Problem

For the function $ f(x) = \frac{1}{4} e^x + e^{-x…

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Problem 38 Hard Difficulty

The arc length function for a curve $ y = f(x) $, where $ f $ is an increasing function, is $ s(x) = \int_0^x \sqrt{3t + 5}\ dt $.
(a) If $ f $ has y-intercept 2, find an equation for $ f $.
(b) What point on the graph of $ f $ is 3 units along the curve from the y-intercept? State your answer rounded to 3 decimal places.

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Calculus: Early Transcendentals

Chapter 8

Further Applications of Integration

Section 1

Arc Length

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Video Transcript

he has clear so in new Marine here. So when we look at the R cling formula, you know that it's one plus the derivative square, so we just make this equal to three t plus five. When we get the derivative to be equal to the square root of three p plus four, we just have to integrate. This FF T is equal to the integral of three t plus four d t. This is equal to to over nine times three T plus four to the three house power plus C. The Y intercept is too, so F zero is equal to two. We just have to plug in. We'll continue here when we got C to be equal to two over nine. Star formula is F of X is equal to 29 turns three acts plus four to the three house power plus two over nine for part B. We're gonna do some calculations, so three is equal to from zero to x square of three T plus five d t, which means that three is equal to 2/3 times 1/3 three T plus five 23 house power from zero to X This gives us 27 over, too. It was 5 to 3. House power to the 2/3 power minus five was equal to three x. So X is equal to 1/3 times 27 over too, plus five three House to the 2/3 power minus five, which is approximately 1.15 8947 34 943 We just have to plug this in to our equation for F of X and for why we get 4.7 6545 297 442 So our points o R. One point 159 calmer A 4.7 65 which are three units from the wine intercept.

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Problem

For the function $ f(x) = \frac{1}{4} e^x + e^{-x…

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Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 8

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Michael Jacobsen

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Michael Jacobsen

Joseph Lentino

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Michael Jacobsen

Joseph Lentino

James Stewart

Calculus: Early Transcendentals