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 2.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8.000 centimeters and the area is 98.000 square centimeters?

Answer 1

Answer:

The base of the triangle is shrinking at a rate of $\frac{131}{32}$ centimeters per minute.

Step-by-step explanation:

The formula of the area of a triangle is given by the following expression:

$A = \frac{1}{2}\cdot b \cdot h$

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

The base of the triangle is:

$b = \frac{2\cdot A}{h}$

If $A = 98000\,cm^{2}$ and $h = 8000\,cm$, the base of the triangle is:

$b = \frac{2\cdot (98000\,cm^{2})}{8000\,cm}$

$b = 24.5\,cm$

The rate of change of the area of the triangle in time, measured in minutes, is obtained after differentiating by rule of chain and using deriving rules:

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

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

The rate of change of the base of the triangle is now cleared:

$2\cdot \frac{dA}{dt} = h\cdot \frac{db}{dt} + b\cdot \frac{dh}{dt}$

$h\cdot \frac{db}{dt} = 2\cdot \frac{dA}{dt}-b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2\cdot \frac{dA}{dt} - b \cdot \frac{dh}{dt} }{h}$

Given that $\frac{dA}{dt} = 2000\,\frac{cm^{2}}{min}$, $b = 24.5\,cm$, $\frac{dh}{dt} = 1500\,\frac{cm}{min}$ and $h = 8000\,cm$, the rate of change of the base of the triangle is:

$\frac{db}{dt} = \frac{2\cdot \left(2000\,\frac{cm^{2}}{min} \right)-(24.5\,cm)\cdot \left(1500\,\frac{cm}{min} \right)}{8000\,cm}$

$\frac{db}{dt} = -\frac{131}{32}\,\frac{cm}{min}$

The base of the triangle is shrinking at a rate of $\frac{131}{32}$ centimeters per minute.