Answer:
[tex]\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}[/tex]
Step-by-step explanation:
We want to find the equation of the line tangent to the graph of:
[tex]\displaystyle y=\sin^{-1}\left(\frac{x}{5}\right)\text{ at } x=\frac{5}{2}[/tex]
So, we will find the derivative of our equation first. Applying the chain rule, we acquire that:
[tex]\displaystyle y^\prime=\frac{1}{\sqrt{1-\left(\dfrac{x}{5}\right)^2}}\cdot\frac{1}{5}[/tex]
Simplify:
[tex]\displaystyle y^\prime=\frac{1}{5\sqrt{1-\dfrac{x^2}{25}}}[/tex]
We can factor out the denominator within the square root:
[tex]\displaystyle y^\prime =\frac{1}{5\sqrt{\dfrac{1}{25}\big(25-x^2)}}[/tex]
Simplify:
[tex]\displaystyle y^\prime=\frac{1}{\sqrt{25-x^2}}[/tex]
So, we can find the slope of the tangent line at x = 5/2. By substitution:
[tex]\displaystyle y^\prime=\frac{1}{\sqrt{25-(5/2)^2}}[/tex]
Evaluate:
[tex]\displaystyle y^\prime=\frac{1}{\sqrt{75/4}}=\frac{1}{\dfrac{5\sqrt{3}}{2}}=\frac{2\sqrt{3}}{15}[/tex]
We will also need the point at x = 5/2. Using our original equation, we acquire that:
[tex]\displaystyle y=\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}[/tex]
So, the point is (5/2, π/6).
Finally, by using the point-slope form, we can write:
[tex]\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}\left(x-\frac{5}{2}\right)[/tex]
Distribute:
[tex]\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}x+\frac{-\sqrt{3}}{3}[/tex]
Isolate. Hence, our equation is:
[tex]\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}[/tex]
