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Question
If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius
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Solution
Let the radius of circle at any time t is r.
Then area of the circle at any time t is A = πr2.
∴ `"d"/"dt" "A" = "d"/"dt"(pi"r"^2)`
⇒ `"dA"/"dt" = 2pi"r" * "dr"/"dt"` ......(i)
Since the area of a circle increases at a uniform rate,
We have `"dA"/"dt"` = k, where k is a constant ......(ii)
From (i) and (ii), we get
`2pi"r" * "dr"/"dt"` = k
⇒ `"dr"/"dt" = "k"/(2pi"r") = "k"/(2pi) * (1/"r")`
⇒ `2pi "dr"/"dt" = "k"/"r"`
⇒ `("d"(2pi"r"))/"dt" = "k"/"r"`
⇒ `"dP"/"dt" = "k"/"r"`, where P = 2πr
⇒ `"dP"/"dt" oo 1/"r"`
Thus perimeter varies inversely as the radius.
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