Find the length of the curve y=(3÷4)x^(4÷3)-(3÷8)x^(2÷3)+5,1

Question

Find the length of the curve y=(3÷4)x^(4÷3)-(3÷8)x^(2÷3)+5,1<=x<=8?

Emma Aldridge · Accepted Answer

#L=99/8# units.

#y=3/4x^(4/3)-3/8x^(2/3)+5# #y'=x^(1/3)-1/4x^(-1/3)# Arc length is given by: #L=int_1^8sqrt(1+(x^(1/3)-1/4x^(-1/3))^2)dx# For ease of reading, apply the substitution #x=u^3#: #L=int_1^2sqrt(1+(u-1/(4u))^2)(3u^2du)# Simplify: #L=3int_1^2u^2sqrt((u+1/(4u))^2)du# Hence: #L=3/4int_1^2(4u^3+u)du# Integrate directly: #L=3/4[u^4+1/2u^2]_1^2# Insert the limits of integration: #L=99/8#

Evelyn Agard · Answer

undefined To find the length of the curve represented by the function $ y = \frac{3}{4}x^{\frac{4}{3}} - \frac{3}{8}x^{\frac{2}{3}} + 5 $ over the interval $ 1 \leq x \leq 8 $, we can use the arc length formula:

$$ L = \int_{1}^{8} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

First, find $ \frac{dy}{dx} $:

$$ \frac{dy}{dx} = \frac{d}{dx} \left(\frac{3}{4}x^{\frac{4}{3}} - \frac{3}{8}x^{\frac{2}{3}} + 5\right) $$

$$ \frac{dy}{dx} = \frac{3}{4} \times \frac{4}{3}x^{\frac{1}{3}} - \frac{3}{8} \times \frac{2}{3}x^{-\frac{1}{3}} $$

$$ \frac{dy}{dx} = x^{\frac{1}{3}} - \frac{1}{4}x^{-\frac{1}{3}} $$

Now, plug $ \frac{dy}{dx} $ into the arc length formula and integrate:

$$ L = \int_{1}^{8} \sqrt{1 + \left(x^{\frac{1}{3}} - \frac{1}{4}x^{-\frac{1}{3}}\right)^2} \, dx $$

$$ L = \int_{1}^{8} \sqrt{1 + x^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{16}x^{-\frac{2}{3}}} \, dx $$

$$ L = \int_{1}^{8} \sqrt{\frac{17}{16} + x^{\frac{2}{3}} + \frac{1}{16}x^{-\frac{2}{3}}} \, dx $$

$$ L = \int_{1}^{8} \sqrt{\left(\frac{\sqrt{17}}{4} + \frac{1}{4}x^{\frac{1}{3}} - \frac{1}{4}x^{-\frac{1}{3}}\right)^2} \, dx $$

$$ L = \int_{1}^{8} \left(\frac{\sqrt{17}}{4} + \frac{1}{4}x^{\frac{1}{3}} - \frac{1}{4}x^{-\frac{1}{3}}\right) \, dx $$

Now integrate this expression over the given interval $1 \leq x \leq 8$.

HIX Tutor · Answer

undefined To find the length of the curve, we can use the formula for the length of a curve given by $y = f(x)$:

$$
L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
$$

Here are the steps to find the length of the curve $y=\left(\frac{3}{4}\right)x^{\frac{4}{3}} - \left(\frac{3}{8}\right)x^{\frac{2}{3}} + 5$ from $x=1$ to $x=8$:

### Step 1: Find the derivative $ \frac{dy}{dx} $

1. Differentiate $y$ with respect to $x$:
   - $ \frac{dy}{dx} = \frac{3}{4} \cdot \frac{4}{3} x^{\frac{4}{3}-1} - \frac{3}{8} \cdot \frac{2}{3} x^{\frac{2}{3}-1} $
   - Simplify:
     - $ \frac{dy}{dx} = x^{\frac{1}{3}} - \frac{1}{4} x^{-\frac{1}{3}} $

### Step 2: Calculate $\left(\frac{dy}{dx}\right)^2$

2. Square the derivative:
   - $ \left(\frac{dy}{dx}\right)^2 = \left(x^{\frac{1}{3}} - \frac{1}{4} x^{-\frac{1}{3}}\right)^2 $

### Step 3: Simplify and expand the squared term

3. Expand:
   - $ = x^{\frac{2}{3}} - 2 \cdot \left(x^{\frac{1}{3}} \cdot \frac{1}{4} x^{-\frac{1}{3}}\right) + \left(\frac{1}{4} x^{-\frac{1}{3}}\right)^2 $
   - $ = x^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{16} x^{-\frac{2}{3}} $

### Step 4: Find $1 + \left(\frac{dy}{dx}\right)^2$

4. Combine:
   - $ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + x^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{16} x^{-\frac{2}{3}} $
   - $ = \frac{1}{2} + x^{\frac{2}{3}} + \frac{1}{16} x^{-\frac{2}{3}} $

### Step 5: Set up the integral for length

5. Set up the integral:
   - Compute:
   $$
   L = \int_{1}^{8} \sqrt{\frac{1}{2} + x^{\frac{2}{3}} + \frac{1}{16} x^{-\frac{2}{3}}} \, dx
   $$

### Step 6: Evaluate the integral

6. The integral can be complex. To solve it, you may need numerical methods or a computer tool.

If you try to evaluate it with a calculator, that will give you the approximate length of the curve between 1 and 8.

### Final Note

If you do not have access to such a tool, I can help guide you through numerical methods. Please let me know!

Find the length of the curve y=(3÷4)x^(4÷3)-(3÷8)x^(2÷3)+5,1&lt;=x&lt;=8?

Find the length of the curve y=(3÷4)x^(4÷3)-(3÷8)x^(2÷3)+5,1<=x<=8?