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Continuous Functions - Setting up word problems
Each side of a square is expanding at 5 cm/sec. What is the rate of change when the length of the sides are 10 cm.
[tex] A = ab [/tex]
[tex] a = 5t, b = 5t [/tex]
and the area is described as [tex] A = ab [/tex]
so the rate of change in the area should be
[tex] \frac {dA}{dt} = a \frac {db}{dt} + b \frac {da}{dt} [/tex]
then [tex] \frac {da}{dt} = 5 [/tex] and [tex] \frac {db}{dt} = 5 [/tex]
so [tex] \frac {dA}{dt} = 5a + 5b [/tex]
then at [tex] t = 2 [/tex]
[tex] \frac {dA}{dt} = 5 *10 + 5 *10 = 50 + 50 = 100 cm^{2}/sec [/tex]
So the area of the square is changing at a rate of 100 sq cm/sec when the sides are at 10 cm
is this correct?
Homework Statement
Each side of a square is expanding at 5 cm/sec. What is the rate of change when the length of the sides are 10 cm.
Homework Equations
[tex] A = ab [/tex]
The Attempt at a Solution
[tex] a = 5t, b = 5t [/tex]
and the area is described as [tex] A = ab [/tex]
so the rate of change in the area should be
[tex] \frac {dA}{dt} = a \frac {db}{dt} + b \frac {da}{dt} [/tex]
then [tex] \frac {da}{dt} = 5 [/tex] and [tex] \frac {db}{dt} = 5 [/tex]
so [tex] \frac {dA}{dt} = 5a + 5b [/tex]
then at [tex] t = 2 [/tex]
[tex] \frac {dA}{dt} = 5 *10 + 5 *10 = 50 + 50 = 100 cm^{2}/sec [/tex]
So the area of the square is changing at a rate of 100 sq cm/sec when the sides are at 10 cm
is this correct?
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