Answer:
The base of the triangle decreases at a rate of 2.262 centimeters per minute.
Step-by-step explanation:
From Geometry we understand that area of triangle is determined by the following expression:
[tex]A = \frac{1}{2}\cdot b\cdot h[/tex] (Eq. 1)
Where:
[tex]A[/tex] - Area of the triangle, measured in square centimeters.
[tex]b[/tex] - Base of the triangle, measured in centimeters.
[tex]h[/tex] - Height of the triangle, measured in centimeters.
By Differential Calculus we deduce an expression for the rate of change of the area in time:
[tex]\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}[/tex] (Eq. 2)
Where:
[tex]\frac{dA}{dt}[/tex] - Rate of change of area in time, measured in square centimeters per minute.
[tex]\frac{db}{dt}[/tex] - Rate of change of base in time, measured in centimeters per minute.
[tex]\frac{dh}{dt}[/tex] - Rate of change of height in time, measured in centimeters per minute.
Now we clear the rate of change of base in time within (Eq, 2):
[tex]\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}[/tex]
[tex]\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt}[/tex] (Eq. 3)
The base of the triangle can be found clearing respective variable within (Eq. 1):
[tex]b = \frac{2\cdot A}{h}[/tex]
If we know that [tex]A = 130\,cm^{2}[/tex], [tex]h = 15\,cm[/tex], [tex]\frac{dh}{dt} = 2.5\,\frac{cm}{min}[/tex] and [tex]\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}[/tex], the rate of change of the base of the triangle in time is:
[tex]b = \frac{2\cdot (130\,cm^{2})}{15\,cm}[/tex]
[tex]b = 17.333\,cm[/tex]
[tex]\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)[/tex]
[tex]\frac{db}{dt} = -2.262\,\frac{cm}{min}[/tex]
The base of the triangle decreases at a rate of 2.262 centimeters per minute.
