At what rate is the base of the triangle changing?

Question

The altitude of a triangle is increasing at a rate of 1.500 centimeters/minute while the area of the triangle is increasing at a rate of 3.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9.500 centimeters and the area is 84.000 square centimetres?

Answer 1

Any time you're asked about rates, you need to figure out a formula that relates the variables involved first. In this case, three variables are involved:

 

A = area of the triangle

h = altitude

b = base

 

The formula that relates the three is the area of a triangle...

 

A = (1/2) b*h

 

Now let's look at the numbers they give us. Note that anything listed as a "rate" is actually a derivative value. So we have...

 

dh (rate of change of altitude) = 1.5 cm/min

dA (rate of change of area) = 3 cm²/min

h = 9.5 cm

A = 84 cm²

 

Now let's take a look at our derivative. Note that both b and h are changing in this situation, so we have to treat them both as variables. This means we have to use the product rule, inserting the derivatives where applicable.

 

dA = (1/2)*db*h + (1/2)b * dh

 

Now let's put in what we know...

 

3 = (1/2) * db * 9.5 + (1/2) b * 1.5

 

They ask for the rate at which the base is changing, which is db. However, looking at this equation, I can't solve for db because I don't know b!! I need b!! Oh no.

 

WELL...we can find b! ...at least for the instant they're asking about. Using the area and altitude given...

 

84 = (1/2) * b * 9.5

 

84 = 4.75 b

 

b = 17.68421

 

Now, looking back at our derivative...

 

3 = (1/2) * db * 9.5 + (1/2) * 17.68421 * 1.5

 

3 = 4.75 db + 13.26316

 

db = -2.16066

 

Now, looking at this, it may seem odd that we're getting a negative number. The base is actually decreasing?? But both the height and the area are increasing...how can the base be decreasing? Well, let's look at a couple points in the area of these numbers.

 

We know that 84 = (1/2) * 17.68421 * 9.5

 

What happens to b if I increase the altitude a bit to 11? They tell us how the area changes as well -- it goes up 3 at the instance h = 9.5, so the area should be around 87.

 

87 = (1/2) * b * 11

87 = 5.5 b

b = 15.81818

 

So in the general vicinity, the base IS getting lower as the area and height increase at approximately those rates. In other words, the change in height is actually more than it needs to be to increase the area that much, so the base's length has to compensate for this by decreasing slightly.

 

 

Answer 2

A = area, b = base, h = altitude,  t = time

A = 1/2(bh)

dA/dt = 1/2(h×db/dt+b×dh/dt)

dA/dt =3cm²/min

dh/dt = 3/2cm/min

h=9.5cm

A=84cm²

These last two give that, with A = 1/2(bh) that

b = 2A/h = 168/9.5

db/dt = (2×dA/dt - b×dh/dt)/h = (2×3-(168/9.5)×3/2)/9.5 = -2.16066482cm/min