the curve with equation y=k cos 3x where x is in radians and k is a constant. the curve cuts the x axis at point A. a) write down the coordinates of point A b) find as a multiple of k, the area of the region R bounded by the x axis , the y axis and the curve. ( there is a sketch of the curve given but is see no way to put it on the internet. however it has no numerical values. It starts of when X=0 and y is a positive value and slopes downwards ....the gradient keeps getting steeper until a point where y=0 and x is a positive value )
The Integral ∫2012-11-18T04:15:57Z
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when it intersects the x-axis, it means that y = 0 ----> so (A , 0) 0 = k cos(3 * A) 0 = cos(3A) π/2 + πn = 3A ---> where n is an integer (π/6) + (πn/3) = A <----- a)
Uhm... properly, in case you have the function A(x) for the section below the curve from a (some consistent) to x (notice that A(a) = 0, because of the fact the section below the curve from a to a is only 0). One only has to tell apart A(x) with know to x, to acquire the unique function. (FTC) as an occasion: the section below some curve from 0 to x could be expressed by using the function g(x) = x ^ 2 + x, so the unique function is f(x) = g'(x) = 2x + a million besides the undeniable fact that, in case you purely understand the fee of the section below the curve from a to b of a few function, you do not have sufficient suggestions to unravel the unique function. e.g: the section below some curve from 0 to a million is two(units), you could have countless opportunities: a million. f(x) = 2 2. f(x) = 4x 3. or maybe piecewise function: f(x) = a million for 0<= x <= 0.5 f(x) = 8x - 3 for x > 0.5 4. ... are you able to get it? :)
y=k cos 3x Since curve cuts x axis then y = 0 k cos(3x) = 0 k is not zero, therefore cos(3x) =0 or 3x = (2n+1)(pi/2) x = (2n+1)(pi/6) ...............Ans Therefor point of A(pi/6,0) ...............ans when the point is on y axis put x = 0 y = k because cos (0) = 1 and cos (6pi) = -1 then y = -k then point on y-axis = B(0,k)