Question

Find an equation of the tangent line to the graph of the function at the given point.y=3arcsinx, (1/2,pi/2)

Answer 1

Given:

$y=3arcsin(x)$

Let's find the equation of the tangent line over the interval:

$(\frac{1}{2},\frac{\pi}{2})$

First find the derivative of the equation:

$y^{\prime}=\frac{3}{\sqrt{1-x^2}}$

Now evaluate the derivative when x = 1/2:

$\begin{gathered} y^{\prime}=\frac{3}{\sqrt{1-(\frac{1}{2})^2}} \\ \\ y^{\prime}=\frac{3}{\sqrt{1-\frac{1}{4}}} \\ \\ y^{\prime}=\frac{3}{\sqrt{\frac{3}{4}}} \\ \\ \end{gathered}$

Solving further:

$\begin{gathered} y^{\prime}=\frac{3}{\frac{\sqrt{3}}{\sqrt{4}}} \\ \\ y^{\prime}=\frac{3\sqrt{4}}{\sqrt{3}} \\ \\ y^{\prime}=\frac{3*2}{\sqrt{3}} \\ \\ y^{\prime}=\frac{6}{\sqrt{3}} \\ \\ y=\frac{6}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}} \\ \\ y=\frac{6\sqrt{3}}{3} \\ \\ y=2\sqrt{3} \end{gathered}$

Now, the slope of the tangent line is 2√3.

Input the value for the slope, then put (1/2, π/2) for x1 and y1 in point-slope form:

$\begin{gathered} y-y1=m(x-x1) \\ \\ y-\frac{\pi}{2}=2\sqrt{3}(x-\frac{1}{2}) \\ \\ y-\frac{\pi}{2}=2\sqrt{3}x-2\sqrt{3}*\frac{1}{2} \\ \\ y-\frac{\pi}{2}=2\sqrt{3}x-\sqrt{3} \end{gathered}$

Now add π/2 to both sides:

$\begin{gathered} y-\frac{\pi}{2}+\frac{\pi}{2}=2\sqrt{3}x-\sqrt{3}+\frac{\pi}{2} \\ \\ y=2\sqrt{3}x-\frac{2\sqrt{3}-\pi}{2} \end{gathered}$

Therefore, the equation of the tangent line is:

$y=2\sqrt{3}x-\frac{2\sqrt{3}-\pi}{2}$

ANSWER:

$y=2\sqrt{3}x-\frac{2\sqrt{3}-\pi}{2}$