Limits
Limit x-->1 of 3x2+x+16
20
Find the derivative of ((x)1/2)+sin(1-x)
(1/2(x)1/2-cos(1-x)
∫60 3x2-2x+1 dx
((6)3- (6)2+6)-((0)3-(0)2+0)
216-36+6-0 = 186
A region is bonded by the x-axis and y=x2+x-6, find the area between the curve.
∫2-3 (x2+x-6) dx = -125/6
dy/dx=cos(x)y
dy/y=cos(x) dx
∫dy/y=∫cos(x) dx
ln(y)=sin(x)+c
y=Cesin(x)
Limit x-->+infinity 1/5x2
Limit x-->+infinity 1/5x2=0
Find the derivative (dy/dx) of x4+2x3+4x2+6x+1+cos(3x)+5xex
dy/dx = 4x3+6x2+8x+6-3sin(3x)+5ex+5xex
S(t) is a differential function that represents the speed of the car in miles, where t is measured in hours. Explain the meaning of S’(20).
The acceleration (rate of change in speed S(t)) in miles/hour2 of the car when t=20 hours.
The base of a solid is the region enclosed by y=x2-10x+21 and the x-axis, find the volume of the solid that has square cross sections perpendicular to the x-axis.
∫73 (x2-10x+21)2 dx = 512/15
dy/dx=3xy+4x2+y
Find d2y/dx2.
d2y/dx2=3(dy/dx)+8x+(dy/dx)
d2y/dx2=3(3xy+4x2+y)+8x+(3xy+4x2+y)
Limit x-->-infinity (5x3+4x2+8x+1)/(3x3+7x2+15x+2)
5/3
Does function f(x) = (x3/3)-3x2+9x+1 have any local extrema?
x2-6x+9 = f’(x)
x2-6x+9 = 0
(x-3)(x-3) = 0
Possible critical point at x = 3
Test numbers: 0,6
f(2) = 1 f(6) = 19
No local extrema.
∫60x2(4)(20x^3)-5 dx
u = 20x3-5
du/dx = 60x2
du/60x2 = dx
∫60x2(4)(u) du/60x2
∫ (4)(u) du
(4(u)/ln(4))+c
(4(20x^3)-5/ln(4))+c
The base of a solid is the region enclosed by f(x)=3sqrt(x) and g(x)=x, find the volume of the solid that has equilateral triangle cross sections perpendicular to the y-axis.
x = (y/3)2
x=y
∫90 (sqrt(3)/4)(y-(y/3)2)2 dy = (27/2)(sqrt(3)/4) = 27(sqrt(3))/8
dy/dx=(4x+ex)/2y
2y dy=(4x+ex) dx
∫2y dy=∫(4x+ex) dx
y2=2x2+ex
y=+/-sqrt(2x2+ex)
Limit x-->1 (x2+3x-4)/(sin(pix))
Limit x-->1 (x2+3x-4)=0
Limit x-->1 (sin(pix))=0
Use L’Hopital’s Rule
Limit x-->1 (2x+3)/((pi)cos(pix))=-5/pi
Given the function f’(x) = (x2+3)/cos(x), and that -4<x<4, at which interval does the function f(x) is concaving down? Explain.
(-4,0) because f’(x) is decreasing in that interval.
∫e2xcos(3x)
v=e2x du=cos(3x)
dv=2e2x u=(sin(3x))/3
∫e2xcos(3x)=((sin(3x))/3)(e2x)-∫(sin(3x))/3) 2e2x dx
v=2e2x du=(sin(3x))/3
dv=4e2x u=(-cos(3x))/9
∫e2xcos(3x)=((sin(3x))/3)(e2x)-((-cos(3x))/9)(2e2x)+∫((-cos(3x))/9)(4e2x) dx
∫e2xcos(3x)=((sin(3x))/3)(e2x)-((-cos(3x))/9)(2e2x)+(-4/9)∫cos(3x)(e2x) dx
(13/9)∫e2xcos(3x)=((sin(3x))/3)(e2x)-((-cos(3x))/9)(2e2x)
∫e2xcos(3x)= (((sin(3x))/3)(e2x)-((-cos(3x))/9)(2e2x))(9/13)
The region R enclosed by y=(4sqrt(x))+1 and x=8, and is rotated around the x-axis, find the volume of the solid that has been formed.
pi∫80 ((4sqrt(x))+1)2 dx = 640.680pi
The rate of change of F is proportional to F, when t=0, F=200, and when t=2, F=500, what is the value of F when t=4?
dF/dt=kF
ln(F)=kt+c
C=200
F=200ekt
500=200ek(2)
k=0.458
F=200e0.458t
F=200e0.458(4)
F=1250
The continuous and differentiable function f is given by f(x)={ax+5 for x less than or equal to 6, 3x2+b for x>6. Find a+b.
ax+5=3x2+b
a=6x
a=36
36x+5=3x2+b
36(6)+5=3(6)2+b
221=108+b
b=113
36+113=149
a+b=149
A cylinder can fill with water has an unknown volume initially, and the water is gradually leaking out from the hole on its bottom side at a rate. The volume of the cylinder is given by the equation V=(pi)r2h. Suppose that the cylinder has a radius of 3 cm, what is the can’s instantaneous rate of change in height at the moment where its instantaneous rate of change in volume is at 4cm3/minute?
dV/dt = (pi)r2(dh/dt)
4 = pi(3)2(dh/dt)
dh/dt = 4/9pi
An object is moving to the right horizontally with an acceleration given by the function a(t) = 6t+4, at t = 1, the object’s velocity is 10 cm/minute, at t = 2, the object’s position is at 28 cm, what is the function that represent the object’s position in p(t)?
a(t) = 6x+4
v(t) = ∫a(t) dt = 3x2+4x+c
v(1) = 3(1)2+4(1)+c = 10 c = 3
v(t) = 3x2+4x+3
p(t) = ∫v(t) dt = x3+2x2+3x+c
p(2) = (2)3+2(2)2+3(2)+c = 28 c = 6
p(t) = x3+2x2+3x+6
The region R enclosed by y=10-x2 and y=5, and is rotated around the x-axis, find the volume of the solid that has been formed.
pi∫2.236068-2.236068 (10-x2)2-(5)2 dx = 208.700pi
Find the time necessary for $1000 to triple if it is invested at a rate of 3.5% compounded daily.
A=1000(1+(0.035/365))365t
3000=1000(1+(0.035/365))365t
t=31.39 years.