How do you find the lengths of the curve #y=intsqrt(t^-4+t^-2)dt# from [1,2x] for the interval #1

Question

How do you find the lengths of the curve #y=intsqrt(t^-4+t^-2)dt# from [1,2x] for the interval #1<=x<=3#?

Emma Aldridge · Accepted Answer

#= 7/3#

Arc length is:

# s= int _{a}^{b} sqrt {1+(y')^{2}} dx qquad triangle#

For function:

#y=int_(1)^(2x) sqrt(t^-4+t^-2)dt#

...use Liebnitz's Rule, specifically tailored here:

#(d)/(dx) int _{u(x)}^{v(x)} \ f(t) \ dt = f(v)\ v'(x) - f(u) \ u'(x)#

#= sqrt((2x)^-4+(2x)^-2)* 2 = 2 sqrt(1/(16x^4)+1/(4x^2)) #

So #triangle# becomes:

# s= int _{1}^{3} sqrt {1+1/(x^2)+1/(4x^4)} dx #

# s= int _{1}^{3} sqrt { (1+1/(2x^2))^2} dx #

# s= [ x - 1/(2x) ]_{1}^{3} = 7/3#

Luke Alvarez · Answer

undefined To find the length of the curve $y = \int \sqrt{t^{-4} + t^{-2}} \, dt$ from $t = 1$ to $t = 2x$ for the interval $1 \leq x \leq 3$, you need to use the arc length formula for a curve defined by a function $y = f(x)$ from $x = a$ to $x = b$:

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

First, you need to find $\frac{dy}{dt}$ and then integrate $\sqrt{1 + \left(\frac{dy}{dt}\right)^2}$ from $t = 1$ to $t = 2x$.

Given $y = \int \sqrt{t^{-4} + t^{-2}} \, dt$, differentiate $y$ with respect to $t$ to find $\frac{dy}{dt}$:

$$ \frac{dy}{dt} = \sqrt{t^{-4} + t^{-2}} $$

Now, you'll substitute this into the arc length formula:

$$ L = \int_{1}^{2x} \sqrt{1 + \left(\sqrt{t^{-4} + t^{-2}}\right)^2} \, dt $$

$$ L = \int_{1}^{2x} \sqrt{1 + t^{-4} + t^{-2}} \, dt $$

$$ L = \int_{1}^{2x} \sqrt{t^{-4} + t^{-2} + 1} \, dt $$

Now, you can integrate this expression with respect to $t$ over the given interval $[1, 2x]$ to find the length of the curve.

How do you find the lengths of the curve #y=intsqrt(t^-4+t^-2)dt# from [1,2x] for the interval #1&lt;=x&lt;=3#?

How do you find the lengths of the curve #y=intsqrt(t^-4+t^-2)dt# from [1,2x] for the interval #1<=x<=3#?