Answer:
The base reduces at 3.75cm/min
Step-by-step explanation:
Given
Let
[tex]h \to altitude[/tex]
[tex]b \to base[/tex]
[tex]A \to Area[/tex]
So:
[tex]\frac{dh}{dt} = 1.5cm/min[/tex]
[tex]\frac{dA}{dt} = 1.5^2cm/min[/tex]
The area of a triangle is:
[tex]A = \frac{1}{2}bh[/tex]
Calculate b when [tex]A =88cm^2; h =8cm[/tex]
[tex]A = \frac{1}{2}bh[/tex]
[tex]88=\frac{1}{2} * b * 8[/tex]
[tex]88 =b * 4[/tex]
Solve for b
[tex]b = 88/4[/tex]
[tex]b = 22[/tex]
We have:
[tex]A = \frac{1}{2}bh[/tex]
Differentiate with respect to time
[tex]\frac{dA}{dt} =\frac{1}{2}(h\frac{db}{dt} + b\frac{dh}{dt})[/tex]
Substitute the following values in the above equation
[tex]\frac{dh}{dt} = 1.5cm/min[/tex] [tex]\frac{dA}{dt} = 1.5^2cm/min[/tex] [tex]b = 22[/tex] [tex]h = 8[/tex]
[tex]1.5 = \frac{1}{2}(8 * \frac{db}{dt} + 22 * 1.5)[/tex]
Multiply both sides by 2
[tex]3 = 8 * \frac{db}{dt} + 22 * 1.5[/tex]
[tex]3 = 8 * \frac{db}{dt} + 33[/tex]
Collect like terms
[tex]8 * \frac{db}{dt} = 3 -33[/tex]
[tex]8 * \frac{db}{dt} = -30[/tex]
Divide both sides by 8
[tex]\frac{db}{dt} = -\frac{30}{8}[/tex]
[tex]\frac{db}{dt} = -3.75[/tex]
