What is the arclength of #(t^2-lnt,lnt)# on #t in [1,2]#?

Question

Isaiah Allen · Accepted Answer

#L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2ln2# units.

#f(t)=(t^2-lnt,lnt)# #f'(t)=(2t-1/t, 1/t)# Arclength is given by: #L=int_1^2sqrt((2t-1/t)^2+1/t^2)dt# Simplify: #L=int_1^2sqrt((2t^2-1)^2+1)/tdt# Apply the substitution #2t^2-1=u#: #L=1/2int_1^7sqrt(u^2+1)/(u+1)du# Rearrange: #L=1/2int_1^7(u^2+1)/(sqrt(u^2+1)(u+1))du# #color(white)(L)=1/2int_1^7(u^2-1+2)/(sqrt(u^2+1)(u+1))du# Factorize: #L=1/2int_1^7((u-1)(u+1)+2)/(sqrt(u^2+1)(u+1))du# Integration is distributive: #L=1/2int_1^7(u-1)/sqrt(u^2+1)du+int_1^7 1/(sqrt(u^2+1)(u+1))du# #color(white)(L)=1/2int_1^7(u/sqrt(u^2+1)-1/sqrt(u^2+1))du+int_1^7 1/(sqrt(u^2+1)(u+1))du# Apply the substitution #u=tan2theta#: #L=1/2[sqrt(u^2+1)-sinh^(-1)u]_1^7+2int(sec2theta)/(tan2theta+1)d theta# Apply the double-angle Trigonometric identities: #L=1/2(sqrt50-sqrt2-sinh^(-1)7+sinh^(-1)1)+2intsec^2theta/(2tantheta+1-tan^2theta)d theta# Apply the substitution #tantheta=v#: #L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+2int_(sqrt2-1)^((5sqrt2-1)/7)1/(2v+1-v^2)dv# Factorize the denominator: #L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+2int_(sqrt2-1)^((5sqrt2-1)/7)1/((sqrt2-1+v)(sqrt2+1-v))dv# Apply partial fraction decomposition: #L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2int_(sqrt2-1)^((5sqrt2-1)/7)(1/(sqrt2-1+v)+1/(sqrt2+1-v))dv# Integrate directly: #L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2[ln|sqrt2-1+v|-ln|sqrt2+1-v|]_(sqrt2-1)^((5sqrt2-1)/7)# Simplify: #L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2[ln|(sqrt2-1+v)/(sqrt2+1-v)|]_(sqrt2-1)^((5sqrt2-1)/7)# Insert the limits of integration: #L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2ln|(12sqrt2-8)/(2sqrt2+8)*2/(2sqrt2-2)|# Simplify: #L=2sqrt2+1/2(sinh^(-1)1-sinh^(-1)7)+1/sqrt2ln2#

Evelyn Agard · Answer

undefined To find the arc length of the curve $ (t^2 - \ln(t), \ln(t)) $ on the interval $[1, 2]$, you use the formula for arc length: $$ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$

First, find the derivatives of $x$ and $y$ with respect to $t$:
$$ \frac{dx}{dt} = 2t - \frac{1}{t} $$
$$ \frac{dy}{dt} = \frac{1}{t} $$

Now, compute the integrand:
$$ \sqrt{\left(2t - \frac{1}{t}\right)^2 + \left(\frac{1}{t}\right)^2} $$

Simplify the expression inside the square root:
$$ \sqrt{4t^2 - 4 + \frac{1}{t^2} + \frac{1}{t^2}} $$
$$ = \sqrt{4t^2 + \frac{2}{t^2} - 4} $$

Now integrate this expression from $t = 1$ to $t = 2$:
$$ L = \int_{1}^{2} \sqrt{4t^2 + \frac{2}{t^2} - 4} \, dt $$