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5 September, 14:56

A) A cuboid with a square x cm and height 2xcm². Given total surface area of the cuboid is 129.6cm² and x increased at 0.01cms-¹. Find the change of the volume the cuboid

b) Given y=2x²+3x, use differentiation to find small change in y when x increased from 4 to 4.02.

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  1. 5 September, 15:27
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    Answer: (given assumed typo corrections)

    (V ∘ X) ' (t) = 0.06 (0.01t+3.6) ^2 cm^3/sec.

    The rate of change of the volume of the cuboid in change of volume per change in seconds, after t seconds. Not a constant, for good reason.

    Part B) y' (x+Δx/2) * Δx gives exactly the same as y (x+Δx) - y (x), 0.3808, since y is quadratic in x so y' is linear in x.

    Step-by-step explanation:

    This problem has typos. Assuming:

    Cuboid has square [base with side] X cm and height 2X cm [not cm^2]. Total surface area of cuboid is 129.6 cm^2, and X [is] increas[ing] at rate 0.01 cm/sec.

    129.6 cm^2 = 2 (base cm^2) + 4 (side cm^2)

    = 2 (X cm) ^2 + 4 (X cm) (2X cm)

    = (2X^2 + 8X^2) cm^2

    = 10X^2 cm^2

    X^2 cm^2 = 129.6/10 = 12.96 cm^2

    X cm = √12.96 cm = 3.6 cm

    so X (t) = (0.01cm/sec) (t sec) + 3.6 cm, or, omitting units,

    X (t) = 0.01t + 3.6

    = the length parameter after t seconds, in cm.

    V (X) = 2X^3 cm^3

    = the volume when the length parameter is X.

    dV (X (t)) / dt = (dV (X) / dX) (X (t)) * dX (t) / dt

    that is, (V ∘ X) ' (t) = V' (X (t)) * X' (t) chain rule

    V' (X) = 6X^2 cm^3/cm

    = the rate of change of volume per change in length parameter when the length parameter is X, units cm^3/cm. Not a constant (why?).

    X' (t) = 0.01 cm/sec

    = the rate of change of length parameter per change in time parameter, after t seconds, units cm/sec.

    V (X (t)) = (V ∘ X) (t) = 2 (0.01t+3.6) ^3 cm^3

    = the volume after t seconds, in cm^3

    V' (X (t)) = 6 (0.01t+3.6) ^2 cm^2

    = the rate of change of volume per change in length parameter, after t seconds, in units cm^3/cm.

    (V ∘ X) ' (t) = (6 (0.01t+3.6) ^2 cm^3/cm) (0.01 cm/sec) = 0.06 (0.01t+3.6) ^2 cm^3/sec

    = the rate of change of the volume per change in time, in cm^3/sec, after t seconds.

    Problem to ponder: why is (V ∘ X) ' (t) not a constant? Does the change in volume of a cube per change in side length depend on the side length?

    Question part b)

    Given y=2x²+3x, use differentiation to find small change in y when x increased from 4 to 4.02.

    This is a little ambiguous, but "use differentiation" suggests that we want y' (4.02) yunit per xunit, rather than Δy/Δx = (y (4.02) - y (4)) / (0.02).

    Neither of those make much sense, so I think we are to estimate Δy given x and Δx, without evaluating y (x) at all.

    Then we want y' (x+Δx/2) * Δx

    y (x) = 2x^2 + 3x

    y' (x) = 4x + 3

    y (4) = 44

    y (4.02) = 44.3808

    Δy = 0.3808

    Δy/Δx = (0.3808) / (0.02) = 19.04

    y' (4) = 19

    y' (4.01) = 19.04

    y' (4.02) = 19.08

    Estimate Δy = (y (x+Δx) - y (x) / Δx without evaluating y () at all, using only y' (x), given x = 4, Δx = 0.02.

    y' (x+Δx/2) * Δx = y' (4.01) * 0.02 = 19.04*0.02 = 0.3808.

    In this case, where y is quadratic in x, this method gives Δy exactly.
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