This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: c with respect to q = derivative of c ! c = 0 . 4 Â· 2 q + 4 = 0 . 8 q + 4 in which we plug in q = 2 and we get c (2) = 0 . 8 Â· 2 + 4 = 1 . 6 + 4 = 5 . 6 As for Î” c Î” q we donâ€™t use the derivative at all! ; what we need is Î” c = c (3)c (2) = (0 . 4 Â· 3 2 + 4 Â· 3 + 5)(0 . 4 Â· 2 2 + 5 Â· 2 + 5) = = 3 . 6 + 12 + 51 . 6105 = 4 Î” q = 32 = 1 and so Î” c Î” q = 4 1 = 4 Â± 36. Proof. (a) rate of change of y with respect to x = derivative of y = y = f ( x ) = (42 x ) =2 (b) relative rate of change= y y =2 42 x (c) plug in x = 3 in the formula in (a): y (3) =2 (itâ€™s a constant! so anything we plug in will give us2) (d) plug in x = 3 in the formula in (b): y y (3) =2 42 Â· 3 =2 46 =22 = 1 (e) just express the result in (d) in percentage form: 1 = 100 100 % = 100% Â±...
View
Full Document
 Spring '08
 Staff
 Math, Calculus, Derivative, Velocity

Click to edit the document details