What is the arclength of #(t-e^(2t),lnt)# on #t in [1,12]#?

Answer 1

The arclength is around #2.64891×10^10#.

First, let's get a sense of how long the arclength might actually be. It will go from

#(1-e^(2*1),ln1)quad~~quad(-6.389,0)#

to

#(12-e^(2*12),ln12)quad~~quad(-2.649×10^10,2.485)#

We can tell that this distance will be very big, so that means the arclength is also very big.

To calculate the actual arclength, we'll need to get an integral in the form of #intsqrt((dx)^2+(dy)^2)#, based on the Pythagorean theorem:

(This link does a great job of explaining arc lengths of parametric curves.)

First, split the parametric function into a function of #t# for #x# and #y#:

#x(t)=t−e^(2t)#

#y(t)=lnt#

Now, differentiate each one and get #dx# and #dy# in terms of #dt#:

#dx=(1-2e^2t)dt#

#dy=1/tdt#

Now, plug these into the aforementioned integral with the appropriate bounds. You'll see that the #dt# can be factored out of the radical:

#color(white)=int_1^12 sqrt( (dx)^2 + (dy)^2 )#

#=int_1^12 sqrt( ((1-2e^2t)dt)^2 + (1/tdt)^2 )#

#=int_1^12 sqrt( (1-2e^2t)^2dt^2 + 1/t^2dt^2 )#

#=int_1^12 sqrt( dt^2((1-2e^2t)^2 + 1/t^2) )#

#=int_1^12 sqrt( (1-2e^2t)^2 + 1/t^2 )# #dt#

At this point, you should probably plug this integral into a calculator because the function probably doesn't have an antiderivative, and even if it does it would be a pain to calculate.

A calculator should spit out something around #2.64891×10^10#, and that's your answer. Hope this helped!

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

To find the arc length of the curve ( (t - e^{2t}, \ln t) ) on the interval ( t \in [1, 12] ), you can use the arc length formula for parametric curves:

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

Given the parametric equations ( x(t) = t - e^{2t} ) and ( y(t) = \ln t ), we need to find ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ) and then evaluate the integral over the given interval.

[ \frac{dx}{dt} = 1 - 2e^{2t} ] [ \frac{dy}{dt} = \frac{1}{t} ]

Now, plug these derivatives into the arc length formula:

[ L = \int_{1}^{12} \sqrt{(1 - 2e^{2t})^2 + \left(\frac{1}{t}\right)^2} , dt ]

You can then integrate this expression over the interval ( t \in [1, 12] ) 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