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Solutions to test # 3 for calculus i, covering topics such as the law of cosines, rates of change, maximum and minimum values, and the mean value theorem. It includes step-by-step calculations and graphical illustrations.
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Test # 3, Calculus I, November 5 th^ 2006, Solutions
Solution: We are going to refer to the next figure for calculations.
We apply the Law Of Cosines for this triangle: a^2 = b^2 +c^2 โ 2 bc cos A. We think of the variables a (measured in meters) and A (measured in radians) as functions of t (time measured in seconds). After differentiating this we obtain: 2 a
da dt
= 2bc sin A
dA dt
So we need to find dAdt (in radians per second) and then substitute for A = ฯ/2 and a =
b^2 + c^2 = 13 m. This I
gives da dt
dA dt
dA dt
To calculate dAdt we use the formula of transforming de- grees in radians:
dA dt
3 ฯ 180
ฯ 60
radians second
Therefore
da dt
ฯ 13
m s
Solution: (a) We denote by x, as in the figure below, the hight of the top of the ladder and by y the distance from the house of the base.
II
(b) If we assume that dydt = v 6 = 0 then dxdt = โ 3 v. We substitute in the related rates equality:
โ 2 x(3v) + 2y(v) = 0
which gives y = 3x. Hence 25^2 = x^2 +y^2 โโ x^2 +9x^2 = 252 which solved for x will give
x =
โ 7. 9 f t,
so when x โ 7. 9 f t the top of this ladder moves three times as fast as the base. โข
Solution: Differentiate the function f to obtain f โฒ(x) = 6(2/3)xโ^1 /^3 โ (5/3)x^2 /^3 or
f โฒ(x) = (
(12/ 5 โ x) โ (^3) x
which gives critical points x = 0 and x = 12/5. Then we need to calculate f (0) = 0, f (12/5) = 3625901 /^3 โ 6 .45, f (โ 1 /2) = 134 21 /^3 โ 4 .09 and f (6) = 0. We conclude that minimum of f on this interval is 0 and the maxi-
mum is
901 /^3 โ 6 .45. The graph of f on [โ 1 / 2 , 6] is
attached:
IV
x
6 5
6
4 3
5
2 1 (^001234)
Solution: Theorem (MTV): Given a function f : [a, b] โ R, (a < b), which is differentiable on (a, b) and continuous on [a, b] there exist a c โ (a, b) such that f (b) โ f (a) = f โฒ(c)(b โ a).
x + 2 โ x^2 + 2
is in-
creasing and where is decreasing.
Solution: Differentiating this function we obtain
gโฒ(x) = (x^2 + 2)โ^1 /^2 + (x + 2)(โ 1 /2)(x^2 + 2)โ^3 /^2 (2x)
or after simplifications
V
Hence there is only one inflection point, at x = 0, since this the only point where the second derivative changes sign around it. The function is concave upward on [0, โ) and concave downward on (โโ, 0]. The graph of f on [โ 1. 2 , 2 .5] and the tangent line at the inflection point, (0, 1) , is:
-1 -0.5 00 1 1.
0.5 x 2 2.
2 1
and it certainly illustrates the indicated behavior.
Proof. 7. Determine which of the graphs below is the graph of f , f โฒ^ and f โฒโฒ.
VII
00
1
-4 2 x
-0.
4
-1.
Solution: The function whose increasing/decreasing be- havior is not reflected in the other graphs must be the graph of f โฒโฒ. This is the tallest graph in the given pic- ture. Its graph is:
0
-4 -2 0
-1.
x 2
1
-0.
4
VIII