Finding Parallel Tangent Planes on a Surface

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In summary, the problem is to find the points on the surface \(4 x^2 + 2 y^2 + 4 z^2 = 1\) at which the tangent plane is parallel to the plane \(4 x - 3 y - 2 z = 0\). To solve this, we take the gradient of both functions and set them equal to each other, resulting in x = 1/2, y = -3/4, z = -1/4. However, we must also consider that the normal vectors should be parallel, not equal, and this can be represented by a constant K. With this understanding, the problem can be solved.
  • #1
EngnrMatt
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Homework Statement



Find the points on the surface \(4 x^2 + 2 y^2 + 4 z^2 = 1\) at which the tangent plane is parallel to the plane \(4 x - 3 y - 2 z = 0\).

Homework Equations



Not sure

The Attempt at a Solution



What I did was take the gradient of both functions, the surface function as f, and the plane as g, so I got ∇f = <8x,4y,8z> and ∇g = <4,-3,-2>. Knowing that these gradients are normal vectors, and the planes would be in parallel, therefore having the same normal vectors, I set the components equal to each other and got x = 1/2, y = -3/4, z = -1/4. However, these are not the points, and I have rune out of ideas on how to work the problem.
 
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  • #2
EngnrMatt said:

Homework Statement



Find the points on the surface \(4 x^2 + 2 y^2 + 4 z^2 = 1\) at which the tangent plane is parallel to the plane \(4 x - 3 y - 2 z = 0\).

Homework Equations



Not sure

The Attempt at a Solution



What I did was take the gradient of both functions, the surface function as f, and the plane as g, so I got ∇f = <8x,4y,8z> and ∇g = <4,-3,-2>. Knowing that these gradients are normal vectors, and the planes would be in parallel, therefore having the same normal vectors, I set the components equal to each other and got x = 1/2, y = -3/4, z = -1/4. However, these are not the points, and I have rune out of ideas on how to work the problem.


The normal vectors should be parallel, not equal. <8x,4y,8z>= K<4,-3,-2>, K is a constant.


ehild
 
  • #3
Alright, I got it. Thank you very much!
 

1. What is a tangential plane?

A tangential plane is a flat surface that is tangent, or just touches, a curved surface at only one point. This point of tangency is perpendicular to the tangent line at that point.

2. What is the tangential planes problem?

The tangential planes problem is a mathematical problem in which the goal is to find the equation of a plane that is tangent to a given curve at a specific point. This problem is commonly encountered in calculus and differential geometry.

3. How is the equation of a tangential plane found?

The equation of a tangential plane can be found by using the derivative of the given curve at the point of tangency. This derivative is used to find the slope of the tangent line, which is then used to determine the coefficients of the equation of the plane.

4. Can the tangential planes problem be solved for any type of curve?

Yes, the tangential planes problem can be solved for any type of curve, including polynomials, trigonometric functions, and exponential functions. However, the method for finding the equation of the tangential plane may differ depending on the type of curve.

5. What are some real-life applications of the tangential planes problem?

The tangential planes problem has many real-life applications, such as in engineering, physics, and computer graphics. For example, it can be used to design curved surfaces for vehicles or to calculate the trajectory of a projectile. In computer graphics, it is used to create smooth and realistic 3D surfaces in video games and animations.

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