Question

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm^2/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 120 cm^2?

Answer 1

Answer:

The base is decreasing at 2 cm/min.

Step-by-step explanation:

The area (A) of a triangle is given by:

$A = \frac{1}{2}bh$   (1)

Where:

b: is the base

h: is the altitude = 10 cm

If we take the derivative of equation (1) as a function of time we have:

$\frac{dA}{dt} = \frac{1}{2}(\frac{db}{dt}h + \frac{dh}{dt}b)$

We can find the base by solving equation (1) for b:

$b = \frac{2A}{h} = \frac{2*120 cm^{2}}{10 cm} = 24 cm$

Now, having that dh/dt = 1 cm/min, dA/dt = 2 cm²/min we can find db/dt:

$2 cm^{2}/min = \frac{1}{2}(\frac{db}{dt}*10 cm + 1 cm/min*24 cm)$

$\frac{db}{dt} = \frac{2*2 cm^{2}/min - 1 cm/min*24 cm}{10 cm} = -2 cm/min$    

         

Therefore, the base is decreasing at 2 cm/min.

               

I hope it helps you!