Hi. Need help with a related rates question?

Question

The altitude of a triangle is increasing at a rate of 1cm/min while the area of the triangle is increasing at a rate of 2cm^2/min. At what rate is the base of the triangle changing when the altitude is 10cm and the area is 100cm^2?

Answer 1

dL/dt = +1 [cm/min]
dA/dt = +2 [cm^2/min]

dB/dt = ? @ L = 10 [cm] and A = 100 [cm^2]

A = (1/2)*B*L

dA/dt = (1/2)[B*(dL/dt) + L*(dB/dt)]
and
100 [cm^2] = (1/2)*B*10 [cm]
B = 20 [cm]

Plug in the values in the dA/dt equation:

2 = (1/2)[ (20 * 1) + (10 *(dB/dt)) ]
4 = 20 + 10(dB/dt)
dB/dt = -1.6 [cm/min]
The base is getting smaller as the altitude and area get larger.

Answer 2

The general equation for the area of a triangle is...
A = (1/2)bh
And we're given the area and altitude, which are not changing, so we can plug them in and solve for the base...
100cm = (1/2)(b)(10 cm)
100cm = 5b
b = 20

Now we can set up the related rate problem...we know that the area is increasing at a rate of 2cm^2/min so that's our dA/dt. The base and height we can plug in. And then solve for db/dt (the rate at which the base is changing)...

dA/dt = (1/2) (b dh/dt + h db/dt)
2 = (1/2) (20*1 + 10 db/dt)
4 = 20 + 10 db/dt
10 db/dt = -16
db/dt = -1.6 cm/min

So the negative sign implies its decreasing, and its doing so at a rate of 1.6 cm/min.

Answer 3

Original base=2*100/10=20
New base=2*102/11=204/11
So, decrease by 16/11.