The base of a triangle is decreasing at the rate of 2 cm/ min and the height is increasing at the rate of
1cm/ min. Find the rate at which the area is changing when the base is 8 cm and the height is 6 cm.
Is the area increasing or decreasing at that moment?
a = 1/2 bh
da/dt = 1/2 (h db/dt + b dh/dt)
Now just plug in your numbers.
Well, well, well, we've got ourselves a triangle with some changing dimensions! Let's see what we can do here.
We know that the rate at which the base is decreasing is 2 cm/min, and the rate at which the height is increasing is 1 cm/min. So, let's call the base of our triangle "b" and the height "h".
To find the rate at which the area is changing, we can use the formula for the area of a triangle: A = 1/2 * b * h. Don't worry, I won't make you do any math! We'll just differentiate this formula with respect to time.
dA/dt = 1/2 * (db/dt * h + b * dh/dt)
Plug in the given values: db/dt = -2 cm/min (since the base is decreasing), b = 8 cm, dh/dt = 1 cm/min (since the height is increasing), and h = 6 cm.
dA/dt = 1/2 * (-2 * 6 + 8 * 1)
dA/dt = -6 + 8
dA/dt = 2 cm^2/min
So, the rate at which the area is changing is 2 cm^2/min. Is the area increasing or decreasing at that moment?
Well, since dA/dt is positive (2 cm^2/min), the area is increasing at that moment. It's like the triangle is growing its area while its dimensions change.
Keep in mind that this analysis is just for that particular moment. The area could increase or decrease at different rates at different times. It's a wild world of triangles out there!
To find the rate at which the area is changing, we can use the formula for the area of a triangle, which is given by:
Area = (1/2) * base * height
Given that the base is decreasing at a rate of 2 cm/min and the height is increasing at a rate of 1 cm/min, we can differentiate the formula with respect to time to find the rate of change of the area:
dA/dt = (1/2) * (d(base)/dt) * height + (1/2) * base * (d(height)/dt)
Substituting the given values, we have:
dA/dt = (1/2) * (-2 cm/min) * 6 cm + (1/2) * 8 cm * (1 cm/min)
Simplifying this equation gives:
dA/dt = -6 cm^2/min + 4 cm^2/min
dA/dt = -2 cm^2/min
The rate at which the area is changing is -2 cm^2/min.
Since the rate is negative, the area is decreasing at that moment.
To find the rate at which the area is changing, we can use the formula for the area of a triangle: A = (1/2) * base * height.
Given that the base is decreasing at a rate of 2 cm/min and the height is increasing at a rate of 1 cm/min, we need to differentiate the formula with respect to time to find the rate of change of the area.
Differentiating A with respect to time (t), we get:
dA/dt = (1/2) * (db/dt) * height + (1/2) * base * (dh/dt)
Substituting the given rates of change:
dA/dt = (1/2) * (-2 cm/min) * 6 cm + (1/2) * 8 cm * (1 cm/min)
Simplifying:
dA/dt = -6 cm²/min + 4 cm²/min
dA/dt = -2 cm²/min
Therefore, the rate at which the area is changing when the base is 8 cm and the height is 6 cm is -2 cm²/min. The negative sign indicates that the area is decreasing at that moment.