For example, 1.) when a perfect clock strikes midnight, this slow clock reads precisely 23 hr 58 minutes,

or, 2.) when this slow clock reaches midnight, the perfect clock reads 0 hr 02 minutes, which is ever so slightly different.

#1 is slower by factor of (23 58/60)/24, while #2 is slower by factor of 24/(24 2/60).

#1 is less perfect a clock than #2 because it is analogous to 2/3's and 3/4's or 9/10's and 10/11's.]]>

If

show that x (dy/dx) = y.]]>

The time of a complete oscillation of a simple pendulum of length l is given by the relation T = 2 π √(l/g) where g is a constant. By what percent should the length be changed in order to correct a loss of 2 minutes per day?

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The volume of a cube is increasing at the rate of 7 cubic centimetres per second. How fast is the surface area increasing when the length of the edge is 12cm?

]]>Differentiate the following with respect to x.

(i) x[sup]x[/sup] (ii) x[sup]sinx[/sup]

]]>The radius of a spherical balloon is increasing at the rate of 5 cm per second when inflated by pumping air. Find the rate of increase of (i) its surface area and (ii) its volume, when the radius is 4 cm.

]]>There is a 3cm margin on each side, so 6 must be added to the width. Similarly, there is a 2cm margin at the bottom, so 2 must be added to the height.

Therefore, the equation that needs to be minimised is (x + 6)(75/x + 2).

Multiplying out of brackets makes this become 75 + 450/x + 2x + 12.

Differentiating gives 2 - 450/x².

The paper is smallest when the differential is equal to 0.

2 - 450/x² = 0

450/x² = 2

450 = 2x²

x² = 225**x = 15cm**

That also had a negative answer, but it was discarded because we are dealing with length.

Anyway, using this value with the original equation shows that the smallest possible paper is 21 * 7 = 147cm² big.

]]>An advertisement is to contain 75 sq. cms. of printed area. There is a 2 cm margin at the bottom, 3 cm margin on each side and no margin at the top. Find the dimensions of the smallest possible paper.

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