Find the arc length of y=((x)/(2))^4+(1)/(2 x^2) over [1,4] . Hint: Show that 1+(y^')^2 is a per

Find the arc length of y=((x)/(2))^4+(1)/(2 x^2) over [1,4]...

Key Concepts

Arc Length Formula

The arc length of a curve defined by a function is calculated using the formula L = ?[a to b] ?(1 + (dy/dx)²) dx. This formula establishes the relationship between the derivative of the function and the length of the curve over a specified interval, and it is a fundamental tool in calculus for measuring distances along curves.

Differentiation

Differentiation involves computing the derivative, dy/dx, of the given function. The derivative represents the instantaneous rate of change of the function with respect to x and is essential in determining the integrand for the arc length formula. Techniques such as the power rule and chain rule are typically used to calculate the derivative of functions involving polynomial or rational expressions.

Algebraic Simplification and Perfect Squares

Once the derivative dy/dx is computed, the next step involves simplifying the expression inside the square root of the arc length formula, which is 1 + (dy/dx)². Recognizing that this expression can be arranged into a perfect square allows one to replace the square root of a square with a simpler algebraic expression. This process of algebraic manipulation simplifies the integration process by reducing a more complicated integrand into a form that is easier to integrate.

Definite Integration

After simplifying the integrand, the next step requires evaluating the definite integral over the given interval [a, b]. This involves applying integration techniques to compute the total arc length. The proper setup of the integral and the evaluation of its bounds are critical steps in arriving at the final solution for the arc length of the curve.

Transcript

00:02 So here we are given a formula of y, and then we want to know the outklence of y over the range 1 to 4.

00:11 So recall the formula for outlance is s equals to the integral from 1 to 4, and then the square root of 1 plus y derivative square, and then d x.

00:28 So this hint is really useful since if we know 1 plus direct of y square is a perfect square, then we can do the integral easier.

00:38 So we have y equals to x to the power of 4 over 16 plus 1 over 2 x squared.

00:50 And the derivative of y is x cubed over 4 plus, sorry, there should be minus x to the power of negative 3.

01:03 And then we have 1 plus y derivative square equals to 1 plus x2 power of 3 over 4 minus x to power of negative 3 square, which is 1 plus x2 power of 6 over 14 plus x2 the power of negative 6.

01:29 And then we get 2 times the first term and the second term.

01:35 With x to the power of negative 3, here we just get negative 1 half.

01:42 And then by this term 1 and negative 1 half, we really get a positive 1 hub.

01:48 And then notice that 1 half is actually equals to 2 times 1 .4 x cubed times x to the power of negative 3.

02:00 The reason for me to do that is we want to know 1 plus derivative of 1 .2 ,000, square to be a perfect square.

02:09 Then we have this expression square actually equals to x3 or 4 square, which come from this term, and then plus x to the power of negative square, three square, come from this term.

02:30 And then since we get one half to be two times this two term, i'll just write it down to be 1 over 4 xx.

02:39 X2x2 the power of negative 3, which yields this expression is really x cubed over 4 plus x to the power of negative 3 square.

02:52 And then we can write us in a nicer way, which is integral from 1 to 4.

03:00 But here we're taking the integral of a square, so we just remain the absolute value of x cube over 4.

03:09 Plus x3 power of negative 3 and then d x.

03:15 But then notice since we are taking x from the range 1 to 4, meaning x is always positive, so this expression is always positive...

Please give Ace some feedback