/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Find an equation of the line tan... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 32

Find an equation of the line tangent to the graph of \(f\) at the given point. $$f(x)=\sin ^{-1}(x / 4) ;(2, \pi / 6)$$

Short Answer

Expert verified
Question: Find the equation of the line tangent to the graph of the function \(f(x) = \sin^{-1}(x/4)\) at the point \((2, \pi/6)\). Answer: The equation of the tangent line to the graph of \(f(x) = \sin^{-1}(x/4)\) at the point \((2, \pi/6)\) is \(y - \frac{\pi}{6} = \frac{1}{\sqrt{3}}(x - 2)\).

Step by step solution

01

Find the derivative of the function

To find the slope of the tangent line, we need to take the derivative of the given function \(f(x) = \sin^{-1}(x/4)\). We will use chain rule, and the derivative also involves the derivative of the inverse sine function (\(\sin^{-1}\)). The derivative of \(\sin^{-1}(u)\) is \(\frac{1}{\sqrt{1 - u^2}}\). Here, \(u = x/4\). Using the chain rule, we obtain: $$f'(x) = \frac{1}{\sqrt{1 - (x/4)^2}} \cdot \frac{d}{dx}(x/4) = \frac{1/4}{\sqrt{1 - x^2 / 16}}$$
02

Evaluate the derivative at the given point

Now we need to find the slope of the tangent line at the point \((2, \pi/6)\) by evaluating the derivative at \(x = 2\). $$m = f'(2) = \frac{1/4}{\sqrt{1 - (2^2) / 16}} = \frac{1/4}{\sqrt{1 - 1/4}} = \frac{1/4}{\sqrt{3/4}} = \frac{1}{\sqrt{3}}$$
03

Use point-slope form to find the equation of the tangent line

Now we have the slope of the tangent line, \(m =\frac{1}{\sqrt{3}}\), and the given point \((2, \pi/6)\). We can now use the point-slope form to find the equation of the tangent line: $$y - y_1 = m(x - x_1)$$ Plugging in the given point \((2, \pi/6)\) and the found slope \(\frac{1}{\sqrt{3}}\), we get: $$y - \frac{\pi}{6} = \frac{1}{\sqrt{3}}(x - 2)$$ This is the equation of the line tangent to the graph of \(f(x) = \sin^{-1}(x/4)\) at the given point \((2, \pi/6)\).

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Most popular questions from this chapter

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