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Calculus I Exam Solutions: November 5th 2006 - Prof. Eugen Ionascu, Exams of Analytical Geometry and Calculus

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.

Typology: Exams

Pre 2010

Uploaded on 08/04/2009

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Test #3, Calculus I, November 5th 2006, Solutions
1.Two sides of a triangle are 5 m and 12 m and the
angle between them is increasing at a rate of 3โ—ฆ/s
Find the rate at which the third side is increasing
when the angle between the sides of fixed length is
90โ—ฆ.
Solution: We are going to refer to the next figure for
calculations.
We apply the Law Of Cosines for this triangle: a2=
b2+c2โˆ’2bc 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:
2ada
dt = 2bc sin AdA
dt .
So we need to find dA
dt (in radians per second) and then
substitute for A=ฯ€/2 and a=โˆšb2+c2= 13 m. This
I
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Test # 3, Calculus I, November 5 th^ 2006, Solutions

  1. Two sides of a triangle are 5 m and 12 m and the angle between them is increasing at a rate of 3 โ—ฆ/s Find the rate at which the third side is increasing when the angle between the sides of fixed length is 90 โ—ฆ.

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 โˆ’ 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

  1. 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 = โˆ’ 3 v. We substitute in the related rates equality:

โˆ’ 2 x(3v) + 2y(v) = 0

which gives y = 3x. Hence 25^2 = x^2 +y^2 โ‡โ‡’ x^2 +9x^2 = 252 which solved for x will give

x =

โ‰ˆ 7. 9 f t,

so when x โ‰ˆ 7. 9 f t the top of this ladder moves three times as fast as the base. โ€ข

  1. Find the maximum and the minimum of the func- tion f (x) = 6x^2 /^3 โˆ’ x^5 /^3 on the interval [โˆ’ 1 / 2 , 6].

Solution: Differentiate the function f to obtain f โ€ฒ(x) = 6(2/3)xโˆ’^1 /^3 โˆ’ (5/3)x^2 /^3 or

f โ€ฒ(x) = (

(12/ 5 โˆ’ x) โˆš (^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‰ˆ 6 .45, f (โˆ’ 1 /2) = 134 21 /^3 โ‰ˆ 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‰ˆ 6 .45. The graph of f on [โˆ’ 1 / 2 , 6] is

attached:

IV

x

6 5

6

4 3

5

2 1 (^001234)

  1. 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 โˆˆ (a, b) such that f (b) โˆ’ f (a) = f โ€ฒ(c)(b โˆ’ a).

  1. Find where the function g(x) =

x + 2 โˆš x^2 + 2

is in-

creasing and where is decreasing.

Solution: Differentiating this function we obtain

gโ€ฒ(x) = (x^2 + 2)โˆ’^1 /^2 + (x + 2)(โˆ’ 1 /2)(x^2 + 2)โˆ’^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, โˆž) and concave downward on (โˆ’โˆž, 0]. The graph of f on [โˆ’ 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 โ€ฒ^ and f โ€ฒโ€ฒ.

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 โ€ฒโ€ฒ. This is the tallest graph in the given pic- ture. Its graph is:

0

-4 -2 0

-1.

x 2

1

-0.

4

VIII