The standard kilogram (fig. 1.1a) is a platinum–iridium cylinder 39.0 mm in height and 39.0 mm in diameter. what is the density

Answer:

E(t) = [ 4cos(t), 2 sin(t) ]

Step-by-step explanation:

The equation for the ellipse can be represented as:

 (x² ÷ a² ) + (y² ÷ b² ) = 1    Here, a and b symbolize the semi-major and semi-minor axes.

In this specific instance, we have    x²/16 + y²/ 4 = 1

This expression indicates that it follows the form of (x/a)² + (y/b)² =1

Specifically in our case,    x²/16  + y²/4 = 1, which resembles                 sin²α + cos²α = 1.

By setting x = 4 cos(t), we can proceed to calculate y.

Utilizing the equation x²/16  + y²/4 = 1, we find that:

x²/16  + y²/4 = 1   ⇒    (x²  +  4y²) ÷ 16  = 1

Solving for y yields: (x²  +  4y²)  = 16    ⇒  y²  = ( 16 - x²) ÷4

Substituting x = 4 cos(t) gives us y²  =  (16 - 16cos²(t)) ÷ 4, leading to y = √4 (1 - cos²(t)).

Consequently, we have y = 2 sin(t).

Thus, the vector parameterization of the ellipse can be stated as:

E(t) = [ 4cos(t), 2 sin(t) ]

Answer:

The tangent plane equation for the hyperboloid

.

Step-by-step explanation:

We have

The ellipsoid's equation is

The equation for the tangent plane at the point

 (Given)

The hyperboloid's equation is

F(x,y,z)=

The tangent plane equation at point

The tangent plane equation for the hyperboloid is

The tangent plane equation

Hence, the required tangent plane equation for the hyperboloid is

the answer is c. I completed the review

Answer:

To summarize the answer:

Step-by-step explanation:

Given:

Here is the graph associated with this question:

The second function, denoted as , does not qualify as a function.

Keep in mind that the g(x) function is the inversion of the f(x) function. Recognizing this pattern indicates a reflection on the Y-axis.

Reflection on the axes:

In the x-axis:

Enhance the function by -1 to illustrate an exponential curve around the x-axis.

In the y-axis:

Decrease the input of the function by -1 to depict the exponential function around the y-axis.