{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

31aPracticeFinal-solutions

8 5(10 points a 6 ft tall man walks away from a 15 ft

This preview shows pages 8–13. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 8 5. (10 points) A 6 ft tall man walks away from a 15 ft lamp post at a speed of 3 ft/sec. At what rate is the length of his shadow increasing? Solution: 15 ft 6 ft X S 15 x + s = 6 s s = 2 3 x ds dt = 2 3 dx dt We are given dx dt = 3 So ds dt = 2 ft per second. 9 6. (10 points) Let f ( x ) = x 2- 5 x- 6 and F ( X ) = R x f ( t ) dt . a ) Find the critical points of F ( x ) and determine whether they are local minima, maxima, or neither. Solution: By the Fundamental Theorem of Calculus F ( x ) = f ( x ) = x 2- 5 x- 6 = ( x- 6)( x + 1). So F ( x ) has critical points at x =- 1 , 6. The second derivative F 00 ( x ) of F ( x ) is just f ( x ) = 2 x- 5. We have F 00 (- 1) =- 7 F 00 (6) = 7 so by the second derivative test, F has a local maximum at x =- 1 and a local minimum at x = 6. b ) Find the points of inflection of F ( x ) and determine the change in concavity (up to down or down to up) of these points. Solution: From the above, F 00 ( x ) = 2 x- 5, which has a zero at x = 5 2 . To the left of this point, F 00 ( x ) is negative, and to the right of this point, F 00 ( x ) is positive. So at x = 5 2 , F ( x ) changes from concave down to concave up. 10 7. (10 points) Find the points on the ellipse 4 x 2 + y 2 = 4 that are farthest away from the point (1 , 0). Solution: The squared distance D between points (1 , 0) and ( s,t ) is given by the distance formula D = ( s- 1) 2 + ( t- 0) 2 = ( s- 1) 2 + t 2 if ( s,t ) also lies on this ellipse, then 4 s 2 + t 2 = 4, so t 2 = 4- 4 s 2 , so D = ( s- 1) 2 + t 2 = ( s- 1) 2 + (4- 4 s 2 ) =- 3 s 2- 2 s + 5 This ellipse has center (0 , 0), with a vertical major axis of length 4 ( a = 2) and a horizontal minor axis of length 2 ( b = 1). Therefore, the possible values for s lie in [- 1 , 1], and we wish to maximize D , as a function of s , on this interval (which means we also need to test the endpoints s =- 1 , 1). We compute dD/ds =- 6 s- 2, which has a zero at s =- 1 / 3. Testing the points we need to test, we get D (- 1) = 4 D (- 1 / 3) = 16 / 3 D (1) = 0 so the points on the ellipse that are furthest away from the point (1 , 0) have x-coordinate equal to- 1 / 3. Plugging this into the equation of the ellipse, we find that there are two such points: 11 t 2 = 4- 4- 1 3 ¶ 2 = 4- 4 9 = 32 9 so t = ± 4 √ 2 3 , and the coordinates of the two points are (- 1 3 , 4 √ 2 3 ) and (- 1 3 ,- 4 √ 2 3 )....
View Full Document

{[ snackBarMessage ]}

Page8 / 16

8 5(10 points A 6 ft tall man walks away from a 15 ft lamp...

This preview shows document pages 8 - 13. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online