What is the arclength of #f(x)=sqrt(x+3)# on #x in [1,3]#?

Question

Paisley Johnson · Accepted Answer

#L=-1/4ln((5-2sqrt16)/(-4+sqrt(17)))+5/2sqrt6-sqrt17~=2.0500#

To solve this problem we should use the formula #L=int_a^b sqrt(1+(f^'(x)^2))*dx#

#f(x)=sqrt(x+3)# #f^'(x)=(1/2)(x+3)^(-1/2)# #f^'(x)=(1/2)(x+3)^(-1)#

So we have the indefinite integral that solves the problem: #F(x)=int sqrt (1-1/4(1/(x+3)))*dx=1/2int sqrt(4-(1/(x+3)))*dx#

Lets begin by using a trigonometrical substitution: #(1/(x+3))=4*tan^2y# #-dx/(x+3)^2*dx=4*2tanysec^2y*dy# #-> -16tan^4y*dx=8tanysec^2y*dy# => #dx=-(sec^2y)/(2tan^3y)dy#

Then the indefinite integral becomes #1/2int cancel(2)secy(-(sec^2y)/(cancel(2)tan^3y))dy=-1/2int sec^3y/(tan^3y)dy# #=-1/2int 1/cancel(cos^3y)(cancel(cos^3y)/sin^3y)*dy=-1/2 int csc^3y*dy#

We find this last integral in the more complete tables. Anyway, I'll solve it: #-1/2int csc^3y*dy=-1/2int cscy*dy-1/2int cscy*cot^2y*dy#

I call this last result as expression 1, and I go on to solve its last term: #int cscy*cot^2y*dy=int (1/(siny))(cos^2y/sin^2y)dy# Making #sin y=z# #cosy*dy=dz# So #=int sqrt(1-z^2)/z^3dz# Using the rule

#int udv=uv-int vdu# #-> u=(1-z^2)^(1/2)# => #du=-2z(1/2)(1-z^2)^(-1/2)=-z(1-z^2)^(-1/2)# #-> dv=dz/z^3# => #v=-1/(2z^2)# we get #=-(1-z^2)^(1/2)/(2z^2)-1/2int dz/(z*sqrt(1-z^2))dz#

Calling the result above as expression 2 and proceeding to solve its last term #-1/2int dz/(z*sqrt(1-z^2))dz=# #z=sin alpha# #dz=cos alpha*dalpha# Therefore #=-1/2int cancel(cos alpha)/(sin alpha*cancel(cos alpha))dalpha=1/2 int csc alpha*dalpha# #=-1/2*ln|csc alpha-cot alpha|=-1/2*ln|1/z-sqrt(1-z^2)/z|=-1/2*ln|(1-sqrt(1-z^2))/z|#

So

#F(x)=-1/4*ln|sqrt(4x+13)-2sqrt(x+3)|+(1/4)2sqrt(x+3)*sqrt(4x+13)# #F(x)= -1/4*ln|sqrt(4x+13)-2sqrt(x+3)|+1/2*sqrt((x+3)(4x+13))#

Finally #L=F(x=3)-F(x=1)# #L=-ln(sqrt 25-2sqrt6)/4+sqrt(6*25)/2-((-ln(sqrt(17)-2sqrt4))/4+sqrt(4*17)/2)# #L=-1/4ln((5-2sqrt16)/(-4+sqrt(17)))+5/2sqrt6-sqrt17~=2.0500#

Axel Alvarez · Answer

undefined To find the arc length of $ f(x) = \sqrt{x + 3} $ on the interval $ x $ in $[1, 3]$, use the arc length formula:

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

1. Calculate the derivative of $ f(x) = \sqrt{x + 3} $ to get $ f'(x) $.

2. Substitute $ f'(x) $ into the arc length formula.

3. Integrate the expression from $ x = 1 $ to $ x = 3 $.

4. Evaluate the integral to find the arc length.

Therefore, compute $ f'(x) $, substitute it into the arc length formula, integrate from $ x = 1 $ to $ x = 3 $, and evaluate the integral to find the arc length of $ f(x) = \sqrt{x + 3} $ on the interval $ x $ in $[1, 3]$.