calculus midterm study
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Apr 3, 2024
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Uploaded by JudgeJellyfishMaster1076
L’hopitals rule Lim as x -> c (f(x)/g(x)) = Lim as x -> c (f’(x)/g’(x))
Implicit diff
Step - 1: Differentiate every term on both sides with respect to x.
Then we get d/dx(y) + d/dx(sin y) = d/dx(sin x).
Step - 2: Apply the derivative formulas to find the derivatives and also apply the chain rule.
(All x terms should be directly differentiated using the derivative formulas; but while differentiating the y terms, multiply the actual derivative by dy/dx)
In this example, d/dx (sin x) = cos x whereas d/dx (sin y) = cos y (dy/dx).
Then the above step becomes:
(dy/dx) + (cos y) (dy/dx) = cos x
Step - 3: Solve it for dy/dx.
Taking dy/dx as common factor:
(dy/dx) (1 + cos y) = cos x
dy/dx = (cos x)/(1 + cos y)
This is the implicit derivative.
riemann sum definite integral steps
Sumnaiton Fundamental theorem of calc part 1
d/dx integral (f(x))dx = f(x)
Part 2
Integral a to b (f(x)) dx = F(b) - F(a)
Where F(x) = integral (f(x)) dx
linearization formula
L(x) = f(a) + f′(a)(x − a) . f(x) ≈ L(x) near x = a.
Area between two curves A=∫ba(upperfunction)−(lowerfunction)dx,a≤x≤b
A=∫dc(rightfunction)−(leftfunction)dy,c≤y≤d
Volume solid revolution
Pi integral a to b (R(x))^2 dx
R(x) = distance between the function and the axis of rotation
a = upper limit
b = lower limit
dx = slides along x
1 Question - mp choice
Let G(x) = integral -1 to sinx (t^2) dt. Then G’(2pi/3) is equal to:
B)−3/8
2 Question - mp choice
The value of lim x-> -inf (xsinh (1/x)) is:
B) 1
2 Question - mp choice
Let f(x) = x^5 + x^3 + 3x − 2. Then (f^−1)′(−2) is equal to:
C) 1/3
Question - Evaluate the following integrals.
a) integral from 0 to cube root (pi) (t^2 cos^2(t^3) sin(t^3) dt)
Pt[u = t^3/du = 3t^2 dt]
= 1/3 integral from 0 to pi (cos^2(u) sin(u) du)
Pt[v = cos(u) / dv = -sin(u) du]
= -1/3 integral from 1 to -1 (v^2 dv) = -1/3 (1/3v^3) from 1 to -1
= -1/9(-1-1) = 2/9
b) integral (dx/sqrt(3 + 6x -9x^2))
= integral (dx/sqrt(4 - (3x -1)^2))
Pt [u = 3x-1/ du = 3dx]
= integral (du / 3sqrt(4-u^2))
= 1/3sin^-1 (u/2) +C
= 1/3sin^-1 (3x-1 / 2) +C
Question - pool 5. Consider a pool shaped like the bottom half of a sphere. The radius of the pool is 10 ft.
Hint: When the water has depth h and the radius of the sphere is r, the volume of
water is π/3 h^2(3r − h).
(a) When the pool is being filled at a rate of 25 ft3/min, find the rate at which the depth of the
water is changing when the water has a depth of 5 ft.
V = pi/3 * h^2(30-h)
=10pih^2 - pi/3 * h^3
So dv/dt = 25
=20pih dh/dt - pih^2 dh/dt
<=>
25 = (100pi - 25 pi) dh/dt
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<=>
dh/dt = 25/75pi
= 1/3pi ft/min
(b) Later when the pool is being drained, if the depth of the water is decreasing at a rate of 1 in/min when its depth is 2 ft, find the rate at which the water is being drained.
Note dh/dt = -1in/min = -1/12 ft/min
dV/dt = 20pih dh/dt - pih^2 dh/dt
= 20pi(2)(-1/12) - pi(2)^2 (-1/12)
= -10pi/3 +pi/3
= -3pi ft^3/min
Question - Find the area of the region bounded by the curves y = 4x−x^2, y = x,
and the positive x-axis.
These two curves intersect at the point (3,3), and have roots at (0,0) and (4,0)
Respectively.
A = integral from 0 to 3 (x) dx + integral from 3 to 4 (4x-x^2) dx
= (1/2 x^2) line from 0 to 3 + (2x^2 - 1/3 x^3) line from 3 to 4
= 9/2 - 0 + 32 - 64/3 - 18 + 9 = 37/6
Question - Find the volume of the solid obtained by rotating the region bounded by y = x^3 + 5/2, y = 9/2 − x^2, and x = 2 about the line x = 3.
Find the intersection point: x^3 + 5/2 = 9/2 - x^2 <-> x^3 + x^2 -2 = 0 <-> (x-1)(x^2+2x+2) = 0
So the intersection point is at (1, 7/2)
Cylindrical shell method: r = 3 − x and h = (x^3 +5/2) − (9/2 − x^2) = x^3 + x^2 − 2 so
V = 2pi * integral from 1 to 2 (3-x)(x^3+x^2-2)dx
= 2pi * integral from 1 to 2 (-x^4 + 2x^3 + 3x^2 +2x - 6) dx
= 2pi * (-1/5x^5 + 1/2x^4 +x^3 + x^2 -6x) line from 1 to 2 = 2pi ((-32/5 + 8 + 8 + 4 -12) - ( -1/5 + 1/2 + 1 + 1 -6))
= 2pi (8/5 + 37/10) = 53pi/5
Question - consider f(x) = sqrt(x) ln(x)
(a) Find the intervals where f (x) is increasing or decreasing.
f’(x) = ln(x)+2/2sqrt(x) which is 0 if x = e^-2, so f(x) is decreasing on (0, e^-2) and increasing on (e^-2, inf)
(b) Find and classify the local extreme values of f(x).
By the first derivative test, f(x) has a local minimum at f(e^−2) = −2/e
(c) Find the intervals where the graph of f (x) is concave up or concave down.
f’’(x) = (2sqrt(x)-(ln(x)+2)1/sqrt(x)) / 4x = -ln(x) / 4x^3/2 which is 0 if x=1 so f(x) is concave up on (0, 1) and concave down on (1, inf)
(d) Find all inflection points (c, f (c)) of f (x).
The only inflection point of f(x) is (1,0).
Question - A rancher will use 600 m of fencing to build a corral in the shape of a semicircle on top of a rectangle (see diagram - dotted line is for emphasis and will not be built using fencing). Find the dimensions that maximize the area of the corral.
The radius of the semicircle is x/2. The perimeter of the corral is x + 2y + pix/2 = 600 <-> y = 300 - (pi+2)x/4
The enclosed area A = 1/2 * pi (x/2)^2 + xy = pix^2/8 + 300x - (pi+2)x^2 / 4 = 300x - (pi+4)x^2 / 8 So A’ = 300 - (pi + 4)x^2/4 = 0 <-> x = 1200/ x+4
Then y = 300 - (pi+2)(1200/pi+4) / 4 = 600/pi+4
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