The two equal sides of an isosceles triangle are each equal to 13 inches. If the third side is
increasing at 24 in/min, at what is the altitude drawn to this side changing when the altitude is
12 inches?
we have the same conclusion and yet the correct answer written on our book is -1/2 in/min?? I don't know how
If the 3rd side has length 2x, then the altitude h is
x^2 + h^2 = 13^2
x dx/dt + h dh/dt = 0
when h=12, you have a 5-12-13 right triangle, so
5 * 24/2 + 12 dh/dt = 0
dh/dt = -60/12 = -5 in/min
To find the rate at which the altitude drawn to the third side of the isosceles triangle is changing, we can use related rates.
Let's denote the length of the third side as "x" and the altitude as "h".
We are given that the two equal sides of the isosceles triangle are each equal to 13 inches, so the triangle has two sides of length 13 and one side of length x.
Using the Pythagorean theorem, we can find the length of the altitude (h) in terms of x:
h^2 = 13^2 - (x/2)^2
h^2 = 169 - (x^2/4)
h^2 = (676 - x^2) / 4
h = √((676 - x^2) / 4)
h = √(676 - x^2) / 2
Now, we can differentiate both sides of this equation with respect to time (t) using the chain rule:
d/dt(h) = d/dt(√(676 - x^2) / 2)
To find d/dt(h), we need to find d/dt(x), which represents the rate at which the length of the third side is changing.
Given that the third side is increasing at a rate of 24 in/min, we have:
d/dt(x) = 24 in/min
Now, let's substitute these values back into our equation:
d/dt(h) = d/dt(√(676 - x^2) / 2)
d/dt(h) = (d(676 - x^2) / dt * (1/2)) / (2 √(676 - x^2)) [Chain Rule]
d/dt(h) = (-2x * dx/dt) / (4 √(676 - x^2))
d/dt(h) = -x dx/dt / (2 √(676 - x^2))
We know that the altitude (h) is 12 inches, so h = 12. Substituting this into our equation:
12 = √(676 - x^2) / 2
24 = √(676 - x^2)
576 = 676 - x^2
x^2 = 676 - 576
x^2 = 100
x = 10 inches
Now we have x = 10 inches and dx/dt = 24 in/min. Substituting these values into our equation:
d/dt(h) = -10 * 24 / (2 √(676 - 10^2))
d/dt(h) = -240 / (2 √(676 - 100))
d/dt(h) = -240 / (2 √(576))
d/dt(h) = -240 / (2 * 24)
d/dt(h) = -240 / 48
d/dt(h) = -5 inches/min
Therefore, the altitude drawn to the third side of the isosceles triangle is changing at a rate of -5 inches/min when the altitude is 12 inches.
To find the rate at which the altitude is changing, we need to use the concept of related rates.
Let's denote the length of the third side of the isosceles triangle as "x" inches and the altitude drawn to this side as "h" inches. We are given that the two equal sides of the triangle are each 13 inches.
To begin, let's draw a diagram of the triangle and label the sides and altitude:
A
/\
/ \
h / \ h
/ \
/________\
x
We are given that the third side, "x", is increasing at a rate of 24 in/min. This means dx/dt = 24 in/min.
We want to find dh/dt, the rate at which the altitude is changing. We are given that when the altitude is 12 inches, which means h = 12 inches. We need to find dh/dt.
Now, let's use the Pythagorean Theorem to relate the side lengths and the altitude:
x^2 = (13/2)^2 + h^2
Differentiating both sides with respect to time (t) using implicit differentiation, we get:
2x(dx/dt) = 2h(dh/dt)
Rearranging the equation and plugging in the given values, we have:
2(13/2)(24) = 2(12)(dh/dt)
Simplifying the equation, we get:
(13)(24) = 12(dh/dt)
312 = 12(dh/dt)
Dividing both sides by 12, we find:
26 = dh/dt
So, when the altitude is 12 inches, the altitude is changing at a rate of 26 in/min.