Now

`f=k/sqrtC`

and substituting our given values, we have that

`920000=k/sqrt(3.5xx10^-12`

and this gives `k = 1.721`.

So

`f=1.721/sqrtC=1.721C^(-1"/"2)`

We need to find `(df)/(dt)`.

`(df)/(dt)=-1.721/2C^(-3"/"2)(dC)/(dt)`

We are told that `C = 3.5\ "pF"` and `(dC)/(dt) =0.3\ "pF/s"`.

So

`(df)/(dt)=-1.721/2(3.5xx10^-12)^(-3"/"2)xx` `(0.3xx10^-12)`

`=-39424.8\ "Hz s"^-1`

So the frequency of the electronic tuner is decreasing at the rate of 39.4 kHz s-1.

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