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marusya05 [52]
3 years ago
5

What is the answer to this

Mathematics
1 answer:
Monica [59]3 years ago
7 0
The answer to this is A.24
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Find the equation of the line tangent to the graph of
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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}\big(\frac{x}{5}\big)\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-(\frac{x}{5})^2}}\cdot\frac{1}{5}

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

We can factor out the denominator within the square root:

\displaystyle y^\prime =\frac{1}{5\sqrt{\frac{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 <em>x</em> = 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}{\frac{5\sqrt{3}}{2}}=\frac{2\sqrt{3}}{15}

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

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

So, a 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}(x-\frac{5}{2})

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}

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