Answer :
The base is decreasing at 1.16 centimeters/minute, when rate is the base of the triangle changing.
Let us assume,
The altitude = h
The base = b
The Area = A
Area = [tex]\frac{1}{2} b*h[/tex]
[tex]\frac{dh}{dt}[/tex] is the rate at which the altitude is increasing
[tex]\frac{dA}{dt}[/tex] is the rate at which the Area is increasing
[tex]\frac{db}{dt}[/tex] is the changing rate of the base
We are to determine [tex]\frac{db}{dt}[/tex] when h = 11cm and A = 98 square centimeters
First we will determine at that time,
A = [tex]\frac{1}{2} b*h[/tex]
b = [tex]\frac{2A}{h}[/tex]
b = [tex]\frac{2*98}{11}[/tex]
b = 196/11
b = 17.8cm
Next, we will differentiate the formula A = [tex]\frac{1}{2} b*h[/tex] with respect to t.
That is,
[tex]\frac{d}{dt} (A)[/tex] = [tex]\frac{d}{dt} (\frac{1}{2} b*h)[/tex]
[tex]\frac{dA}{dt}[/tex] = [tex]\frac{1}{2} \frac{db}{dt} *h+\frac{1}{2} \frac{dh}{dt} *b[/tex]
Now, we can find [tex]\frac{db}{dt}[/tex]
[tex]\frac{dA}{dt}[/tex] = 2.5 square centimeters/minute
[tex]\frac{dh}{dt}[/tex] = 1 centimeters/minute
b = 17.8cm
h = 11 centimeters
Hence,
2.5 = [tex]\frac{1}{2} \frac{db}{dt} (11)+\frac{1}{2} (1)(17.8)[/tex]
2.5 = 5.5 [tex]\frac{db}{dt}[/tex] + 8.9
2.5 - 8.9 = 5.5 [tex]\frac{db}{dt}[/tex]
-6.4 = 5.5 [tex]\frac{db}{dt}[/tex]
[tex]\frac{db}{dt}[/tex] = -6.4/5.5
[tex]\frac{db}{dt}[/tex] = -1.16 [tex]\frac{cm}{min}[/tex]
This means the base is decreasing at 1.16 [tex]\frac{cm}{min}[/tex].
Therefore,
The base is decreasing at 1.16 centimeters/minute, when rate is the base of the triangle changing.
To learn more about information visit Triangle area :
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