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In Exercises 49-53 sketch the parametric curve and find its length.

x=5+2t,y=et+e-t,tโˆˆ[0,1]

Short Answer

Expert verified

The graphical representation of the points (5,2)(7,3,07)is as follows,

The length of the curve is equal to localid="1654248352139" e-1e

Step by step solution

01

Given information

The parametric curve is x=5+2t,y=et+e-t,tโˆˆ[0,1]

02

Calculation

Consider the parametric equations x=5+2t,y=e'+e-t,tโˆˆ[0,1].

The objective is to draw the parametric curve and find the arc length of the curve.

The formula to find the arc length of the curve is,

Length of the curve =โˆซabf'(t)2+g'(t)2dt

First, find the derivative of the parametric equations x=5+2t,y=e'+e-t.

Takex=5+2t

That is f(t)=5+2t

The derivative with respect to t is written as follows,

f'(t)=5+2t

On simplifying.

f'(t)=ddt(5+2t)[sinced(uv)=udv+vdu]f'(t)=ddt5+ddt2tf'(t)=0+2f'(t)=2

Take y=e'+e-t

That isg(t)=e'+e-t

The derivative with respect to t is written as follows,

ddtg(t)=ddte'+e-tddtg(t)=ddte'+ddte-t

Thus,

g'(t)=et-e-t

Now by using the Length of the curve =โˆซabf'(t)2+g'(t)2dt.

Length of the curve =โˆซ01(2)2+e'-e-t2dt

=โˆซ014+e2t+e-2t-2e2tยทe-2tdt

On further simplification,

Length of the curve =e1-e-1-e1+e'

=โˆซ01e'+e-t2dt=โˆซ01e'+e-tdt

Thus,

Length of the curve=โˆซ01e'dt+โˆซ01e-tdt

Length of the curve =e'-e-101

By substituting the limits,

Length of the curve =e1-e-1-e0+e0

Then

Length of the curve =e1-e-1-e1+e'

=e1-e-1-e1+e'

=e1-e-1=e-1e

To draw the curve for the parametric equations, first, find the points when tโˆˆ[0,1].

Given that tโˆˆ[0,1]

When t=0,

(x,y)=5+2t,e'+e-t(x,y)=5+2ยท0,e0+e-0(x,y)=(5,2)

When t=1,

Then

(x,y)=5+2t,e'+e-t(x,y)=5+2ยท1,e'+e-1(x,y)=(7,3.07)

The graphical representation of the points (5,2)(7,3,07)is as follows,

Therefore, the length of the curve is equal to localid="1654248367060" e-1e

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