Problem 9 - Calculus Early Transcendentals

Question

Find an equation of the plane tangent to the following surfaces at the given points. $$x^{2}+y+z=3 ;(1,1,1) \text { and }(2,0,-1)$$

Step-by-Step Solution

Answer

The equations of the tangent planes to the given surface at the specified points are: At point (1, 1, 1): \(2x + y + z = 4\) At point (2, 0, -1): \(4x + y + z = 7\)

Step 1: Find the partial derivatives of the function

Let's find the partial derivatives of the given surface equation \(x^2 + y + z = 3\). \(\frac{\partial f}{\partial x} = 2x\) \(\frac{\partial f}{\partial y} = 1\) \(\frac{\partial f}{\partial z} = 1\)

Step 2: Find the gradient at the given points

We now find the gradient of the surface at the given points by replacing x, y, and z coordinates in the partial derivatives. For point \((1, 1, 1)\): \(\nabla f (1,1,1) = (2(1), 1, 1) = (2, 1, 1)\) For point \((2, 0, -1)\): \(\nabla f (2,0,-1) = (2(2), 1, 1) = (4, 1, 1)\)

Step 3: Find the normal vector of the tangent plane

Since the gradient at a point on the surface is perpendicular to the tangent plane at that point, we can use the gradient that we found in Step 2 as the normal vector of the tangent plane. For point \((1, 1, 1)\): Normal vector (N) = \((2, 1, 1)\) For point \((2, 0, -1)\): Normal vector (N) = \((4, 1, 1)\)

Step 4: Find the equation of the tangent plane

To find the equation of the tangent plane, we use the equation of the plane given by \(ax + by + cz = d\) and the normal vector components as a, b, and c. Then, we substitute the coordinates of the given point to get the equation of the tangent plane. For point \((1, 1, 1)\): The tangent plane equation is \(2x + y + z = d\). Substituting the point coordinates, we have \(2(1) + 1 + 1 = d\). Hence, d = 4 and the equation for the tangent plane is \(2x + y + z = 4\). For point \((2, 0, -1)\): The tangent plane equation is \(4x + y + z = d\). Substituting the point coordinates, we have \(4(2) + 0 + (-1) = d\). Hence, d = 7 and the equation for the tangent plane is \(4x + y + z = 7\).

Step 5: Final Answer

The equations of the tangent planes to the given surface at the specified points are: At point (1, 1, 1): \(2x + y + z = 4\) At point (2, 0, -1): \(4x + y + z = 7\)

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes as we vary one of its input variables, while keeping the others constant.
They are particularly useful in multivariable calculus, as they allow us to study the surface's behavior in specific directions.
In the problem, we're dealing with the surface equation \(x^2 + y + z = 3\). To find the partial derivatives, we'll differentiate with respect to each variable independently:

• Partial derivative with respect to \(x\), \(\frac{\partial f}{\partial x} = 2x\), tells us how the surface changes as \(x\) changes, holding \(y\) and \(z\) constant.
• Partial derivative with respect to \(y\), \(\frac{\partial f}{\partial y} = 1\), shows the rate of change in the direction of the \(y\)-axis.
• Partial derivative with respect to \(z\), \(\frac{\partial f}{\partial z} = 1\), indicates the change along the \(z\)-axis.

By calculating these derivatives, we establish a foundation for further analysis of the surface's local behavior at specific points.

Gradient Vector

The gradient vector combines all partial derivatives into a vector that points in the direction of the steepest ascent from a point on the surface.
It’s a powerful tool in vector calculus because it not only tells us how much the surface changes but also in which direction.
For a function \(f\), the gradient \(abla f\) is given by \((\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})\).When we compute this for the given surface at specific points:

• At \((1, 1, 1)\), the gradient is \((2, 1, 1)\).
• At \((2, 0, -1)\), the gradient is \((4, 1, 1)\).

These gradient vectors are vital as they serve as normal vectors to the tangent planes at these points. Understanding this concept helps in visualizing the directional behavior of the surface.

Normal Vector

A normal vector is perpendicular to the surface at a given point. In the context of tangent planes, it's used to define the plane's orientation.
The key takeaway is that the gradient vector of a function at a point is always normal (perpendicular) to the surface at that point.
Thus, we use the gradient calculated earlier to serve directly as our normal vector.For example:

• At \((1, 1, 1)\), the normal vector \((N) = (2, 1, 1)\).
• At \((2, 0, -1)\), the normal vector \((N) = (4, 1, 1)\).

These vectors are crucial in drafting the tangent plane equations as they determine the orientation and position of the plane in three-dimensional space.

Surface Equation

The surface equation describes a set of points that form a surface in space.
In the exercise, we began with \(x^2 + y + z = 3\), a simple implicit surface equation.
Surface equations provide a framework for analyzing and understanding the geometry of surfaces.When calculating tangent planes to this surface at specific points, we derive equations from:

• The general plane equation \(ax + by + cz = d\).
• The normal vector gives the coefficients \(a\), \(b\), and \(c\) directly from its components.
• The coordinates of the specific points help solve for \(d\).

For example:

• For point \((1, 1, 1)\), using normal vector \((2, 1, 1)\), the equation is \(2x + y + z = 4\).
• For point \((2, 0, -1)\), with normal vector \((4, 1, 1)\), the equation is \(4x + y + z = 7\).

These equations define planes that are tangent to the surface at the specified points, giving insight into the surface's local geometry.