What is the arc length of #f(t)=(sqrt(t-1),2-8t) # over #t in [1,3]#?

Question

What is the arc length of #f(t)=(sqrt(t-1),2-8t)  # over #t in [1,3]#?

Aurora Alvarado · Accepted Answer

#L=3/2sqrt114+1/32ln(16sqrt2+3sqrt57)# units.

#f(t)=(sqrt(t−1),2−8t)# #f'(t)=(1/(2sqrt(t−1)),−8)# Arc length is given by: #L=int_1^3sqrt(1/(4(t−1))+64)dt# Apply the substitution #t-1=u^2#: #L=int_0^sqrt2sqrt(1/(4u^2)+64)(2udu)# Simplify: #L=int_0^sqrt2sqrt(1+256u^2)du# Apply the substitution #16u=tantheta#: #L=1/16intsec^3thetad theta# This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals: #L=1/32[secthetatantheta+ln|sectheta+tantheta|]# Reverse the last substitution: #L=[1/2usqrt(1+256u^2)+1/32ln|16u+sqrt(1+256u^2)|]_0^sqrt2# Insert the limits of integration: #L=3/2sqrt114+1/32ln(16sqrt2+3sqrt57)#

Charles Anctil · Answer

undefined To find the arc length of the curve defined by the function $ f(t) = (\sqrt{t-1}, 2-8t) $ over the interval $ t \in [1,3] $, we use the arc length formula:

$$ L = \int_{a}^{b} \sqrt{[f'(t)]^2 + [g'(t)]^2} dt $$

Where $ f(t) = f(t) $ and $ g(t) = 2-8t $. Then, we find the derivatives of $ f(t) $ and $ g(t) $ with respect to $ t $.

$$ f'(t) = \frac{1}{2\sqrt{t-1}} $$
$$ g'(t) = -8 $$

Now, plug these derivatives into the arc length formula:

$$ L = \int_{1}^{3} \sqrt{\left(\frac{1}{2\sqrt{t-1}}\right)^2 + (-8)^2} dt $$

$$ L = \int_{1}^{3} \sqrt{\frac{1}{4(t-1)} + 64} dt $$

$$ L = \int_{1}^{3} \sqrt{\frac{1+256(t-1)}{4(t-1)}} dt $$

$$ L = \int_{1}^{3} \sqrt{\frac{256t-251}{4(t-1)}} dt $$

$$ L = \int_{1}^{3} \frac{\sqrt{256t-251}}{2\sqrt{t-1}} dt $$

$$ L = \frac{1}{2} \int_{1}^{3} \frac{\sqrt{256t-251}}{\sqrt{t-1}} dt $$

This integral can be solved using various techniques, such as substitution or integration by parts. After evaluating the integral, you'll find the arc length of the curve.