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15.6 The Tangent Plane and Linear Approximation

NOT on exam 1. HW due Thu. 2/16

recall if \( y = f(x) \), we can find the linear approx. of \( f(x) \) near \( x = a \)

2D plot showing a curve y=f(x) and its tangent line at x=a.
\[ L \approx f(x) = f(a) + f'(a)(x-a) \]

as long as \( x \) is "near" \( a \), \( L \approx f(x) \)

\( z = f(x,y) \) is a surface and at \( (x_0, y_0, z_0) \), can we find the equivalent of the tangent line approx.?

\( \rightarrow \) tangent plane

3D coordinate system showing a surface z=f(x,y) with a tangent plane at (x0, y0, z0).

the surface is made up of infinitely curves, each having a tangent vector at \( (x_0, y_0, z_0) \)

the tangent plane contains ALL of these tangent vectors

\( \rightarrow \) find vector \( \perp \) to ALL tangent vectors

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the surface is made up of infinitely many curves \( \rightarrow \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \)

each having tangent vector \( \vec{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle \)

find a vector \( \perp \) to these

Surface: \( z = f(x,y) \)

it contains many \( \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \)

so, \( z = f(x,y) \rightarrow z(t) = f(x(t), y(t)) \)

defined a new function \( F(x,y,z) = f(x,y) - z = 0 \)

really a function of \( t \)
Tree diagram for the chain rule: F depends on x, y, and z, which each depend on t.
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by the Chain Rule,

\[ \frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial y} \frac{dy}{dt} + \frac{\partial F}{\partial z} \frac{dz}{dt} = 0 \]

because \( F = f - z = 0 \)

rewrite:

\[ \underbrace{\langle \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \rangle}_{\text{gradient of } F \, \vec{\nabla} F} \cdot \underbrace{\langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \rangle}_{\text{tangent vectors on surface}} = 0 \]

so, \( \vec{\nabla} F = \langle F_x, F_y, F_z \rangle \) is \( \perp \) to all tangent vectors (which form the tangent plane)

the gradient of \( F = f - z \) is the normal vector of the tangent plane

\[ \langle F_x, F_y, F_z \rangle \cdot \langle x-x_0, y-y_0, z-z_0 \rangle = 0 \]

or

\[ F_x(x-x_0) + F_y(y-y_0) + F_z(z-z_0) = 0 \]

where \( F = f - z \)

Geometric sketch of a tangent plane at point (x0, y0, z0) with normal vector grad F.
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if we prefer to use \( z = f(x,y) \) explicitly, then make a minor tweak

\( F = f - z \) so \( F_x = f_x \), \( F_y = f_y \), \( F_z = -1 \)

\[ f_x(x-x_0) + f_y(y-y_0) - (z-z_0) = 0 \]

or

\[ z - z_0 = f_x(x-x_0) + f_y(y-y_0) \]

example

\( z = f(x,y) = \sqrt{x^2 + y^2} \) cone (upper half)

find the tangent plane at \( (3, 4, 5) \)

define \( F = f(x,y) - z = \sqrt{x^2 + y^2} - z \)

\[ \vec{\nabla} F = \langle F_x, F_y, F_z \rangle = \langle \frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}, -1 \rangle \]

\( \vec{\nabla} F(3,4,5) = \langle \frac{3}{5}, \frac{4}{5}, -1 \rangle \)

tangent plane:

\[ \langle \frac{3}{5}, \frac{4}{5}, -1 \rangle \cdot \langle x-3, y-4, z-5 \rangle = 0 \]
\[ \frac{3}{5}(x-3) + \frac{4}{5}(y-4) - (z-5) = 0 \]
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near \( (3, 4, 5) \) we expect the tangent plane \( \approx \) true surface function

Tp: \( z = 5 + \frac{3}{5}(x-3) + \frac{4}{5}(y-4) \)

which should be close to true \( z = f(x,y) = \sqrt{x^2 + y^2} \)

let's try \( x = 3.01, y = 3.99 \)

TP approx: \( z = 5 + \frac{3}{5}(0.01) + \frac{4}{5}(-0.01) = 4.998 \)

true value of \( z \): \( f(3.01, 3.99) = \sqrt{(3.01)^2 + (3.99)^2} = 4.99802 \)

they are close because \( \Delta x, \Delta y \) are small

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Differentials and Tangent Planes

Tangent Plane Equation

\[z - z_0 = f_x(x - x_0) + f_y(y - y_0)\]

Let \(x - x_0 = \Delta x = dx\) (very small change), \(y - y_0 = dy\), and \(z - z_0 = dz\).

Then:

\[dz = f_x dx + f_y dy\]
This tells us the change in \(z\) given changes in \(x\) and \(y\).

Example

\(z = f(x, y) = x^2 y\)

  • \(x\) starts at 1 and ends at 1.01 \(\rightarrow dx = 0.01\)
  • \(y\) starts at 3 and ends at 2.91 \(\rightarrow dy = -0.09\)

Find approx change in \(z \rightarrow dz = ?\)

\[dz = f_x dx + f_y dy\]
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\[dz = (2xy) dx + (x^2) dy\]
Use the starting \(x, y\) because that's where the tangent plane is formed.
\[= (2 \cdot 1 \cdot 3)(0.01) + (1)^2(-0.09) = -0.03\]

This is the estimate of change in \(z = x^2 y\) given \(dx = 0.01\) and \(dy = -0.09\).

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Example: Volume of a Cone

The volume of a cone is \(V = \frac{1}{3} \pi r^2 h\).

A sketch of a cone with a vertical arrow labeled h for height and a horizontal arrow labeled r for radius.

If \(r\) is increased by 1% and \(h\) is decreased by 3%, what is the approx. % change in volume \(V\)?

\[V = \frac{1}{3} \pi r^2 h\]
\[dV = \frac{\partial V}{\partial r} dr + \frac{\partial V}{\partial h} dh\]

(\(dz = f_x dx + f_y dy\))

\[dV = (\frac{2}{3} \pi r h) dr + (\frac{1}{3} \pi r^2) dh\]

Refer to absolute change, NOT relative change.

Do NOT put in 1% for \(dr\) and -3% for \(dh\).

Divide the eq. by \(V = \frac{1}{3} \pi r^2 h\):

\[\frac{dV}{V} = \frac{\frac{2}{3} \pi r h}{\frac{1}{3} \pi r^2 h} dr + \frac{\frac{1}{3} \pi r^2}{\frac{1}{3} \pi r^2 h} dh = 2 \left( \frac{dr}{r} \right) + \left( \frac{dh}{h} \right)\]

Where \(\frac{dV}{V}\) is the relative change in \(V\), and \(\frac{dr}{r}\), \(\frac{dh}{h}\) are relative (%) changes.

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Now we get:

\[\frac{dV}{V} = 2(0.01) + (-0.03) = -0.01 = 1\% \text{ decrease}\]

Note: 0.01 corresponds to a 1% increase, and -0.03 corresponds to a 3% decrease.