Find the points on the curve at which the tangent is horizontal?
y=(cosx)/(2+ sinx) y' = (-sin x (2+sin x) - cos x cos x) / (2+ sinx)² = (-2 sin x - sin² x - cos² x)/ (2+ sinx)² = ( -2 sin x - 1)/ (2+ sinx)² the tangent is horizontal means y' = 0 0 = ( -2 sin x - 1)/ (2+ sinx)² 0 = -2 sin x - 1 -1/2 = sin x x = 7π/6, 11π/6 y = (cos x)/(2+ sin x) You can find the two points.
Find the points on the curve y = cos x/(2 + sin x) at which the tangent is horizontal.?
Recall that the derivative of a function evaluated at a given point yields the slope of the tangent line at that point, so at the point(s) of horizontal tangency, the derivative is zero (since the slope of any horizontal line is zero). Using the Quotient Rule, y = cos(x)/[2 + sin(x)] has derivative: dy/dx = {[cos(x)]'[2 + sin(x)] - cos(x)[2 + sin(x)]'}/[2 + sin(x)]^2 = {-sin(x)[2 + sin(x)] - cos^2(x)}/[2 + sin(x)]^2 = [-2sin(x) - sin^2(x) - cos^2(x)]/[2 + sin(x)]^2 = -[2sin(x) + sin^2(x) + cos^2(x)]/[2 + sin(x)]^2, by factoring out the negative = -[2sin(x) + 1]/[2 + sin(x)]^2, since sin^2(x) + cos^2(x) = 1. Then, dy/dx = 0 when: -[2sin(x) + 1]/[2 + sin(x)]^2 = 0 ==> 2sin(x) + 1 = 0, which solves to give sin(x) = -1/2. From the unit circle, sin(x) = -1/2 has solutions x = π/6 and x = 11π/6 on (0, 2π). Due to sine's periodicity, all other solutions can be found by adding a multiple of 2π to these solutions. Thus, the values of x where horizontal tangency occurs are: x = π/6 + 2πk and x = 11π/6 + 2πk, where k is an integer. When x = π/6 + 2πk (with k being an integer), we see that: y = cos(π/6 + 2πk)/[2 + sin(π/6 + 2πk)] = cos(π/6)/[2 + sin(π/6)], due the periodicity of sine and cosine = √3/5. When x = 11π/6 + 2πk (with k being an integer): y = cos(11π/6 + 2πk)/[2 + sin(11π/6 + 2πk)] = cos(11π/6)/[2 + sin(11π/6)] = √3/3. Therefore, the required points are: (π/6 + 2πk, √3/5) and (11π/6 + 2πk, √3/3), where k is an integer. I hope this helps!
Find the point on the curve y=(cos x)/(2+sin x) at which the tangent line is horizontal?
-0.5219...,-6.805576..., several other places also
Find the points on the curve y=(cos x) / (2+ sin x) where the tangent is horizontal?
The slope of the point will be horizontal if the derivative at that point is zero. Assuming you did your math right: 0 = -(cos^2 x + ( sin x )*( sinx + 2))/(2 + sin x)^2 multiply both sides by -(2 + sinx)^2 to get: 0 = cos^2 x + (sin x)*( sinx + 2) 0 = cos^2 x + sin^2 x +2sin x (cos^2 x + sin^2 x = 1) 0 = 1 + 2sin x sin x = -1/2 x = -pi/6 or -30deg.
How do i find the points on y = (cos x)/(2 + sin x) at which the tangent line is horizontal?
y = cos x / ( 2 + sin x) . . . . use derivative of fraction y ' = [ cos x ( cosx) - ( 2 + sin x)(- sin x ) ] / ( 2 + sin x)^2 = 0 cos^2 x + 2 sin x + sin^2 x = 0 2 sin x = -1 . . . . . . . . . . . . since . . . sin^2 x + cos^2 = 1 sin x = - 1/2 x = - 30 or 330 deg . . . . and . . . x = 210 deg
Find the point on the curve y=sinx closest to the point (1,1)?
d=sq rt of (x-1)^2 + (y-1)^2 let y=Sin x d=sq rt of (x-1)^2 + (Sinx-1)^2 differentiate d'=[2(x-1)+2Cosx(Sinx-1)]/2sq.Rt[(x-1)... + (Sinx-1)^2] set d'=0 0=[2(x-1)+2Cosx(Sinx-1)]/2sq.Rt[(x-1)^... + (Sinx-1)^2] 0=2(x-1)+2Cosx(Sinx-1) 0=x-1+Cosx(Sinx-1) 0=x-1+SinxCosx-Cosx solving the equation weh have x=1.062 solve for y y=Sinx y=Sin1.062 y=0.873 so closest point to (1,1) is (1.062,0.873) thatz b.
What is the area bounded by curve y=sinx, y=cosx, x=0 and x=pi/2?
Graph the two functions. Note the intersection point within the stated interval. Set the two functions equal to each other:[math]\sin(x)\, =\, \cos(x)[/math]…and solve for the intersection point. Between zero and the intersection point, the cosine wave will be above the sine wave; between the intersection point and the given end-point, the sine wave will be above the cosine wave.Set up the integrals for each of the sub-intervals. (You should get the two integrals that you were given in an earlier reply. They are correct.)Integrate, using the trig derivatives that you’ve memorized.[math]\displaystyle \int\, \cos(x)\, dx\, =\, \sin(x)\, +\, C[/math][math]\displaystyle \int\, \sin(x)\, \dx\, =\, -\cos(x)\, +\, C[/math](Since you’re doing definite integrals, you can ignore the constants, of course.)Evaluate each integrated expression over its interval, using the trig special-angle values you memorized back in trigonometry.Then sum the two values to obtain the total value of the indicated area.
At what point on the curve y=cos(x) would the 99th derivative of the curve be 0?
y' = -sin(x) y'' = -cos(x) y''' = sin(x) y'''' = cos(x) = y Note that after 4th derivative, it returns to y. 99 = 4*24+3 So, the 99th derivative of y = y''' = sin(x) Solve sin(x) = 0 for x, x = n*pi, where n can be any integer.
Why isn't the area bounded by the curve y=cosx between the points x=0 and x=2pi not numerically equal to the definite integral of cosx from 0 to 2pi?
The area bounded by a curve given by [math]y = f(x)[/math], the x axis and the values [math]x = a, x = b[/math] is the integral[math]\int_{a}^{b} |f(x)| dx[/math]So in your example, the area is the integral of [math]|\cos(x)|[/math], and because you take the absolute value, you get a nonzero answer. If you just take the integral of [math]\cos(x)[/math], the integrals will cancel themselves out over a period of [math]2\pi[/math] and you will get 0 as the answer.