How do you find the arc length of the curve #y=lnx# over the interval [1,2]?

Answer 1

Apply the arc length formula.

#y=lnx#
#y'=1/x#

Arc length is given by:

#L=int_1^2sqrt(1+1/x^2)dx#

Rearrange:

#L=int_1^2sqrt(x^2+1)/xdx#
Multiply numerator and denominator by #sqrt(x^2+1)#:
#L=int_1^2(x^2+1)/(xsqrt(x^2+1))dx#

Integration is distributive:

#L=int_1^2x/sqrt(x^2+1)dx+int_1^2 1/(xsqrt(x^2+1))dx#
Apply the substitution #x=tantheta#:
#L=[sqrt(x^2+1)]_ 1^2+intsectheta/tanthetad theta#

Simplify:

#L=sqrt5-sqrt2+intcscthetad theta#

Integrate directly:

#L=sqrt5-sqrt2-[ln|csctheta+cottheta|]#
Rewrite in terms of #tantheta# and #sectheta#:
#L=sqrt5-sqrt2-[ln|(1+sectheta)/tantheta|]#

Reverse the substitution:

#L=sqrt5-sqrt2-[ln((1+sqrt(x^2+1))/x)]_1^2#

Insert the limits of integration:

#L=sqrt5-sqrt2-ln((1+sqrt5)/(2(1+sqrt2)))#

Rearrange for clarity:

#L=sqrt5-sqrt2+ln2-ln((1+sqrt5)/(1+sqrt2))#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

Length = #sqrt5+1/2ln((sqrt5-1)/(sqrt5+1))-sqrt2-1/2ln((sqrt2-1)/(sqrt2+1))~~1.222#

The arc length of a function #y=f(x)# over the interval #[a,b]# is given by #L=int_a^bsqrt(1+(dy/dx)^2)dx#. #y=lnx->dy/dx=1/x# #:.L=int_1^2sqrt(1+(1/x)^2)dx# #L=int_1^2sqrt(1+1/x^2)dx# I will use this solution #intsqrt(1+1/x^2)dx=sqrt(x^2+1)+1/2ln|(sqrt(x^2+1)-1)/(sqrt(x^2+1)+1)|+C# courtesy of mason m (https://tutor.hix.ai) to find that #L=sqrt(x^2+1)+1/2ln|(sqrt(x^2+1)-1)/(sqrt(x^2+1)+1)|]_1^2# #:.L=sqrt5+1/2ln((sqrt5-1)/(sqrt5+1))-sqrt2-1/2ln((sqrt2-1)/(sqrt2+1))~~1.222#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To find the arc length of the curve y = ln(x) over the interval [1,2], you use the arc length formula:

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

For the curve y = ln(x), you find (\frac{dy}{dx}), which is (\frac{1}{x}). Then plug this into the formula and integrate from 1 to 2:

[ L = \int_{1}^{2} \sqrt{1 + \left(\frac{1}{x}\right)^2} , dx ]

Solve the integral to find the arc length.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7