# (¡)find the equation of the curve(¡¡)find the equation of the normal to the curve at the point whre x=3p¡¡/4 A curve is such that dy/dx=2cos(2x-p¡¡/2). The curve passes thru the point...

(¡)find the equation of the curve

(¡¡)find the equation of the normal to the curve at the point whre x=3p¡¡/4

A curve is such that dy/dx=2cos(2x-p¡¡/2). The curve passes thru the point (p¡¡/2,3).

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### 1 Answer

You should multiply by d`x` the given equation such that:

`dy = 2cos(2x- pi/2)dx`

You need to integrate both sides such that:

`int dy = int 2cos(2x- pi/2)dx => y = int 2cos(2x- pi/2)dx`

You need to expand `cos(2x- pi/2)` using the formula `cos(a-b)=` `cos a cos b + sin a sin b` such that:

`cos(2x- pi/2) = cos 2x cos (pi/2) + sin 2x sin (pi/2)`

You need to substitute 0 for `cos (pi/2)` and 1 for `sin (pi/2)` such that:

`cos(2x- pi/2) = sin 2x`

`y = int 2sin 2x dx y => y = 2 int sin 2x dx => y = 2*(-cos 2x)/2 + c`

`y = -cos 2x + c`

Since the curve passes through `(pi/2, 3), ` hence, you should substitute 3 for y and `pi/2` for x such that:

`3 = -cos 2(pi/2) + c => 3 = -cos pi + c `

Since `cos pi = -1` yields:

`3 = 1 + c => c = 3 - 1 => c = ` 2

**Hence, evaluating the equation of the curve, under the given conditions yields `y = -cos 2x + 2.` **

You need to remember that the normal to the curve is perpendicular to tangent to the curve, hence, the product of slopes of normal and tangent lines is -1.

You need to evaluate the derivative of the function `y = -cos 2x + 2` , at `x = 3pi/4` such that:

`(dy)/(dx) = 2sin 2x => (dy)/(dx)|_(x=3pi/4) = 2sin 2(3pi/4) = 2sin(3pi/2) = -2`

Hence, the slope of tangent line at `x = 3pi/4` is m = -2, hence, the slope of the normal to the curve is `m = 1/2` .

You need to write the equation of the normal to the curve such that:

`y - y|_(x=3pi/4) = (1/2)(x - 3pi/4)`

`y|_(x=3pi/4) = -cos2*(3pi/4) + 2 = 0 + 2 = 2`

`y - 2 = (1/2)(x - 3pi/4)`

**Hence, evaluating the equation of the normal to the curve, under the given conditions, yields `y - 2 = (1/2)(x - 3pi/4).` **