Could you help me with one question in calculus?

Question

The altitude (i.e., height) of a triangle is increasing at a rate of 2 cm/minute while the area of the triangle is increasing at a rate of 1.5 square cm/minute. At what rate is the base of the triangle changing when the altitude is 8.5 centimeters and the area is 100 square centimeters?

The base is changing at ?

Answer 1

The formula for the area of a triangle in terms b and h: A = (1/2)bh.

Those measures are all changing with respect to time.

dh/dt = 2 cm/min

dA/dt = 3/2 cm²/min

Find db/dt when A = 100 cm² and h = 17/2 cm. We will have to determine the measure of the base at that point in time. We will see why later.

Differentiating the formula for the area of a triangle with respect to t gives:

dA/dt = 1/2 [b dh/dt + h db/dt] Now we see why we need to know the measure of b.

100 = (1/2)b(17/2)

b = 400/17

You can solve for db/dt in terms of dA/dt, dh/dt, b, and h, then substitute OR, substitute then solve for db/dt.

Let's do the latter.

3/2 cm²/min = 1/2 [ 400/17 cm 2cm/min + 17/2 cm db/dt cm/min]

3 cm²/min = 800/17 cm²/min + 17/2 cm db/dt [multiply both sides by 2]

102 cm²/min = 1600 cm²/min + 289 cm db/dt [multiply every term by 34]

289cm db/dt = [102 - 1600]cm²/min

db/dt = [102 - 1600]/289 cm/min (note that cm²/min divided by cm = cm/min}

db/dt ≈ -5.18 cm/min at the time when A = 100 and h = 8.5.

Probably you would not bother carrying the units through these calculations, but just apply the correct units at the end.

Answer 2

Let h be the height, b be the base and A be the area.

Given:

dh/dt = 2 cm/min

dA/dt = 1.5 cm²/min

A=bh/2

dA/dt = (bdh/dt+hdb/dt)/2

db/dt = (2dA/dt-bdh/dt)/h

when h = 8.5

100 = 8.5b/2

b = 200/8.5

db/dt = (2(1.5)-(200/8.5)(2))/8.5

I'll let you finish the calculation.