Sketch the surface z = 3y^2 - 2x^2. The traces in the vertical planes x = k are

Question

Sketch the surface z = 3y^2 - 2x^2. The traces in the vertical planes x = k are the parabolas z = which open upward. The traces in y = k are the parabolas z = which open downward. The horizontal traces are = k, a family of hyperbolas. We draw the family of traces in Figure 2, and we show how the traces appear when placed in their correct planes in Figure 3. In Figure 1 we fit together the terms to form the surface z = 3y^2 - 2x^2, a hyperbolic paraboloid. Notice that the shape of the surface near the origin resembles that of a saddle. This surface will be investigated further in a later section when we discuss saddle points.

Explanation / Answer

Given z = 3y2 - 2x2

First, x = k

z = 3y2 - 2k2

(z + 2k2)/3 = y2

Which is a parabola in z-y plane which opens upward.

second,. y = k

   z = 3k2 - 2x2

.   z = -2x2 + 3k2

. (z - 3k2)/2 = -x2

which is a parabola in z-x plane which open downward.

Third, to become a hyperbolic equation, z has to be constant.

So z = k

so that 3y2 - 2x2 = k could be an equation of hyperbola.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.