Question

1. A ladder 26 feet long is leaning against the wall of a building. The ladder is pulled away from
the base of the wall at the rate of 3.5 feet per second.
a) How fast is the top moving down the wall when the base of the ladder is 10 feet from the wall?
b) Find the rate at which the angle between the ladder and the wall of the building is changing when the
base of the ladder is 10 feet from the wall.

2) Find the both absolute extrema for

f(x) = 145x2 – 4| on (-2,2].

please show work

Answer 1

A Here x and y are functions to t. TAC is the ladder of length 26 feet. AB i i the wall Them Pythagorean Theuns x2 + y2 = (26)2 – 0 Defferenbating both sides of 0 wst it. 2x dx + 2y dy = 0 - per sec. ec E dt . ladder is being pulled away at the rate of 3.5 feet le de : 3.5 feet/sec. When ladded is 10 feet from the Wall lexz10. from 0, y = (26) ² - (10) ² 576 y = 24 g -24 y can't be -ve so y = 24.

| Substitutionen der in substituting doze - 35 , x= 10,4+24 in ), 2x10x 3.5+ 2x 24 x dy zo dt 70 + 48 dy zo dt -70 48 ve sign indicates that yis decreasing. When the ladder is lo fect from the wall, top of adder is falling at the rate of 35 feet/sec. Now let o be the angle b/w the ladder and the wall sin (e) - Prependicular Hypotenuse Differentiating on both sides,

From faguse a cose non Coso dt We know that de - 3.5 and Y: 24, when x = lofest :dt .de - 21. 3.5 1 3.5 = 35 t = I dt 1 24 to X24 ton z do dt 48 7radians per sec. for f(x) z 145x2_41 f'(x) = 1 e 190x} .. (45x2-4) 45x²_41 Phone non (458°-4) 90 (45x3 4x) 145x²_4 |

f'(2) - 0 + 45x3 - 4x = 0 x 145x2_4) 20 x = 0, 45x², 4 322 2 420, xz 2 x 2-2 315 are 3 butical points, - 315 Monotone Interwals are - 100(232) lz35*250), 10 << 315 35 (2) __ loe We will check behavious of f'(x) on these intervals. f'(2) < 0 When at (-os, -2 315) le fle) is Decreasing on this Interval when te lizano), f't) 70 f'(x) = 90. 45x lx-2 315 315 tve f(t) is Ancreasing on 420) le - 2 is a point of local Minima.

When x E 10, 2 f (x) to . Tie flx) is Decreasing on 10, 2 ) 315) lie x20 is a local Maxima ( f (x) 70 befor x=0$ f'(x)<o after azo when a E ( 20) , f (2) 70. le xz 2 315 is a local Minima Max & Min, we will find Now to find absolute f(2), fl-2) f(2)z 145X4-4) - Le 176 f1-2) : 14584-4) = 176 f(0) = 10-4)2 4 (local Max.) f ) - 458 let) -41 -0 (local Min). f(-2): 145x ( - 2 ) ² - 4 o (Local Mini) f(2) > f(4 lie 20= 2 IS ABSOLUTE MAXIMA $ R=4 are ABSOLUTE MINIMA, 355 315