using integration find the length of the curve y=3-x from (-1,4) to (3,0)
Answers
Step-by-step explanation:
What is the length of the curve y=3-x from (-1,4) to (3,0) using integration?
You should use the distance formula. No matter what your school wants you to use, the best way to solve this problem is to use the distance formula.
L=[3−(−1)]2+[0−4]2−−−−−−−−−−−−−−−−−√
⟹L=42–√
The point of asking you to solve it by integration is probably just to get you used to the method.
The length of a curve C in the Cartesian plane with the equation y=f(x) is given by the following formula.
L=∫C1+(dydx)2−−−−−−−−−−√dx
The derivative of the ‘curve’ y=3−x is a constant.
[math]\dfrac{\mathrm{d}y}{\mathrm{d}[/math]
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I will use vector calculus. Let us parametrize the equation.
Let x=t,y=3−t
Therefore, any point on given line will be given by,
r(t)=ti+(3−t)j
Differentiating,
r′(t)=i−j
r′.r′=1×1+(−1)(−1)=2
Now, calling the arc length function,
s(t)=∫3−1r′.r′−−−−√dt
=[2–√t]3−1=2–√(3−(−1))=42–√
answer: 2-1 please check and explain that what's the correct information