a^2=w^2+g^2
take the derivative
0=2w dw/dt + 2g dg/dt
or w dw/dt=-gdg/dt
given dg/dt=4ft/sec
20^2=w^2+11^2
w=sqrt(400/121)=20/11
when base g is 11 ft.
dw/dt=-11/(20/11) (4)
Area=2wg
dA/dt=2g dw/dt + 2w dg/dt
dArea/dt=22*dw/dt + 2*20/11*5
put in dw/dt from above, and you have it.
A ladder a = 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of b = 4 feet/second. Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changed when the base of the ladder is 11 feet from the wall.
Is this 33.41 ft^2/sec?
4 answers
let the height be h
then h^2 + b^2 = 20^2
h = (400-b^2)^(1/2)
Area = (1/2)(bh)
= (1/2)(b)(400 - b^2)^(1/2)
d(Area)/dt = (b/2)(1/2)(400-b^2)^(-1/2) (-2b db/dt) + (1/2)(db/dt)(400 - b^2)^(1/2)
= -b^2 (db/dt)/√(400-b^2) + (1/2)(db/dt)√(400-b^2)
when b = 11 and db/dt = 4
d(Area)/dt = -121(4)/√279 + (1/2)(4)√279
= appr 4.43
check my arithmetic, my method is correct
then h^2 + b^2 = 20^2
h = (400-b^2)^(1/2)
Area = (1/2)(bh)
= (1/2)(b)(400 - b^2)^(1/2)
d(Area)/dt = (b/2)(1/2)(400-b^2)^(-1/2) (-2b db/dt) + (1/2)(db/dt)(400 - b^2)^(1/2)
= -b^2 (db/dt)/√(400-b^2) + (1/2)(db/dt)√(400-b^2)
when b = 11 and db/dt = 4
d(Area)/dt = -121(4)/√279 + (1/2)(4)√279
= appr 4.43
check my arithmetic, my method is correct
x^2+y^2 = 20^2
a = xy/2 = x√(400-x^2)/2
da/dt = (200-x^2)/√(400-x^2) dx/dt
at x=11,
da/dt = (200-121)/√279 * 4 = 18.92 ft^2/s
It would have been polite to show your work, so we could find the mistake.
Or, of course, check my work to be sure I haven't goofed!
a = xy/2 = x√(400-x^2)/2
da/dt = (200-x^2)/√(400-x^2) dx/dt
at x=11,
da/dt = (200-121)/√279 * 4 = 18.92 ft^2/s
It would have been polite to show your work, so we could find the mistake.
Or, of course, check my work to be sure I haven't goofed!
there you have three answers. Take your pick...