Related Rates Problem
An isosceles triangle with a base of 20root3 cm long. If the length of the leg decreases at rate 3 cm/h, find the rate of decreasing of the area of the triangle in the instant at which the triange becomes equilateral.
To solve this related rates problem, we need to find the rate of change of the area of the triangle with respect to time.
First, let's assume the base of the isosceles triangle is represented by the variable 'b' and the length of each leg is represented by the variable 'l', both measured in centimeters. We're given that the base 'b' has a constant length of 20√3 cm.
The area of an isosceles triangle can be calculated using the formula A = (1/2) * b * l, where A represents the area and b and l represent the respective base and leg lengths.
Since we're given that the length of the leg is changing over time, we can express the change in area in terms of the change in leg length. We'll use the derivative notation, where dA/dt represents the rate of change of the area with respect to time (t), and dl/dt represents the rate of change of the leg length with respect to time.
Now, in order to find the exact rate of decrease of the area, we need to determine the relationship between the base 'b' and the length of each leg 'l' at the point when the triangle becomes equilateral.
In an equilateral triangle, all sides are of equal length. Therefore, the base 'b' of the isosceles triangle will also be equal to the length of each leg 'l' at that particular instant.
Let's denote the length of each side as 's'.
Thus, at the instant the triangle becomes equilateral, we have b = l = s.
Now we can differentiate both sides of the equation with respect to time:
db/dt = dl/dt = ds/dt
Since the base and each leg length are equal, we can say db/dt = dl/dt. Therefore, the rate at which the base length changes is equal to the rate at which each leg length changes.
Now, let's differentiate the area formula A = (1/2) * b * l with respect to time:
dA/dt = (1/2) * (db/dt * l + b * dl/dt)
Since the base length and leg length are equal, we can substitute b for l:
dA/dt = (1/2) * (dl/dt * l + l * dl/dt)
Simplifying, we get:
dA/dt = l * dl/dt
Substituting the values we know:
dA/dt = (20√3 * 3) cm²/h
Calculating:
dA/dt = 60√3 cm²/h
Therefore, the rate at which the area of the triangle is decreasing at the instant when the triangle becomes equilateral is 60√3 cm²/h.