Question

Show that the surfaces \(z=x^{2} y\) and \(y=\frac{1}{4} x^{2}+\frac{3}{4}\) intersect at \((1,1,1)\) and have perpendicular tangent planes there.

Short answer

The surfaces intersect at (1,1,1) and have perpendicular tangent planes there.

Step-by-step solution

Verify the Point of Intersection

To find if the point \((1,1,1)\) lies on the surfaces, substitute \(x=1\), \(y=1\) into both surface equations.**For the first surface:**Substitute into \(z = x^2y\):\(z = 1^2 \times 1 = 1\)**For the second surface:**Substitute into \(y = \frac{1}{4}x^2 + \frac{3}{4}\):\(1 = \frac{1}{4} \times 1^2 + \frac{3}{4} = 1\).Since both calculations satisfy the equations, the point \((1,1,1)\) is indeed on both surfaces.

Find the Gradient of Surface Equations

The gradient is used to find the normal vector of the tangent plane.**For the surface \(z = x^2y\):**The implicit form is \(f(x,y,z) = x^2y - z = 0\).Calculate the gradient, \(abla f = (f_x, f_y, f_z)\):\[f_x = 2xy, \quad f_y = x^2, \quad f_z = -1\].**For the surface \(y = \frac{1}{4}x^2 + \frac{3}{4}\):**The implicit form is \(g(x,y) = y - \frac{1}{4}x^2 - \frac{3}{4} = 0\).Calculate the gradient, \(abla g = (g_x, g_y)\): \[g_x = -\frac{1}{2}x, \quad g_y = 1\].

Evaluate the Gradients at the Point (1,1,1)

**For the surface \(z = x^2y\) at \((1,1,1)\):**\[abla f(1,1,1) = (2 \times 1 \times 1, 1^2, -1) = (2, 1, -1)\].**For the surface \(y = \frac{1}{4}x^2 + \frac{3}{4}\) at \((1,1,1)\):**Since \(z\) is not part of this surface, augment the gradient by 0 for the \(z\) dimension:\[abla g(1,1) = (-\frac{1}{2} \times 1, 1, 0) = (-\frac{1}{2}, 1, 0)\].

Check if the Normal Vectors are Perpendicular

To determine if the surfaces have perpendicular tangent planes at \((1,1,1)\), check if the dot product of their normal vectors is zero.Calculate the dot product of \(abla f\) and \(abla g\):\[(2, 1, -1) \cdot \left(-\frac{1}{2}, 1, 0\right) = 2 \left(-\frac{1}{2}\right) + 1(1) + (-1)(0)\].This simplifies to:\[-1 + 1 + 0 = 0\].Since the dot product is zero, the normal vectors are perpendicular, indicating the tangent planes are also perpendicular at the point.

Key concepts

Intersection of Surfaces

The intersection of surfaces refers to the points where two or more geometric surfaces meet or cross each other. In this context, we're dealing with two specific surfaces given by equations. The equation \( z = x^2 y \) represents one surface, whereas \( y = \frac{1}{4} x^2 + \frac{3}{4} \) represents the other. To determine if they intersect at a certain point, we substitute the coordinates of that point into both equations.
For instance, the point \((1,1,1)\) is tested by plugging \( x = 1 \), \( y = 1 \), and \( z = 1 \) into both surface equations. If both equations hold true (i.e., are satisfied), then the point \((1,1,1)\) indeed lies on the intersection of the surfaces. Such intersections can reveal important behavioral or characteristic similarities between the surfaces at that point. This forms the basis for further geometric explorations, such as analyzing tangent planes.

Gradient Vectors

Gradient vectors are essential in analyzing surfaces as they provide the necessary information about the slope of the surface at any given point.
In mathematics, the gradient vector is represented as \( abla f \), where \( f \) is a function defining the surface.

• For a surface expressed as \( f(x,y,z) = 0 \), the gradient vector \( abla f \) is derived from its partial derivatives: \( (f_x, f_y, f_z) \).
• The gradient vector gives insight into the direction of the steepest ascent of the surface from the point.

To calculate the gradient vectors of the surfaces \( z = x^2y \) and \( y = \frac{1}{4}x^2 + \frac{3}{4} \), we derive the partial derivatives with respect to \( x, y, \) and \( z \) as necessary. These gradients tell us how quickly our surfaces change in each direction, which is crucial in understanding their interactions, especially at specific intersections.

Dot Product

The dot product is a fundamental tool in vector calculus, particularly when working with vectors to determine angles and orthogonality. For two vectors \( \mathbf{a} \) and \( \mathbf{b} \), their dot product is calculated as \( \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \).
In the case of normal vectors obtained from gradient vectors of the surfaces, the dot product helps establish if the planes are perpendicular. When the dot product of two vectors equals zero, the vectors are said to be orthogonal or perpendicular to each other.
In our specific problem, the gradient vectors evaluated at \((1,1,1)\) resulted in the dot product being zero, confirming perpendicularity of the tangent planes to the surfaces at this point.

Normal Vectors

Normal vectors are vectors that are perpendicular to a surface at a given point and are derived from gradient vectors. They are crucial in studying the orientation and slope of surfaces at points of interest.
For a surface described by a function \( f \), the normal vector at a point is the gradient vector \( abla f \).
These vectors are vital in determining the perpendicularity of tangent planes of intersecting surfaces. If two surfaces have normal vectors that are perpendicular to each other (as indicated by a dot product of zero), their tangent planes at the intersection point are also perpendicular. This perpendicularity can be important in applications like physics and engineering, where such properties affect balance, stability, and force interactions.