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The solutions to test #4 in a calculus i course. Topics covered include curve sketchting of a function, finding extrema and points of inflection, calculating limits, finding the volume of a cone, determining velocity and acceleration, and optimizing the dimensions of a wooden beam. The document also includes the use of maple for finding derivatives and graphs.
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1.a Sketch the graph of f (x) = (^) x 2 x+1 , showing the maxima, minima, inflection points, and horizontal asymptotes. You may use Maple to find the graph, and the first and second derivatives.
-0.
-0.
0
-10 -8 -6 -4 -2 (^2 4) x 6 8 10
1.b What equations did you solve to find the extrema, and the points of inflec- tion? f^0 (x) = d dx
μ x x^2 + 1
1 − x^2 (x^2 + 1)^2
Set
1 − x^2 = 0 x = − 1 , 1 f(x) = −
0
1
-10 -8 -6 -4 -2 (^2 4) x 6 8 10
derivative
min at (− 1 , −^12 ),max at (1, 12 )
f 0 0 (x) = d dx
μ 1 − x^2 (x^2 + 1)^2
= 2x x^2 − 3 (x^2 + 1)^3
Set
x
x^2 − 3
x = 0 ,
f(x) = 0 ,
found to 325 ft with a possible error of 10 feet. The volume of a cone is given by
πr^2 h 3
2.a Give a formula for the volume of a pile of sand in terms of the diameter of the base. Use the formula to calculate the volume of sand. Let D be the diameter and C the circumference Then
r =
, h =
π and so V = π 3
μ D 2
π 30
The volume of the sand is about π 30
(325/π)^3 = 1. 1594 × 105 cu.ft.
2.b What is the maximum possible error in calculation of the volume of sand. Since dV = π 10 D^2 dD
and dD = (^) π^1 dC, Then
dV = π 10
dC π
dC.
Thus |dV | =
|dC| ≤ (325/π)^2 10
A 3 kg mass sitting at rest at the origin is subjected to the force F (t) = 6t − 3 t^2 newtons from t = 0 to t = 2 seconds. Determine the velocity and displacement of the function after 2 seconds have elapsed.
a(t) = F (t) m
6 t − 3 t^2 3 = 2t − t^2.
Then v(t) =
a(t)dt =
2 t − t^2
dt = t^2 −
t^3 + C,
and
0 = v(0) = 0 + 0 + C, C = 0 , so v(t) = t^2 −
t^3.
Next
s(t) =
v(t)dt =
Z μ t^2 −
t^3
dt =
t^3 −
t^4 + C
and
0 = s(0) = 0 + 0 + C, C = 0 , so s(t) =
t^3 −
t^4.
We were asked for v(2) = 4 − 83 = 43 , and s(2) = 83 − 1612 = 43 = 43. The graphs are:
0
1
0.2 0.4 0.6 0.8 (^1) t 1.2 1.4 1.6 1.8 2
A wooden beam has a rectangular cross-section of height h and width w. The strength S of the beam, measured in appropriate units, is 200 times the product
and h =
Aletrnative.Let θ the acute angle between a horizontal radius and the radial line from the centre of a beam and a corner. Then
w = 24 cos θ, h = 24 sin θ.
Thus
S = 200 (24 cos θ) (24 sin θ)^2 = 2764 800 cos θ sin^2 θ
Here is the graph
0
00000
00000
00000
00000
1e+
0.2 0.4 0.6 0.8 1 1.2 1.
Taking the derivative of S we get
dS dθ
d dθ 115 200 cos θ sin^2 θ = 345 600 cos^2 θ sin θ − 115 200 sin θ 115 200 (sin θ)
3 cos^2 θ − 1
We must chose
cos θ =
, sin θ =
and so
w = 24
h = 24
