Expressing h in terms of x Cone/Sphere non-calc

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In summary, to express h in terms of x for the volume of a cone and sphere, the equation is h = 32x. This is obtained by equating the volumes of the two objects and solving for h in terms of x. The volumes are equal when h = 32x.
  • #1
thomas49th
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How do I do:
Question21002.jpg


the formula for volume of a cone is [tex]\frac{1}{3} \pi r^{2}h[/tex]
and for sphere volume is [tex]\frac{4}{3} \pi r^{3}[/tex]

It says express h in terms of x, so the equation will look somthing like

x =

Can someone give me help on how to rearrange this formula?
Thx :biggrin:
 
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  • #2
equate the volumes and replace r with appropriate values
 
  • #3
thomas49th said:
How do I do:

the formula for volume of a cone is [tex]\frac{1}{3} \pi r^{2}h[/tex]
and for sphere volume is [tex]\frac{4}{3} \pi r^{3}[/tex]

It says express h in terms of x, so the equation will look somthing like

x =

Can someone give me help on how to rearrange this formula?

You don't have an equation to rearrange yet! You need to express the volumes of the cone and sphere in terms of the variables you are given in this question. Note that "r" for the cone is not equal to "r" for the sphere in this case. What is each one equal to?

Once you have found this, you know that the volumes of the two objects are equal. This should enable you to obtain an equation.
 
  • #4
[tex]v = \frac{1}{3} \pi x^{2}h[/tex]


[tex]v = \frac{4}{3} \pi (2x)^{3}[/tex]
 
  • #5
Ok, now the volumes are the same, so can you combine these into one equation?
 
  • #6
what u need to do now is to equal the two right sides, like doing v=v, just substitute the both expressions for v. then try to multiply by 3 and then divide by (pi*x^2) and you will get h in terms of x.
 
  • #7
thomas49th said:
It says express h in terms of x, so the equation will look somthing like

x =


Thx :biggrin:

and the equation will look something like

h=
 
  • #8
[tex]\frac{4}{3} \pi x^{3}[/tex] = [tex]\frac{1}{3} \pi x^{2}h[/tex]


[tex]\frac{4}{3} \pi 8x^{3}[/tex] = [tex]\frac{1}{3} \pi x^{2}h[/tex]

divide by [tex]\frac{1}{3} \pi x^{2}[/tex]

gives

[tex]h = 32x[/tex]

is that right?
 
  • #9
I did the math and that's what I got too.
 
  • #10
yeah i guess that's right. good job
 
  • #11
how come I can't edit these posts anymore... the edit button is gone. I can do it with other threads. I want to edit post 8, change [tex]\frac{4}{3} \pi x^{3}[/tex]
to [tex]\frac{4}{3} \pi 2x^{3}[/tex]

EDIT: I can edit post, but not my previous
 
  • #12
thomas49th said:
how come I can't edit these posts anymore... the edit button is gone. I can do it with other threads. I want to edit post 8, change [tex]\frac{4}{3} \pi x^{3}[/tex]
to [tex]\frac{4}{3} \pi 2x^{3}[/tex]

EDIT: I can edit post, but not my previous

You can only edit upto 24 hours after posting.
 

1. How do you express h in terms of x for a cone?

To express h in terms of x for a cone, use the formula h = (x/r) * h, where r is the radius of the base of the cone.

2. What is the formula for expressing h in terms of x for a sphere?

The formula for expressing h in terms of x for a sphere is h = √(r^2 - x^2), where r is the radius of the sphere and x is the distance from the center of the sphere to the point on the surface where the height h is measured.

3. Can I use this formula for both right and oblique cones?

Yes, the formula for expressing h in terms of x for a cone can be used for both right and oblique cones. However, for oblique cones, r represents the radius of the base projected onto the plane perpendicular to the axis of the cone.

4. How do you find the radius of a cone or sphere?

The radius of a cone or sphere can be found by measuring the distance from the center of the base (for a cone) or the center of the sphere to any point on the surface. Alternatively, you can use the given dimensions of the cone or sphere to calculate the radius using a specific formula.

5. Why is it important to express h in terms of x for cones and spheres?

Expressing h in terms of x allows us to find the height of a cone or sphere at any given point on its surface, which is useful in various applications such as engineering, architecture, and physics. It also makes it easier to calculate volume and surface area for these shapes.

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