11. Partial Derivatives and Tangent Planes

c. Geometric Interpretation of Partial Derivatives

We would like to have a geometrical understanding of partial derivatives. Since a partial derivative is the derivative with respect to one variable while holding the other fixed, it is the slope of the tangent line to the curve obtained by letting one variable vary while holding the other fixed. This curve is called a trace.

Traces and their Tangent Lines

Consider the graph of a function of two variables: $z=f(x,y)$

$x$ -Trace, its Tangent Line and the $x$ -Partial Derivative
Fix a value of $y$ , say $y=b$ . This determines a vertical plane parallel to the $xz$ -plane that contains every point whose $y$ -coordinate is $b$ . The graph of the function $f(x,y)$ intersects this plane in a curve called the $x$ -trace with $y=b$ . Since $y=b$ , the $x$ -trace is: $z=f(x,b)$ The $x$ -partial derivative of $f$ at $x=a$ with $y$ held fixed at $y=b$ , denoted $f_x(a,b)$ , is the to the $x$ -trace at $y=b$ . So the tangent line to the $x$ -trace at $(a,b)$ is: $z=f(a,b)+f_x(a,b)(x-a)$

Recall that the slope of a function $z=f(x)$ at $x=a$ is $f'(a)$ and the tangent line is $z=f(a)+f'(a)(x-a)$ . For a function of $2$ variables, the $y$ coordinate $y=b$ just comes along for the ride.

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Similarly:

$y$ -Trace, its Tangent Line and the $y$ -Partial Derivative
Fix a value of $x$ , say $x=a$ . This determines a vertical plane parallel to the $yz$ -plane that contains every point whose $x$ -coordinate is $a$ . The graph of the function $f(x,y)$ intersects this plane in a curve called the $y$ -trace with $x=a$ . Since $x=a$ , the $y$ -trace is: $z=f(a,y)$ The $y$ -partial derivative of $f$ at $y=b$ with $x$ held fixed at $x=a$ , denoted $f_y(a,b)$ , is the slope of the tangent line to the $y$ -trace at $x=a$ . So the tangent line to the $y$ -trace at $(a,b)$ is: $z=f(a,b)+f_y(a,b)(y-b)$

$x$ -Trace and its Tangent Line
The plot at the right shows the graph of the function: $f(x,y)=-x^2-y^2$ Complete the below steps in order. (You will see a as you complete each one.)

Select the $xz$ -Slice checkbox below the plot, to show a vertical plane parallel to the $xz$ plane. Move the $y$ -slider, so the $xz$ -Slice is at $y=5$ .
Select the $x$ -Trace checkbox to show the intersection of the $xz$ -Slice and the graph of $z=f(x,y)$ which is called the $x$ -Trace at $y=5$ .
Move the $x$ -slider, so the dot on the $x$ -Trace is at $x=6$ .
Select the $x$ -Tangent checkbox to show a small tangent line to the $x$ -Trace at $y=5$ . The slope of this tangent line is the $x$ -partial derivative, $f_x(6,5)$ .
Click on Play $x$ to animate the point moving along the $x$ -Trace.
Note: You can rotate the entire plot with your mouse.
Find the equation of the tangent line at $x=6$ to the $x$ -trace with $y=5$ .

xz

-Slice

x

-Trace

x

-Tangent Play

x

x=

y=

Hint

Find $f(6,5)$ and $f_x(6,5)$ . Then use the formula: $z=f(6,5)+f_x(6,5)(x-6)$

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Answer

$z=-12x+11$

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Solution

Let's find $f(6,5)$ and $f_x(6,5)$ . We make a table: $\begin{aligned} f(x,y)&=-x^2-y^2 \quad &f(6,5)&=-6^2-5^2=-61 \\ f_x(x,y)&=-2x \quad &f_x(6,5)&=-12 \end{aligned}$ So, the tangent line to the $x$ -trace at the point $(6,5)$ is: $\begin{aligned} z&=f(6,5)+f_x(6,5)(x-6) \\ &=-61-12(x-6) \\ &=-12x+11 \end{aligned}$

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$y$ -Trace and its Tangent Line
The plot at the right again shows the graph of the function: $f(x,y)=-x^2-y^2$

Select the $yz$ -Slice checkbox below the plot, to show a vertical plane parallel to the $yz$ plane. Move the $x$ -slider, so the $yz$ -Slice is at $x=6$ .
Select the $y$ -Trace checkbox to show the intersection of the $yz$ -Slice and the graph of $z=f(x,y)$ which is called the $y$ -Trace at $x=6$ .
Move the $y$ -slider, so the dot on the $y$ -Trace is at $y=5$ .
Select the $y$ -Tangent checkbox to show a small tangent line to the $y$ -Trace at $x=6$ . The slope of this tangent line is the $y$ -partial derivative, $f_y(6,5)$ .
Click on Play $y$ to animate the point moving along the $y$ -Trace.
Find the equation of the tangent line at $y=5$ to the $y$ -trace with $x=6$ .

yz

-Slice

y

-Trace

y

-Tangent Play

y

x=

y=

Hint

Find $f(6,5)$ and $f_y(6,5)$ . Then use the formula: $z=f(6,5)+f_y(6,5)(y-5)$

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Answer

$z=-10y-11$

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Solution

Let's find $f(6,5)$ and $f_y(6,5)$ . We make a table: $\begin{aligned} f(x,y)&=-x^2-y^2 \quad &f(6,5)&=-6^2-5^2=-61 \\ f_y(x,y)&=-2y \quad &f_y(6,5)&=-10 \end{aligned}$ So, the tangent line to the $y$ -trace at the point $(6,5)$ is: $\begin{aligned} z&=f(6,5)+f_y(6,5)(y-5) \\ z&=-61-10(y-5) \\ z&=-10y-11 \end{aligned}$

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At a point $(a,b)$ the tangent plane to the surface $z=f(x,y)$ is the plane containing the tangent lines to the $x$ -trace and $y$ -trace at $(a,b)$ . This is discussed geometrically on the next page and algebraically on the following page.

11. Partial Derivatives and Tangent Planes

c. Geometric Interpretation of Partial Derivatives

Traces and their Tangent Lines

Hint

Answer

Solution

Hint

Answer

Solution

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