Problem 19
Question
Find an equation of the plane tangent to the following surfaces at the given points. $$z=e^{x y} ;(1,0,1) \text { and }(0,1,1)$$
Step-by-Step Solution
Verified Answer
Question: Find the equations of the tangent planes to the surface \(z = e^{xy}\) at the points \((1, 0, 1)\) and \((0, 1, 1)\).
Answer: The tangent plane equation at the point \((1, 0, 1)\) is \(z = y+1\), and the tangent plane equation at the point \((0, 1, 1)\) is \(z = x+1\).
1Step 1: Find the partial derivatives of the given function
To find the tangent plane, we need the partial derivatives of the function \(z = e^{xy}\). Differentiating the function with respect to \(x\) we get:
$$\frac{\partial z}{\partial x} = y e^{xy}$$
and with respect to \(y\), we get:
$$\frac{\partial z}{\partial y} = x e^{xy}$$
2Step 2: Evaluate the partial derivatives at the given points
Now that we have the partial derivatives, we need to evaluate them at the given points \((1, 0, 1)\) and \((0, 1, 1)\).
For point \((1, 0, 1)\):
$$f_x(1, 0) = 0 \cdot e^{(1)(0)} = 0$$
$$f_y(1, 0) = 1 \cdot e^{(1)(0)} = 1$$
For point \((0, 1, 1)\):
$$f_x(0, 1) = 1 \cdot e^{(0)(1)} = 1$$
$$f_y(0, 1) = 0 \cdot e^{(0)(1)} = 0$$
3Step 3: Find the equations of the tangent planes
Lastly, we need to use the partial derivatives and tangent plane equation to find the desired tangent planes at the given points.
For point \((1, 0, 1)\):
$$z = f(1, 0) + f_x(1, 0)(x-1) + f_y(1, 0)(y-0)$$
$$z = 1 + 0 \cdot (x-1) + 1 \cdot y$$
The tangent plane equation at the point \((1, 0, 1)\) is:
$$z = y+1$$
For point \((0, 1, 1)\):
$$z = f(0, 1) + f_x(0, 1)(x-0) + f_y(0, 1)(y-1)$$
$$z = 1 + 1 \cdot x + 0 \cdot (y-1)$$
The tangent plane equation at the point \((0, 1, 1)\) is:
$$z = x+1$$
Key Concepts
Partial DerivativesSurface EquationEvaluating Derivatives at a Point
Partial Derivatives
In calculus, partial derivatives are used to analyze how a function of multiple variables changes as one of the variables is adjusted, while keeping the others constant. This is crucial for understanding the behavior of multivariable functions.
Imagine a function represented as a surface floating in a 3D space. By taking partial derivatives, we are essentially slicing this surface along one of the coordinate axes to see how it changes at that slice. For example, when we compute the partial derivative of the function \(z = e^{xy}\) with respect to \(x\), denoted as \(\frac{\partial z}{\partial x}\), we are observing how \(z\) changes for small changes in \(x\), assuming \(y\) remains constant. This results in \(\frac{\partial z}{\partial x} = y e^{xy}\). Similarly, the partial derivative with respect to \(y\) yields \(\frac{\partial z}{\partial y} = x e^{xy}\).
These partial derivatives are foundational tools for finding tangent planes and analyzing the local behavior of surfaces.
Imagine a function represented as a surface floating in a 3D space. By taking partial derivatives, we are essentially slicing this surface along one of the coordinate axes to see how it changes at that slice. For example, when we compute the partial derivative of the function \(z = e^{xy}\) with respect to \(x\), denoted as \(\frac{\partial z}{\partial x}\), we are observing how \(z\) changes for small changes in \(x\), assuming \(y\) remains constant. This results in \(\frac{\partial z}{\partial x} = y e^{xy}\). Similarly, the partial derivative with respect to \(y\) yields \(\frac{\partial z}{\partial y} = x e^{xy}\).
These partial derivatives are foundational tools for finding tangent planes and analyzing the local behavior of surfaces.
Surface Equation
The surface equation is your starting point in problems involving geometry and calculus in three dimensions. It represents a surface in the \(xy\)-plane, meaning it defines every point on this surface in 3D space.
In our example, the surface is given by \(z = e^{xy}\). This equation maps out how \(z\) values change based on different \(x\) and \(y\) inputs. This is vital for visualizing the surface's shape, which can be intricate and multidimensional.
By dissecting this equation using partial derivatives, we can derive more valuable insights, such as the equation of the tangent plane. The tangent plane provides a linear approximation of the surface at a specific point, effectively "flattening" the surface locally. Understanding the surface equation is key to navigating the complexities of multivariable calculus.
In our example, the surface is given by \(z = e^{xy}\). This equation maps out how \(z\) values change based on different \(x\) and \(y\) inputs. This is vital for visualizing the surface's shape, which can be intricate and multidimensional.
By dissecting this equation using partial derivatives, we can derive more valuable insights, such as the equation of the tangent plane. The tangent plane provides a linear approximation of the surface at a specific point, effectively "flattening" the surface locally. Understanding the surface equation is key to navigating the complexities of multivariable calculus.
Evaluating Derivatives at a Point
Evaluating derivatives at a specific point is like honing in on a precise location on a vast landscape to understand the immediate terrain.
For our exercise, we need to assess the partial derivatives of the surface \(z = e^{xy}\) at particular points to define the tangent planes. For instance, at the point \((1, 0)\), we find that \(\frac{\partial z}{\partial x} = 0\) and \(\frac{\partial z}{\partial y} = 1\). These results tell us how \(z\) changes with small adjustments in \(x\) and \(y\), exactly at this point, considered alongside the constant point \((1,0)\).
This calculation is essential, as it gives us the slopes we need for constructing the tangent plane. Without evaluating the derivatives at the given points, we would lack the specific information necessary to map the precise nature of the surface at those points. Thus, it's a crucial step in bridging the gap between abstract partial derivatives and their tangible applications in calculus.
For our exercise, we need to assess the partial derivatives of the surface \(z = e^{xy}\) at particular points to define the tangent planes. For instance, at the point \((1, 0)\), we find that \(\frac{\partial z}{\partial x} = 0\) and \(\frac{\partial z}{\partial y} = 1\). These results tell us how \(z\) changes with small adjustments in \(x\) and \(y\), exactly at this point, considered alongside the constant point \((1,0)\).
This calculation is essential, as it gives us the slopes we need for constructing the tangent plane. Without evaluating the derivatives at the given points, we would lack the specific information necessary to map the precise nature of the surface at those points. Thus, it's a crucial step in bridging the gap between abstract partial derivatives and their tangible applications in calculus.
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