Fazle R. A 13-ft ladder is leaning against a house. the distance from the bottom of the ladder to the house is 7-ft less than the distance from the top of the ladder to the ground .how far is the bottom of the ladder from the house. More
1 Expert Answer
(x-7)^2 +x^2 = 13^2 x^2 -14x +49 +x^2 = 169 2x^2 -14x -120 = 0 x^2 -7x- 60 =0
(x-12)(x+5) = 0
x =12 feet
x-7= 5 feet from the bottom of the ladder to the house
the ladder forms a 5-12-13 right triangle
5^2 +12^2 = 13^2
25+144= 169
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Solution : `x^(2)+y^(2)=169` <br> `rArr " " 2x(dx)/(dt)+2y(dy)/(dt)=0` <br> `rArr "" ((dy)/(dt))=((dx)/(dt))(x-x)/(y)=5 xx(-12)/(5)=-12 ft//s` <br> `A=(1)/(2) xy` <br> `rArr " " (dA)/(dt)=(y)/(2)(dx)/(dt)+(x)/(2)(dy)/(dt)=(5)/(2) xx 5+ (12)/(2) xx-12` <br> `=(25-144)/(2)=-59.5 ft^(2)//s` <br> ` y=x tan theta` <br> `rArr " " (dy)/(dt)=(tan theta)(dx)/(dt)+x sec^(2) theta (d theta)/(dt)` <br> `rArr " " -12 = 5 xx (5)/(12) + 12 xx (169)/(12^(2)) (d theta)/(dt)` <br> `rArr " " (169)/(12) (d theta)/(dt) =-12 -(25)/(12)` <br> ` rArr " " (d theta)/(dt)=(144)/(169)-(25)/(169)=-1 "rad"//s`
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a. How fast is the top of the ladder sliding down the wall then?
b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?
c. At what rate is the angle between the ladder and the ground changing then?
Answer by addingup(3677) (Show Source):
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Let x be the distance between the base of the wall and the base of the ladder. Let y be the height the top of the ladder is off the ground. The given is
that dx/dt = +5 at the moment x = 12. The length of the ladder,
13 feet, is a constant, since its length doesn't change.
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(a) How fast is the top of the ladder sliding down the wall then?
x^2+y^2 = 13^2
Solve for dy/dt in terms of x,y, and dx/dt:
use your calculator and you get y = 5. Therefore:
dy/dx = -x*dx/y*dt = (-12/5)*5 = -12
Note that the minus sign indicates the ladder is sliding down. So, the top of the ladder is sliding down at 12 ft/second
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b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing.
Here the unknown is the rate of change of the area, dA/dt
dA/dt = 1/2 (x*dy/dt+y*dx/dt)
Substitute with x = 12 and dx/dt = 5 and y=5 dy/dt = -12:
dA/dt = 1//2(12*(-12)5*5)= -144+25/2 = -119/2 = 59.5 ft^2/sec
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c. At what rate is the angle between the ladder and the ground changing then?
ormula that relates the angle θ with the sides x and y of a right triangle:
Answer
1 rad/sec