Derivatives
Find the first derivative of the equation: x²+4x+7
42x^6+6x²+148x
∫e2xdx
1/2e2x+C
A group of people are taking a road trip in a car such that their distance from their starting location can be represented in miles by f(t)=2t3+7t2+8t for any time t, in hours.
a) How fast is the car moving when t=3 hours?
b) When is the car stopped?
a) At t=3 the speed is 38mi/hr
b) T=-1,-4/3
6p = 4w2 +17
The value of w is increasing at a rate of 5 inches per hour. Find the rate at which p is changing when w is 3/2 inches
5/12
Evaluate:
limx->1 lnx/x2
0
Find the first derivative of the equation: 6x7 -2x3+74x2
42x6-6x2+148x
∫6/(1-2x)3dx
3/2(1-2x)2+C
The position of a particle moving along the x-axis is modeled by the function: s(t)=t2-6t2+9t and t≥0
a) Find the acceleration at each time the velocity is zero
b) Find the speed each time the acceleration is zerp
c) Find the displacement over the first 6 seconds
d) Find the total distance traveled in the first 2 seconds
a) 3 and 1
b) 3=s
c) 54
d) 30
q = 1/4b3
The value of q is decreasing at a rate of 30 cubic inches per second. Find the rate at which b is changing when b is 6 inches
30/8
Evaluate:
limx->5 x2-25/x-5
10
Find the Second Derivative: 1/3x3+2/7x7+x5
2x+12x5+20x3
∫Sin(1/x)/3x2dx
1/3cos(1/x)+C
Starting with position, consider the following: A particle moves along a horizontal line so that its position is given by x(t) = t3+ 3/2t2 - 18t + 4, t >0
a. Find v(t) and a(t)
b. When is the particle speeding up?
c. What is the displacement of the particle over [0,4]?
a. v(t) = 3t2 + 3t - 18, a(t) = 6t + 3
b. t > 2
c. 16 units right
Last night, a runner was jogging away from a lamppost 37 feet high at a rate of 9 feet per second. How fast is the shadow of the runner lengthening?
45/32
Evaluate:
limx->∞ x3+2x-1/3-5x2-2x3
1/2
Find the First Derivative:
csc-1(3x)
-1/⟦x⟧sqrt(9x2-1)
Write as a definite integral
limn->∞ ∑ni=1[1-3(-2+3i/n)]3/n
∫1-2 (1-3x)dx
-Particle P moves along the x -axis such that, for time t > 0, its position is given by x t = 6 4e−t P ( )
-Particle Q moves along the y -axis such that, for time t > 0, its velocity is given by 1 vq ( t) = 2 t
- At time t = 1, the position of particle Q is yq (1) = 2
a) Find vP (t), the velocity of particle P at time t
b) Write the integral of the position particle at t
c) As t → ∞, which particle will eventually be farther from the origin?
a) Vp(t)=dy/dx xp(t)=4e-t
b) yq(t)=yq(t)+∫t1 1/s2 ds
c) Yes, the particle will eventually be farther from the origin
4w2 + 5h2 = 536
The value of w is decreasing at 15 miles per hour. Find the rate at which h is changing when h is 10 miles and w is 3 miles.
36/10
limx->pi/2sin(2x)/cos(3x)
-2/3
Find the Second Derivative:
3x3+cot-1(3x)
First Derivative:
9x2 -3x2/(1+x6)
Second Derivative:
18x -18x7(1+x6)-2+6x(1+x6)-1
Find the exact value of ∫50 (x2+x)dx
Solutions
∑ni=1c=cn ∑ni=1i=n(n+1)/2 ∑ni=1i2=n(n+1)(2n+1)/6
∑ni=1i3=n2(n+1)2/4
325/6
Starting with the position particle: x(t)=3t2-12x+2
a) At what time is the particle stopped?
b) During what time interval is the particle moving to the right?
c) Is the particle moving away or towards the origin at t=1?
d) What is the average velocity of the particle over (0,5)?
a) t stops at 2
b) (2,∞)
c) Towards
d) 3ft/sec
Water is flowing into a tank in the form of an inverted cone having an height of 16 meters and a radius of a 4 meters at the rate of 2 meters/3min. How fast is the radius changing at the instant the radius of the tank is 2.5?
2/25π meters per min
Find the value of K for which the following limit exists then rewrite the equation
limx->3 4x2+kx+7k-6/22-5x-3
k=-3
(4x+0)(x-3)/(x-3)(2x+1)