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?
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.
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
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!
there you have three answers. Take your pick...
To find the rate at which the area of the triangle is changing, we can use the formula for the area of a triangle:
Area = (1/2) * base * height
In this case, the base of the triangle is given as the distance between the wall and the ladder's base, which is changing at a rate of 4 feet per second. Let's call this rate of change "db/dt". So, db/dt = 4 ft/s.
The height of the triangle is the length of the ladder, which remains constant at 20 feet. So, h = 20 ft.
Now, we need to find the rate at which the area of the triangle is changing, which is given by dA/dt. To find dA/dt, we need to differentiate the area formula with respect to time:
dA/dt = (1/2) * (db/dt) * h
Substituting the given values, we have:
dA/dt = (1/2) * (4 ft/s) * (20 ft)
= 40 ft^2/s
So, the rate at which the area of the triangle is changing when the base of the ladder is 11 feet from the wall is 40 ft^2/s, not 33.41 ft^2/s.
It seems that there may have been an error in the calculation or in the given values. Please double-check your work or the problem statement to make sure that all the given values and calculations are accurate.