with polar equation
r = β(6) cos (2ΞΈ) , 0β€ ΞΈ β€ Ο/4.
The straight line l is parallel to the initial line and is a tangent to C .
Find an equation of l , giving the answer in the form r =f (ΞΈ )
Answer: r= 2/3cosecΞΈ
Evaluate β« x3 - ((e-x - 4)/e-x) dx
(1/4)x4 - x + 4ex + c
f(x)={5x-1 x<1, 10-2x xβ₯ 1}
Does f'(1) exist and is MVT satisfied on (-2,2)
fβ(1) = -2 and no, discontinuity -> theorem not satisfied
Given that x=-2, y=1, and x'=-4, determin y' for the following equation.
6y2+x2=2-x3e4-4y
y'=8/11
4. Let π be the function given by π(π₯) = cos(2π₯). Write the first four nonzero terms and the general term of the Taylor series for π about π₯ = 0.
π β ((ππ)2)/ π! + ((ππ)4)/π! β ((ππ)6)/π! + β― + (βπ)n ((ππ)2n)/(ππ)!
The following polar equations are given:
r1=cosΞΈ, 0 β€ ΞΈ β€ Ο
r2= 1/(cosΞΈ - sin ΞΈ), (-ΒΌ)Οβ€ ΞΈ β€(5/4)Ο
Find, in exact simplified form, the area of the smaller of the two finite regions, bounded by r1 and r2.
Answer: (3Ο+ 2)/16
Evaluate β«49 (βx + (1/3βx) dx
40/3 = 13.333
There are two graphs:
f(x) is going from the bottom to up with a limit of -3 at x=2. From up to bottom, it has a limit of 4 at x=2.
g(x) is approaching from the bottom on both sides to 2 when x=2.
What is the limit of f(g(x)) at x approaches 2?
Limit of f(g(2)) is -3
g(x)=(ln(x2+1)-tan-1(6x))10
g'(x)=10(ln(x2+1)-tan-1(6x)^9 ((2x/(x2+1)-(6/(36x2+1)
5. At tβ₯0, a particle moving in the xy plane has the velocity vector given by v(t)= <4e-t, sin(1+β(t)). What is the total distance the particle moves between t=1 and t=3?
1.861
r2= 2cos2ΞΈ, 0β€ΞΈ< Ο/4
The finite region R is bounded by C , L and the half line with equation ΞΈ = Ο/2 (y-axis).
Find the Area of region R.
Answer: (β )(3β(3) -4)
Determine h(t) given that h''(t) = 24t2 - 48t + 2, h(1) = -9 and h(-2) = -4
h(t) = 2t4 - 8t3 + t2 + (100/3)t - (112/3)