Two sides of a triangle are 12m and 15m in length and the angle between them is increasing at a rate of 2 deg/min. Find the rate at which the area of the triangle is increasing when the angle between their sides of fixed length is 60 deg?
My answer is 155m^2/min i just wanted to confirm if im doing it right.
A = (1/2)(12)(15)sin x , where x is the angle in radians
dA/dt = 90 cosx dx/dt
given: dx/dt = 2π/180 = π/90
so when x = π/3
dA/dt = 90(cos π/3)(π/90)
= (90)(1/2)(π/90) = π/2
How did you get 155 ? Remember, you have to use radians.
To find the rate at which the area of the triangle is increasing, we need to use the formula for the area of a triangle:
Area = 0.5 * base * height
In this case, the base of the triangle is the side length of 12m, and the height can be found using trigonometry. Let's call the angle between the sides of fixed length "θ". We can use the sine function to find the height:
sin(θ) = height / 15
Rearranging the equation, we get:
height = 15 * sin(θ)
Now, let's find the rate at which the area is increasing with respect to time. We can differentiate both sides of the area formula with respect to time:
d(Area) / dt = 0.5 * (d(base) / dt) * height + 0.5 * base * (d(height) / dt)
Since the base length is fixed at 12m, the rate of change of the base is zero, so d(base) / dt = 0. Therefore, the equation simplifies to:
d(Area) / dt = 0.5 * 0 * height + 0.5 * 12 * (d(height) / dt)
d(Area) / dt = 6 * (d(height) / dt)
Now, let's find the rate at which the height is changing with respect to time. Since the angle between the sides is increasing at a rate of 2 deg/min, we differentiate the height equation with respect to time:
d(height) / dt = 15 * cos(θ) * (d(θ) / dt)
We know that d(θ) / dt = 2 deg/min, and we need to convert it to radians per minute to match the units of cos(θ). Since there are 180 degrees in π radians, we have:
(2 deg/min) * (π / 180 deg) = (π / 90) rad/min
Substituting this value into the equation, we get:
d(height) / dt = 15 * cos(θ) * (π / 90) rad/min
Now, we can plug this expression for d(height) / dt back into the equation for d(Area) / dt:
d(Area) / dt = 6 * (15 * cos(θ) * (π / 90)) = 2π * cos(θ)
To find the rate at which the area of the triangle is increasing when θ = 60 degrees, we substitute θ = 60 degrees into the equation:
d(Area) / dt = 2π * cos(60 degrees)
The value of cos(60 degrees) is 0.5, so the rate at which the area of the triangle is increasing when θ = 60 degrees is:
d(Area) / dt = 2π * 0.5 = π square units per minute.
Therefore, the rate at which the area of the triangle is increasing when the angle between their sides of fixed length is 60 degrees is π square units per minute.
To find the rate at which the area of the triangle is increasing, we can use the formula for the area of a triangle:
Area = (1/2) * base * height
In this instance, the base and height of the triangle will change as the angle between the sides increases. Let's denote the angle between the sides as θ, the base as b, and the height as h.
We have a fixed side of length 12m (b) and a side of length 15m (h). The angle θ is increasing at a rate of 2 deg/min.
To find the height (h), we can use trigonometry. The side opposite to the angle θ is the height, so we can use the sine function:
sin θ = h /15
Rearranging that equation, we have:
h = 15 * sin θ
Now let's differentiate both sides of the equation with respect to time (t):
d(h)/dt = d(15 * sin θ) / dt
The derivative of sin θ is cos θ, and since θ is increasing at a rate of 2 deg/min, we have:
d(h)/dt = d(15 * sin θ) / dt = 15 * cos θ * d(θ)/dt
Substituting d(θ)/dt = 2 deg/min, we get:
d(h)/dt = 15 * cos(θ) * 2
Now we need to find the base (b). The side opposite to the angle θ is the hypotenuse of a right triangle, so we can use the cosine function:
cos θ = b /12
Rearranging that equation, we have:
b = 12 * cos θ
Again, let's differentiate both sides of the equation with respect to time (t):
d(b)/dt = d(12 * cos θ) / dt
The derivative of cos θ is -sin θ, so we have:
d(b)/dt = -12 * sin(θ) * d(θ)/dt
Substituting d(θ)/dt = 2 deg/min, we get:
d(b)/dt = -12 * sin(θ) * 2
Now that we have expressions for d(h)/dt and d(b)/dt, we can find the rate at which the area of the triangle is increasing.
The area of the triangle can be calculated using the formula:
Area = (1/2) * base * height
Substituting the expressions for base (b) and height (h), we get:
Area = (1/2) * (12 * cos θ) * (15 * sin θ) = 90 * cos θ * sin θ
To find the rate at which the area is increasing, we can differentiate both sides of the equation with respect to time (t):
d(Area)/dt = d(90 * cos θ * sin θ) / dt
Using the product rule and the chain rule, we have:
d(Area)/dt = 90 * ( cos θ * d(sin θ)/dt + sin θ * d(cos θ)/dt )
Substituting the values of d(sin θ)/dt and d(cos θ)/dt from our previous calculations, we get:
d(Area)/dt = 90 * ( cos θ * 15 * cos(θ) * 2 + sin θ * (-12 * sin(θ) * 2) )
Now, let's substitute the value of θ = 60 degrees into this equation to find the rate at which the area is increasing when the angle between the sides of fixed length is 60 degrees:
θ = 60 degrees
d(Area)/dt = 90 * ( cos(60) * 15 * cos(60) * 2 + sin(60) * (-12 * sin(60) * 2) )
Calculating this expression will give us the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is 60 degrees.