Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

5. Graphical Applications of Derivatives

Finding Global Extrema

Calculus

5. Graphical Applications of Derivatives

Finding Global Extrema

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Multiple Choice

Find the exact length of the curve $y = 2 \sqrt 3 x^{\frac{3}{2}}$ for $0 \leq x \leq 6$ .

56

36 + 12 \sqrt 19

74 + 12 \sqrt 37

74

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Verified step by step guidance

Step 1: Recall the formula for the arc length of a curve y = f(x) over the interval [a, b]: $ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx $. This formula calculates the exact length of the curve by integrating the square root of 1 plus the square of the derivative of the function.

Step 2: Compute the derivative of the given function $ y = 2 \sqrt{3} x^{3/2} $. Using the power rule, $ \frac{dy}{dx} = 2 \sqrt{3} \cdot \frac{3}{2} x^{1/2} = 3 \sqrt{3} x^{1/2} $.

Step 3: Substitute $ \frac{dy}{dx} $ into the arc length formula. The integrand becomes $ \sqrt{1 + \left( 3 \sqrt{3} x^{1/2} \right)^2} = \sqrt{1 + 27x} $. The arc length formula now reads $ L = \int_0^6 \sqrt{1 + 27x} \, dx $.

Step 4: Solve the integral $ \int_0^6 \sqrt{1 + 27x} \, dx $. Use a substitution method: let $ u = 1 + 27x $, so $ du = 27 \, dx $. Adjust the limits of integration accordingly: when $ x = 0 $, $ u = 1 $; when $ x = 6 $, $ u = 163 $. The integral becomes $ \frac{1}{27} \int_1^{163} \sqrt{u} \, du $.

Step 5: Evaluate $ \frac{1}{27} \int_1^{163} \sqrt{u} \, du $. The antiderivative of $ \sqrt{u} $ is $ \frac{2}{3} u^{3/2} $. Apply the limits of integration: $ \frac{1}{27} \left[ \frac{2}{3} u^{3/2} \right]_1^{163} $. Simplify the expression to find the exact length of the curve.