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Test # 3, Calculus I, November 5 th^ 2006, Solutions
- Two sides of a triangle are 5 m and 12 m and the angle between them is increasing at a rate of 3 โฆ/s Find the rate at which the third side is increasing when the angle between the sides of fixed length is 90 โฆ.
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
- A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at the rate of 1. 5 feet per second. (a) How fast is the top moving down the wall when the base of the ladder is 20 feet from the wall? (b) How high is the top of the ladder when it moves three times as fast as the base?
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
- Find the maximum and the minimum of the func- tion f (x) = 6x^2 /^3 โ x^5 /^3 on the interval [โ 1 / 2 , 6].
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)
- State the Mean Value Theorem.
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).
- Find where the function g(x) =
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
