How Do You Model a Downward-Opening Cone in Multivariable Calculus?

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SUMMARY

The formula for a downward-opening cone with its vertex at the origin in multivariable calculus is given by the equation z = -(x^2 + y^2). This equation effectively describes the cone's shape, ensuring that the domain of z is restricted to negative values. To achieve circular cross-sections, one can set z to a constant value, resulting in the equation of a circle in the x-y plane. The discussion confirms that the transformation of the standard cone equation does not alter its fundamental properties.

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Homework Statement


Give a formula for a function whose graph is described. Sketch it using a computer or a calculator.

A cone of circular cross-section opening downward and with its vertex at the origin.

Homework Equations


I know how to draw the cone and the properties of a cone, so that's not the issue here.

I know the formula of a cone is x^2 + y^2 = z^2 but the problem asks for the cone opening downward with the vertex at the origin (so the domain of z would be [0,-inf)). So would the formula simply be -(x^2 + y^2) = -z^2? Furthermore, since they want circular cross-sections, would I insert a constant (i.e. "a") in the formula such that it's something like -(a^2 +y^2) = -z^2?

The Attempt at a Solution


I'm thinking that the formula is -(x^2 + y^2) = -z^2 since they want the cone opening downward with the vertex at the origin.
 
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x2+y2=z2 gives the full cone opening both upwards and downwards. Multiplying both sides by -1 doesn't change that. What you need to do is solve for z and only take negative values.

[Edit] I didn't notice your cross section question. The cross sections are obtained by setting z = constant.
 
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