Question

The altitude of a triangle is increasing at a rate of 1.5 1.5 centimeters/minute while the area of the triangle is increasing at a rate of 1.5 1.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 8 8 centimeters and the area is 88 88 square centimeters

Answer 1

Answer:

The base reduces at 3.75cm/min

Step-by-step explanation:

Given

Let

$h \to altitude$

$b \to base$

$A \to Area$

So:

$\frac{dh}{dt} = 1.5cm/min$

$\frac{dA}{dt} = 1.5^2cm/min$

The area of a triangle is:

$A = \frac{1}{2}bh$

Calculate b when $A =88cm^2; h =8cm$

$A = \frac{1}{2}bh$

$88=\frac{1}{2} * b * 8$

$88 =b * 4$

Solve for b

$b = 88/4$

$b = 22$

We have:

$A = \frac{1}{2}bh$

Differentiate with respect to time

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

Substitute the following values in the above equation

$\frac{dh}{dt} = 1.5cm/min$        $\frac{dA}{dt} = 1.5^2cm/min$      $b = 22$     $h = 8$

$1.5 = \frac{1}{2}(8 * \frac{db}{dt} + 22 * 1.5)$

Multiply both sides by 2

$3 = 8 * \frac{db}{dt} + 22 * 1.5$

$3 = 8 * \frac{db}{dt} + 33$

Collect like terms

$8 * \frac{db}{dt} = 3 -33$

$8 * \frac{db}{dt} = -30$

Divide both sides by 8

$\frac{db}{dt} = -\frac{30}{8}$

$\frac{db}{dt} = -3.75$