Here is the assignment and i am having trouble showing the correct work for this

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Here is the assignment and i am having trouble showing the correct work for this assignment. Please show work to help me to understand this assignment!

A trough in the shape of a triangular prism. The trough is 12 ft. long and 3 ft. across. Its ends are isosceles triangles with altitudes of 3 ft. The water level is rising at a rate of 3/8 ft. per minute when it is 2 ft. deep. Determine the rate at which the water is being pumped into the trough. Find the value of the function y = + sin x-0.5, correct to 3 decimal places, when x = 0.05

Explanation / Answer

1) for the first one, we derive a relation between the volume of water in the trough and the height of water.

So,

Volume = Area of triangular cross-section * length of the trough

Let at any time 't' the height be 'h' feet

So, from the figure, in the triangular cross section the height = the base of the triangle

So, area of triangular cross-section = 0.5 * base * height = 0.5 *h *h = (0.5)h^2

Volume = Area of triangle * length of trough = 0.5h^2 * 12 = 6h^2

V(h) = 6h^2

Differentiate w.r.t 't',

dV/dt = (dV/dh) * dh/dt

dh/dt = 3/8 ft/sec (given)

dV/dh = 12 h = 12 * 2 = 24 ft (Given that h=2 at that instant0

So, dV/dt = 24 * 3 / 8 = 3 *3 = 9ft^3 / sec is the rate at which trough is being filled.

2) We use the basic definition of differentiation,

f'(x) = lim (h--->0) [f(x+h) - f(x)] / h

Here, f(x) = sqrt(x+1) + sin (x) - 0.5

h = 0.05, x = 0

f'(x) = 1/2sqrt(x+1) + cos (x) - 0.5

f(x+h) = f(0 + 0.05) = f(x) + hf'(x) = f(0) + 0.05f'(0)

f(0.05) = f(0) + 0.05f'(0)

f(0) = 1 - 0.5 = 0.5

f'(0) = 0.5 + 1 -0.5 = 1

So, f(0.05) = 0.5 + 0.05(1) = 0.5 + 0.05 = 0.550

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