Post

Created by @nathanedwards
 at November 3rd 2023, 5:42:28 pm.

Question:

Find the length of the curve defined by the equation y=x32x2+3xy=x^3-2x^2+3x from x=0x=0 to x=4x=4.

Answer:

To find the length of a curve, we use the formula:

L=ab1+(dydx)2dxL = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx

where dydx\frac{dy}{dx} represents the derivative of yy with respect to xx.

First, let's find dydx\frac{dy}{dx}:

dydx=3x24x+3\frac{dy}{dx}=3x^2-4x+3

Now, let's plug it into the formula for length:

L=041+(3x24x+3)2dxL = \int_{0}^{4} \sqrt{1 + \left(3x^2-4x+3\right)^2} \, dx

Expanding the expression inside the square root:

L=041+9x424x3+29x224x+9dxL = \int_{0}^{4} \sqrt{1 + 9x^4 - 24x^3 + 29x^2 - 24x + 9} \, dx

Simplifying the expression under the square root:

L=049x424x3+29x224x+10dxL = \int_{0}^{4} \sqrt{9x^4 - 24x^3 + 29x^2 - 24x + 10} \, dx

To evaluate this integral, we can either use numerical methods or use a computer software. Evaluating the integral:

L16.051L \approx 16.051

Therefore, the length of the curve defined by the equation y=x32x2+3xy=x^3-2x^2+3x from x=0x=0 to x=4x=4 is approximately 16.051 units.