<nobr id="r1bpv"><listing id="r1bpv"><menuitem id="r1bpv"></menuitem></listing></nobr>
<pre id="r1bpv"></pre>
      <address id="r1bpv"></address>

      <em id="r1bpv"><sub id="r1bpv"><video id="r1bpv"></video></sub></em> <em id="r1bpv"><address id="r1bpv"></address></em>
        <th id="r1bpv"><noframes id="r1bpv">

          <meter id="r1bpv"></meter>
            Paul's Online Notes
            Paul's Online Notes
            Home / Calculus III / Applications of Partial Derivatives / Tangent Planes and Linear Approximations
            Show Mobile Notice Show All Notes Hide All Notes
            Mobile Notice
            You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

            Section 3-1 : Tangent Planes and Linear Approximations

            Earlier we saw how the two partial derivatives \({f_x}\) and \({f_y}\) can be thought of as the slopes of traces. We want to extend this idea out a little in this section. The graph of a function \(z = f\left( {x,y} \right)\) is a surface in \({\mathbb{R}^3}\)(three dimensional space) and so we can now start thinking of the plane that is “tangent” to the surface as a point.

            Let’s start out with a point \(\left( {{x_0},{y_0}} \right)\) and let’s let \({C_1}\) represent the trace to \(f\left( {x,y} \right)\) for the plane \(y = {y_0}\) (i.e. allowing \(x\) to vary with \(y\) held fixed) and we’ll let \({C_2}\) represent the trace to \(f\left( {x,y} \right)\) for the plane \(x = {x_0}\) (i.e. allowing \(y\) to vary with \(x\) held fixed). Now, we know that \({f_x}\left( {{x_0},{y_0}} \right)\) is the slope of the tangent line to the trace \({C_1}\) and \({f_y}\left( {{x_0},{y_0}} \right)\) is the slope of the tangent line to the trace \({C_2}\). So, let \({L_1}\) be the tangent line to the trace \({C_1}\) and let \({L_2}\) be the tangent line to the trace \({C_2}\).

            The tangent plane will then be the plane that contains the two lines \({L_1}\) and \({L_2}\). Geometrically this plane will serve the same purpose that a tangent line did in Calculus I. A tangent line to a curve was a line that just touched the curve at that point and was “parallel” to the curve at the point in question. Well tangent planes to a surface are planes that just touch the surface at the point and are “parallel” to the surface at the point. Note that this gives us a point that is on the plane. Since the tangent plane and the surface touch at \(\left( {{x_0},{y_0}} \right)\) the following point will be on both the surface and the plane.

            \[\left( {{x_0},{y_0},{z_0}} \right) = \left( {{x_0},{y_0},f\left( {{x_0},{y_0}} \right)} \right)\]

            What we need to do now is determine the equation of the tangent plane. We know that the general equation of a plane is given by,

            \[a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right) + c\left( {z - {z_0}} \right) = 0\]

            where \(\left( {{x_0},{y_0},{z_0}} \right)\) is a point that is on the plane, which we have. Let’s rewrite this a little. We’ll move the \(x\) terms and \(y\) terms to the other side and divide both sides by \(c\). Doing this gives,

            \[z - {z_0} = - \frac{a}{c}\left( {x - {x_0}} \right) - \frac{b}{c}\left( {y - {y_0}} \right)\]

            Now, let’s rename the constants to simplify up the notation a little. Let’s rename them as follows,

            \[A = - \frac{a}{c}\hspace{0.25in}B = - \frac{b}{c}\]

            With this renaming the equation of the tangent plane becomes,

            \[z - {z_0} = A\left( {x - {x_0}} \right) + B\left( {y - {y_0}} \right)\]

            and we need to determine values for \(A\) and \(B\).

            Let’s first think about what happens if we hold \(y\) fixed, i.e. if we assume that \(y = {y_0}\). In this case the equation of the tangent plane becomes,

            \[z - {z_0} = A\left( {x - {x_0}} \right)\]

            This is the equation of a line and this line must be tangent to the surface at \(\left( {{x_0},{y_0}} \right)\) (since it’s part of the tangent plane). In addition, this line assumes that \(y = {y_0}\) (i.e. fixed) and \(A\) is the slope of this line. But if we think about it this is exactly what the tangent to \({C_1}\) is, a line tangent to the surface at \(\left( {{x_0},{y_0}} \right)\) assuming that \(y = {y_0}\). In other words,

            \[z - {z_0} = A\left( {x - {x_0}} \right)\]

            is the equation for \({L_1}\) and we know that the slope of \({L_1}\) is given by \({f_x}\left( {{x_0},{y_0}} \right)\). Therefore, we have the following,

            \[A = {f_x}\left( {{x_0},{y_0}} \right)\]

            If we hold \(x\) fixed at \(x = {x_0}\) the equation of the tangent plane becomes,

            \[z - {z_0} = B\left( {y - {y_0}} \right)\]

            However, by a similar argument to the one above we can see that this is nothing more than the equation for \({L_2}\) and that it’s slope is \(B\) or \({f_y}\left( {{x_0},{y_0}} \right)\). So,

            \[B = {f_y}\left( {{x_0},{y_0}} \right)\]

            The equation of the tangent plane to the surface given by \(z = f\left( {x,y} \right)\) at \(\left( {{x_0},{y_0}} \right)\) is then,

            \[z - {z_0} = {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)\]

            Also, if we use the fact that \({z_0} = f\left( {{x_0},{y_0}} \right)\) we can rewrite the equation of the tangent plane as,

            \[\begin{align*}z - f\left( {{x_0},{y_0}} \right) & = {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)\\ z & = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)\end{align*}\]

            We will see an easier derivation of this formula (actually a more general formula) in the next section so if you didn’t quite follow this argument hold off until then to see a better derivation.

            Example 1 Find the equation of the tangent plane to \(z = \ln \left( {2x + y} \right)\) at \(\left( { - 1,3} \right)\).
            Show Solution

            There really isn’t too much to do here other than taking a couple of derivatives and doing some quick evaluations.

            \[\begin{align*}f\left( {x,y} \right) & = \ln \left( {2x + y} \right)\hspace{0.25in}& {z_0}& = f\left( { - 1,3} \right) = \ln \left( 1 \right) = 0\\ {f_x}\left( {x,y} \right) & = \frac{2}{{2x + y}}\hspace{0.25in} &{f_x}\left( { - 1,3} \right) & = 2\\ {f_y}\left( {x,y} \right) & = \frac{1}{{2x + y}}\hspace{0.25in} & {f_y}\left( { - 1,3} \right) & = 1\end{align*}\]

            The equation of the plane is then,

            \[\begin{align*}z - 0 & = 2\left( {x + 1} \right) + \left( 1 \right)\left( {y - 3} \right)\\ z & = 2x + y - 1\end{align*}\]

            One nice use of tangent planes is they give us a way to approximate a surface near a point. As long as we are near to the point \(\left( {{x_0},{y_0}} \right)\) then the tangent plane should nearly approximate the function at that point. Because of this we define the linear approximation to be,

            \[L\left( {x,y} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)\]

            and as long as we are “near” \(\left( {{x_0},{y_0}} \right)\) then we should have that,

            \[f\left( {x,y} \right) \approx L\left( {x,y} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\left( {x - {x_0}} \right) + {f_y}\left( {{x_0},{y_0}} \right)\left( {y - {y_0}} \right)\]
            Example 2 Find the linear approximation to \(z = 3 + \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9}\) at \(\left( { - 4,3} \right)\).
            Show Solution

            So, we’re really asking for the tangent plane so let’s find that.

            \[\begin{align*}f\left( {x,y} \right) & = 3 + \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9}\hspace{0.25in} & f\left( { - 4,3} \right) & = 3 + 1 + 1 = 5\\ {f_x}\left( {x,y} \right) & = \frac{x}{8}\hspace{0.25in} & {f_x}\left( { - 4,3} \right) & = - \frac{1}{2}\\ {f_y}\left( {x,y} \right) & = \frac{{2y}}{9}\hspace{0.25in} & {f_y}\left( { - 4,3} \right) & = \frac{2}{3}\end{align*}\]

            The tangent plane, or linear approximation, is then,

            \[L\left( {x,y} \right) = 5 - \frac{1}{2}\left( {x + 4} \right) + \frac{2}{3}\left( {y - 3} \right)\]

            For reference purposes here is a sketch of the surface and the tangent plane/linear approximation.

            This is a graph with the standard 3D coordinate system.  The positive z-axis is straight up, the positive x-axis moves off to the left and slightly downward and positive y-axis move off the right and slightly downward.  This is a cup shaped object that is centered on the z-axis, starts at z=3 and opens upwards.  At the point (-4, 3, 5) there is a plane that is tangent to the cup shaped object.
            天天干夜夜爱 天天色播 天天射天天舔 <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <蜘蛛词>| <文本链> <文本链> <文本链> <文本链> <文本链> <文本链>