Find, correct to four decimal places, the length of the curve of intersection of the cylinder 16x2 + y2 = 16 and the plane x + y + z = 1

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.

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;

Now use the arc length formula:

Hence, the length of the curve is 24.08 units.

To know more about the Parametric equations click the link given below.

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Let the curve C be the intersection of the cylinder  

and the plane

The projection of C on to the x-y plane is the ellipse

To see clearly that this is an ellipse, le us divide through by 16, to get

or  

,

We can write the following parametric equations,

for  

Since C lies on the plane,

it must satisfy its equation.

Let us make z the subject first,  

This implies that,

We can now write the vector equation of C, to obtain,

The length of the curve of the intersection of the cylinder and the plane is now given by,

But  

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.

i will pick c put check it over to be sure

step-by-step explanation:

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: