the two bases of an isosceles trapezoid are 12 and 20 feet long, respectively. find the rate at which the area is changing when the equal sides are 5 feet long and are increasing at the rate of 2 feet per minute.
Make a sketch
I dropped perpendiculars at the ends of the 12 ft side to the 20 ft side and let the height be h ft
It is easy to see that
Area = 16h
also we know x^2 = 4^2 + h^2
h = √(x^2 - 16)
when x = 5, h = 3, and dx/dt = 2
Area = 16h = 16(x^2 - 16)^(1/2)
d(Area)/dt = 8(x^2 - 16)^(-1/2) (2x) dx/dt
which for the given data
= 8(25-16)^(-1/2)(10)(2)
= 8/√9 (20)
= 160/3
At that moment, the area is increasing at 160/3 ft^2 / min
check my arithmetic
Where did 16 come from?
To find the rate at which the area of the isosceles trapezoid is changing, we can use the formula for the area of a trapezoid:
Area = (base1 + base2) / 2 * height
Let's denote the length of the equal sides as x feet and the height as h feet. We are given that the length of the bases are 12 feet and 20 feet, respectively.
According to the problem, x (the length of the equal sides) is increasing at a rate of 2 feet per minute. Therefore, dx/dt (the rate of change of x) is 2 ft/min.
To find dh/dt (the rate at which the height is changing), we can use similar triangles. The height h is related to the length of the equal sides x and the bases of the trapezoid. We can set up the following proportion:
h / (20 - 12) = x / (20 - x)
Simplifying the equation gives:
h = 8x / (20 - x)
Now, differentiate both sides of this equation with respect to time t:
dh/dt = (8 * dx/dt * (20 - x) - 8x * d(20 - x)/dt) / (20 - x)^2
Since we are asked to find the rate at which the area is changing, we differentiate the area formula with respect to time t:
dA/dt = (1/2) * [(d(base1)/dt + d(base2)/dt) * h + (base1 + base2) * dh/dt]
Since the bases of the trapezoid are not changing with time, d(base1)/dt and d(base2)/dt are both 0. Substituting the values for the bases and the equation we derived for dh/dt:
dA/dt = (1/2) * [(0 + 0) * h + (12 + 20) * ((8 * dx/dt * (20 - x) - 8x * d(20 - x)/dt) / (20 - x)^2)]
Simplifying further:
dA/dt = 4 * ((dx/dt * (20 - x) - x * d(20 - x)/dt) / (20 - x)^2)
Now we can substitute the given values: dx/dt = 2 ft/min, x = 5 ft, d(20 - x)/dt = -2 ft/min.
dA/dt = 4 * ((2 * (20 - 5) - 5 * (-2)) / (20 - 5)^2)
dA/dt = 4 * ((30 + 10) / 15^2)
dA/dt = 40 / 225
Simplifying the fraction gives:
dA/dt = 8 / 45
Therefore, the rate at which the area of the isosceles trapezoid is changing when the equal sides are 5 feet long and increasing at a rate of 2 feet per minute is 8/45 square feet per minute.
To find the rate at which the area of the isosceles trapezoid is changing, we can first start by drawing a diagram to better understand the problem. Let's label the longer base of the trapezoid as 20 feet, and the shorter base as 12 feet.
Next, let's denote the equal sides of the trapezoid as "x" feet and the rate at which they are increasing as "dx/dt," which is given as 2 feet per minute. We are interested in finding the rate at which the area of the trapezoid is changing, which can be expressed as "dA/dt."
The formula for the area of a trapezoid is:
A = (1/2) * (b1 + b2) * h
Where,
A = Area of the trapezoid
b1 = Length of the longer base
b2 = Length of the shorter base
h = Height of the trapezoid
In our case, b1 = 20 feet and b2 = 12 feet. The height of the trapezoid can be found using the Pythagorean theorem since the trapezoid is isosceles. By drawing an altitude from one vertex to the base, we get a right triangle with the hypotenuse as the equal side (x), one leg as the height (h), and the other leg as half the difference between the longer and shorter bases.
Using the Pythagorean theorem, we have:
x^2 = h^2 + (1/2 * (b1 - b2))^2
Now, let's differentiate both sides of this equation with respect to time (t) to find the rate of change of x with respect to t (dx/dt) since we know the rate at which x is changing:
2x * (dx/dt) = 2h * (dh/dt) + (1/2 * (b1 - b2))^2 * (dh/dt)
Now, let's solve this equation for dh/dt, which represents the rate at which the height of the trapezoid is changing.
dh/dt = (2x * (dx/dt)) / (2h + (1/2 * (b1 - b2))^2)
Since we know the value of x and dx/dt (5 and 2 feet per minute, respectively), we can substitute these values into the equation.
dh/dt = (2 * 5) / (2h + (1/2 * (20 - 12))^2)
Now, we need to find the value of h. To do that, we can use the Pythagorean theorem again, considering the initial values of x and the bases.
5^2 = h^2 + (1/2 * (20 - 12))^2
Solving this equation will give us the value of h.
Once we have the values of h and dx/dt, we can substitute them into the equation we found earlier to calculate dh/dt.
Finally, to find the rate at which the area of the trapezoid is changing (dA/dt), we can substitute the values of h and dh/dt into the formula for the area of a trapezoid:
A = (1/2) * (b1 + b2) * h
We can differentiate both sides of this equation with respect to time (t) to find the rate of change of A with respect to t.
dA/dt = (1/2) * (b1 + b2) * (dh/dt) + (1/2) * (db1/dt)
Since we know the values of b1, b2, dh/dt, and db1/dt (from the information given in the problem), we can substitute these values into the equation to find the rate at which the area of the trapezoid is changing (dA/dt).