Two sides of a triangle have lengths 15 m and 18 m. The angle between them is decreasing at a rate of 0.08 rad/s. Find the rate at which the area of the triangle is changing when the angle between the sides of fixed length is 120°.
Using the formula:
area = (1/2) ab sinØ , where a and b are the sides and Ø is the contained angle
A = (1/2)(15(18)sinØ
A = 135 sinØ
dA/dt = 135 cosØ dØ/dt
given: dØ/dt = -.08 , and cos 120° = -.5
dA /dt = 135(-.5)(-.08) = 5.4 m^2
when the angle is 120°, the area is increasing at 5.4 m^2/s
0.9
To find the rate at which the area of the triangle is changing, we need to use the formula for the area of a triangle:
Area = (1/2) * base * height
Let's denote the side of length 15 m as 'a' and the side of length 18 m as 'b'. The angle between these sides is denoted as θ. The area of the triangle can also be expressed using the trigonometric function sine:
Area = (1/2) * a * b * sin(θ)
Now, let's take the derivative of the area with respect to time (t) using the chain rule:
d(Area)/dt = (1/2) * [(da/dt) * b * sin(θ) + a * (db/dt) * sin(θ) + a * b * cos(θ) * (dθ/dt)]
The rates of change we know are:
- (da/dt) = 0 (since 'a' is fixed)
- (db/dt) = 0 (since 'b' is fixed)
- (dθ/dt) = -0.08 rad/s (given in the problem)
Substituting the known values into the derivative equation, we can solve for d(Area)/dt:
d(Area)/dt = (1/2) * [0 * 18 * sin(120°) + 15 * 0 * sin(120°) + 15 * 18 * cos(120°) * (-0.08 rad/s)]
To simplify further, we'll use the fact that sin(120°) = sqrt(3)/2 and cos(120°) = -1/2:
d(Area)/dt = (1/2) * [0 + 0 + 15 * 18 * (-1/2) * (-0.08 rad/s)]
d(Area)/dt = (1/2) * [15 * 18 * 0.04 rad/s]
d(Area)/dt = (1/2) * [10.8 * 0.04 rad/s]
d(Area)/dt = (1/2) * 0.432 rad/s
d(Area)/dt = 0.216 rad/s
Therefore, the rate at which the area of the triangle is changing when the angle between the sides of fixed length is 120° is 0.216 square radians per second.