How do you find an equation of the tangent line to the curve at the given point #y=3arccos(x/2) # and #(1,pi)#?

Question

Christopher Agresta · Accepted Answer

#y=-sqrt3*x+sqrt3+pi#

From the given curve #y=3*arccos(x/2)# and point #(1, pi)#

Find the first derivative by the formula

#d/dx(arccos u)=(-1)/sqrt(1-u^2)*d/dx(u)#

#(dy)/dx=d/dx(3*arccos(x/2))=3*(-1)/sqrt(1-(x/2)^2)*d/dx(x/2)#

#(dy)/dx=-3/(1/2*sqrt(4-x^2))*(1/2)#

at #x=1#

#(dy)/dx=-3/(1/2*sqrt(4-1^2))*(1/2)#

#(dy)/dx=-3/sqrt3=-sqrt3""""" "the" "slope"#

The Tangent line:

By using point-slope formula with slope #m=-sqrt3# and given point #(x_1, y_1)=(1, pi)#

#y-y_1=m*(x-x_1)#

#y-pi=-sqrt3*(x-1)#

#y=-sqrt3*x+sqrt3+pi#

I hope the explanation is useful...God bless..

William Abdalla · Answer

undefined To find the equation of the tangent line to the curve y = 3arccos(x/2) at the point (1, π), we need to find the derivative of the function and evaluate it at x = 1.

The derivative of y = 3arccos(x/2) can be found using the chain rule.

dy/dx = 3 * d(arccos(x/2))/dx

To find d(arccos(x/2))/dx, we differentiate arccos(x/2) with respect to x.

d(arccos(x/2))/dx = -1 / sqrt(1 - (x/2)^2)

Substituting this back into the derivative equation, we have:

dy/dx = 3 * (-1 / sqrt(1 - (x/2)^2))

Now, we can evaluate dy/dx at x = 1:

dy/dx = 3 * (-1 / sqrt(1 - (1/2)^2))
       = 3 * (-1 / sqrt(1 - 1/4))
       = 3 * (-1 / sqrt(3/4))
       = -4 / sqrt(3)

So, the slope of the tangent line at x = 1 is -4 / sqrt(3).

Using the point-slope form of a line, we can write the equation of the tangent line:

y - π = (-4 / sqrt(3))(x - 1)

Simplifying this equation, we get:

y = (-4 / sqrt(3))(x - 1) + π

Therefore, the equation of the tangent line to the curve y = 3arccos(x/2) at the point (1, π) is y = (-4 / sqrt(3))(x - 1) + π.

Axel Alvarez · Answer

undefined To find the equation of the tangent line to the curve $ y = 3\arccos\left(\frac{x}{2}\right) $ at the point $ (1, \pi) $, follow these steps:

1. Find the derivative of the function $ y = 3\arccos\left(\frac{x}{2}\right) $ with respect to $ x $.
2. Evaluate the derivative at the given point $ x = 1 $ to find the slope of the tangent line.
3. Use the point-slope form of the equation of a line to find the equation of the tangent line.

First, find the derivative $ \frac{dy}{dx} $ of the function $ y = 3\arccos\left(\frac{x}{2}\right) $:

$$ \frac{dy}{dx} = \frac{d}{dx}\left[3\arccos\left(\frac{x}{2}\right)\right] $$

Apply the chain rule:

$$ \frac{dy}{dx} = 3\left(\frac{d}{du}\arccos(u)\right)\left(\frac{d}{dx}\frac{x}{2}\right) $$

$$ \frac{dy}{dx} = 3\left(-\frac{1}{\sqrt{1-u^2}}\right)\left(\frac{1}{2}\right) $$

$$ \frac{dy}{dx} = -\frac{3}{2\sqrt{1-\left(\frac{x}{2}\right)^2}} $$

Now, evaluate the derivative at $ x = 1 $ to find the slope of the tangent line:

$$ \left.\frac{dy}{dx}\right|_{x=1} = -\frac{3}{2\sqrt{1-\left(\frac{1}{2}\right)^2}} $$

$$ \left.\frac{dy}{dx}\right|_{x=1} = -\frac{3}{2\sqrt{1-\frac{1}{4}}} $$

$$ \left.\frac{dy}{dx}\right|_{x=1} = -\frac{3}{2\sqrt{\frac{3}{4}}} $$

$$ \left.\frac{dy}{dx}\right|_{x=1} = -\frac{3}{2\sqrt{\frac{3}{4}}} = -\frac{3}{2}\cdot\frac{2}{\sqrt{3}} = -\frac{3\sqrt{3}}{3} = -\sqrt{3} $$

Now that we have the slope of the tangent line, which is $ -\sqrt{3} $, and the point $ (1, \pi) $, we can use the point-slope form of the equation of a line:

$$ y - y_1 = m(x - x_1) $$

$$ y - \pi = -\sqrt{3}(x - 1) $$

$$ y - \pi = -\sqrt{3}x + \sqrt{3} $$

$$ y = -\sqrt{3}x + \sqrt{3} + \pi $$

Therefore, the equation of the tangent line to the curve $ y = 3\arccos\left(\frac{x}{2}\right) $ at the point $ (1, \pi) $ is $ y = -\sqrt{3}x + \sqrt{3} + \pi $.