Respuesta :
Let the curve C be the intersection of the cylinder
[tex]16x^2+y^2=16[/tex]
and the plane
[tex]x+y+z=1[/tex]
The projection of C on to the x-y plane is the ellipse
[tex]16x^2+y^2=16[/tex]
To see clearly that this is an ellipse, le us divide through by 16, to get
[tex]\frac{x^2}{1}+ \frac{y^2}{16}=1[/tex]
or
[tex]\frac{x^2}{1^2}+ \frac{y^2}{4^2}=1[/tex],
We can write the following parametric equations,
[tex]x=cos(t), y=4sin(t)[/tex]
for
[tex]0\le t \le 2\pi[/tex]
Since C lies on the plane,
[tex]x+y+z=1[/tex]
it must satisfy its equation.
Let us make z the subject first,
[tex]z=1-x-y[/tex]
This implies that,
[tex]z=1-sin(t)-4cos(t)[/tex]
We can now write the vector equation of C, to obtain,
[tex]r(t)=(cos(t),4sin(t),1-cos(t)-4sin(t)) [/tex]
The length of the curve of the intersection of the cylinder and the plane is now given by,
[tex]\int\limits^{2\pi}_0 {|r'(t)|} \, dt[/tex]
But
[tex]r'(t)=(-sin(t),4cos(t),sin(t)-4cos(t))[/tex]
[tex]|r'(t)|=\sqrt{(-sin(t))^2+(4cos(t))^2+(sin(t)-4cos(t))}[/tex]
[tex]\int\limits^{2\pi}_0 {\sqrt{2sin^2(t)+32cos(t)-8sin(t)cos(t)} }\, dt=24.08778184[/tex]
Therefore the length of the curve of the intersection intersection of the cylinder and the plane is 24.0878 units correct to four decimal places.


The length of the curve is 24.08 units.
Given
The equation of a cylinder, a natural choice for a parameterization would be one utilizing cylindrical coordinates.
[tex]\rm 16x^2 + y^2 = 16\\\\x^2+\dfrac{y^2}{4^2}=1\\\\x+y+z=1[/tex]
What is the parametric equation?
A parametric equation for the curve that is traced out by varying the values of the parameter t.
The parametric equation x = cost and y = 4sint (0 ⩽ t ⩽ 2π) traces out a circle.
From equation 1;
[tex]\rm x+y+z=1\\\\z=1-x-y\\\\z=1-4sint-cost[/tex]
Now use the arc length formula:
[tex]\rm Length=\int\limits^{2\pi }_0 { \sqrt{\left (\dfrac{dx}{dt}\right)^2+\left (\dfrac{dy}{dt}\right)^2+\left (\dfrac{dz}{dt}\right)^2 dt}}}\\\\ Length=\int\limits^{2\pi }_0 { \sqrt{sin2t+16cos^2t+(sint-4cost)^2}}\\\\Length = 24.08[/tex]
Hence, the length of the curve is 24.08 units.
To know more about the Parametric equations click the link given below.
https://brainly.com/question/9056657