Question

A5.5 m ladder is leaning against a house when its base starts to slide away. By the time the base is 44 m from the house, the base is moving away at the rate of 6.6 m/s. a. What is the rate of change of the height of the top of the ladder? b. At whatrate is the area of the triangle formed by the ladder, wall, and ground changing then? P 6. At whatrate is the angle between the ladder and the ground changing then? a. The rate of change of the height of the top of the ladder in m/s. (Simplify your answer.)

Answer 1

I am unable to read the picture clearly. I am not able to read what is the rate of base is given? Is it 6.6 m/s???? Is it correct? or is it 6.8 or 8.8 ??? I have solved for 6.6 m/s. If it is different, please tell me the correct one, I will upload solution for that! Thank you!

ck Su s m Blka - Let at any time t, the base of ladder is am far from the house and top of ladder is ym high. then by Pythagoras theorem n ДАВе и (5.5)² = y2 + x² Differentiating with respect to time to

Canth咒(6) dt ⇒ o 金(9+ → 売りです。 - 29+ ". → htx = =

we have been given that base of ladder is morning at the rate of 6o 6 m/s lie. dx = 66 at

s dt = 6.10 when base (a) is 4.4m from hause, then height & y q top of ladder is (5:5)= y²+x2 (55)+= 9+(4.4)2 g² (55)² - (9.432 = y2 = 10.89 y = 303m. ②

Putting x=404m and y=3.3m en dy =-6.6*4.4 at 3} dy =-&&m's at Negative sige shows y is decreasing with time.

So the rate of change of height of top of the ladder is — 8.8 m/s

b) Area (A) 9 S ABC at some time t is A = 1 / 2 ay - from phythagoras theorem in BABC : (5.5)² = y²742 5 y? (5.5)?x2 3 y = IN(5.5)=x2 > y = N(5, 5²-24 F'y is length So, y is positive]

- - peeting this in A = my JA= {xw|5:5)?-n2 Different ating with respect to time t da conte 1 X N45.5 *=xt] se ze 124 5:57- xt yle] *4 = [(4 53- xeyli come + ne ate (59) -x58]

d A = at 11155=x23 Where to X, 4[15:55-x237 (55)x2] s et = { [30.15-x230content . (30.25–2354240-2297 A = *[V 50-25-+ y do at 15-x2 2010.0522 (Demoty ovome do 14 8391572 - ,25-42 - dar 1 2 dt 2 /30.25-02 J

we have, de pe = 6.6 for X= 4.4. So, dA = 6.6 | Ž 130. 25-(4.472 - (4.4)2 2530-25-(4-4) = 6.6 [ 1 x 3o 3 - (4.4) 27 2x303 aA at

9 A = 6.6 (3.3)–(4-4)41 at 1 2X3.3. that 6.5 [- 8.47] DA = -8.47 m25. at Megative sign shows area is decreasing with time. So, rate of change of Area is - 8.47 m/s.

%. Let o be - the ground time t. the angle between and ladder, at so sm IKN

coso=2 55 Differentiating with respect to t (esse) Estos at :.5-bino e se o e ass durch den for a=4.4m, we have dze - 6.6 cgiven) y = 3.3 (from ② by: Phythagoras theorem) 이 t

o sino 5.5 = sino = 3.33 So, from 6 : a menesis som det as a FREE X 66 — а 53 x 303 Suf 35 - Come

dt sale =-2 rad) sec. Negative sign show angle between ladder and ground is decreasing with time So, Rate of change of علويه منه -2rad/s