How Does the Base of a Triangle Change as Its Altitude and Area Increase?

[hobo_ken]

Hi, I'm having trouble with this one problem. I've been struggling to do it for hours! Ok maybe mins, but I still can't get the right answer.

The altitude of a triangle is increasing at a rate of 3 centimeters/minute while the area of the triangle is increasing at a rate of 3.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9.5 centimeters and the area is 99 square centimeters?

The answer I got was 7/3.

 

[mathman]

A=area, b=base, h=altitude.

b=2A/h. Therefore b'=2(hA'-Ah')/h².

You do the arithmetic. (I did not get 7/3).

 

[M98Ranger]

5 but that doesn't seem right.

Hi there,

I understand your frustration with this problem. It can be tricky to figure out at first, but with some practice, you'll get the hang of it.

To solve this problem, we will use the formula for the area of a triangle: A = (1/2)bh, where A is the area, b is the base, and h is the altitude.

First, we need to find the rate at which the base is changing. We can do this by taking the derivative of the area formula with respect to time: dA/dt = (1/2)(db/dt)h + (1/2)b(dh/dt)

Since we know that the altitude is increasing at a rate of 3 cm/min, we can substitute that into the equation: dA/dt = (1/2)(db/dt)(9.5) + (1/2)b(3)

Next, we can plug in the values we know. We know that at this particular moment, the area is increasing at a rate of 3.5 square centimeters/minute, so we can substitute that in for dA/dt. We also know that the altitude is 9.5 centimeters, so we can substitute that in for h. Finally, we can solve for db/dt:

3.5 = (1/2)(db/dt)(9.5) + (1/2)b(3)
3.5 = 4.75(db/dt) + 1.5b
db/dt = (3.5 - 1.5b)/4.75

Now, we need to find the value of b when the area is 99 square centimeters. We can do this by plugging in 99 for A in the area formula and solving for b:

99 = (1/2)b(9.5)
99 = 4.75b
b = 20.84 cm

Finally, we can plug in this value for b into our derivative equation to find the rate at which the base is changing:

db/dt = (3.5 - 1.5(20.84))/4.75
db/dt = 1.26 cm/min

So, the base of the triangle is changing at a rate of 1.26 cm/min when the altitude is 9.5 cm and the area