Question

Find the equation of the line tangent to the graph of
y = sin^-1 (x/5) at x = 5/2. Show your work.

Answer 1

Answer:

$\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}$

Step-by-step explanation:

We want to find the equation of the line tangent to the graph of:

$\displaystyle y=\sin^{-1}\left(\frac{x}{5}\right)\text{ at } x=\frac{5}{2}$

So, we will find the derivative of our equation first. Applying the chain rule, we acquire that:

$\displaystyle y^\prime=\frac{1}{\sqrt{1-\left(\dfrac{x}{5}\right)^2}}\cdot\frac{1}{5}$

Simplify:

$\displaystyle y^\prime=\frac{1}{5\sqrt{1-\dfrac{x^2}{25}}}$

We can factor out the denominator within the square root:

$\displaystyle y^\prime =\frac{1}{5\sqrt{\dfrac{1}{25}\big(25-x^2)}}$

Simplify:

$\displaystyle y^\prime=\frac{1}{\sqrt{25-x^2}}$

So, we can find the slope of the tangent line at x = 5/2. By substitution:

$\displaystyle y^\prime=\frac{1}{\sqrt{25-(5/2)^2}}$

Evaluate:

$\displaystyle y^\prime=\frac{1}{\sqrt{75/4}}=\frac{1}{\dfrac{5\sqrt{3}}{2}}=\frac{2\sqrt{3}}{15}$

We will also need the point at x = 5/2. Using our original equation, we acquire that:

$\displaystyle y=\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$

So, the point is (5/2, π/6).

Finally, by using the point-slope form, we can write:

$\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}\left(x-\frac{5}{2}\right)$

Distribute:

$\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}x+\frac{-\sqrt{3}}{3}$

Isolate. Hence, our equation is:

$\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}$