An isosceles triangle is inscribed inside a circle of radius 12 metres. Its height
is increasing at a rate of 2 m/sec. At the instant when the height of the
triangle is 18 m, at what rate is:
a) its area changing?
b) its perimeter changing?
help me plz~ Thank you so much!
To determine the rates of change, we need to find the relationship between the different variables involved in the problem. Let's start by drawing a diagram and labeling the triangle's dimensions:
Let ABC be the isosceles triangle, with AB = AC as the base and BC as the height. Let O be the center of the circle, and let M be the midpoint of AB.
First, we need to find the length of the base AB. Since the triangle is isosceles and AC is the radius of the circle, AC = 12 meters. The base AB is twice the radius, so AB = 2 * AC = 24 meters.
a) To find the rate at which the area of the triangle is changing, we need to differentiate the area formula with respect to time and then substitute the given values.
The area A of an isosceles triangle can be calculated using the formula A = (1/2) * base * height. In this case, the base is AB = 24 meters, and the height BC is changing with time t. So, we can write A = (1/2) * 24 * BC.
To differentiate A with respect to time t, we will apply the chain rule, considering BC as a function of t:
dA/dt = (1/2) * 24 * d(BC)/dt.
Given that d(BC)/dt = 2 m/sec, we can substitute this value into the equation:
dA/dt = (1/2) * 24 * 2 = 24 m^2/sec.
Therefore, the area of the triangle is changing at a rate of 24 m^2/sec when the height is 18 m.
b) To find the rate at which the perimeter of the triangle is changing, we need to differentiate the perimeter formula with respect to time and then substitute the given values.
The perimeter P of an isosceles triangle can be calculated using the formula P = 2 * base + height. In this case, the base AB is 24 meters, and the height BC is changing with time t. So, we can write P = 2 * 24 + BC.
To differentiate P with respect to time t, we will apply the chain rule, considering BC as a function of t:
dP/dt = d(BC)/dt.
Given that d(BC)/dt = 2 m/sec, we can substitute this value into the equation:
dP/dt = 2 m/sec.
Therefore, the perimeter of the triangle is changing at a constant rate of 2 m/sec, regardless of the height value.