{"id":1124,"date":"2025-10-15T14:31:05","date_gmt":"2025-10-15T14:31:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1124"},"modified":"2025-10-27T21:14:39","modified_gmt":"2025-10-27T21:14:39","slug":"tangent-planes-and-linear-approximations-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/tangent-planes-and-linear-approximations-learn-it-4\/","title":{"raw":"Tangent Planes and Linear Approximations: Learn It 4","rendered":"Tangent Planes and Linear Approximations: Learn It 4"},"content":{"raw":"

Differentials<\/h2>\r\n

You first encountered differentials when studying linear approximations in single-variable calculus. The differential of [latex]y[\/latex], written [latex]dy[\/latex], is defined as [latex]f'(x)dx[\/latex]. We use differentials to approximate [latex]\\Delta y = f(x+\\Delta x)-f(x)[\/latex], where [latex]\\Delta x=dx[\/latex].<\/p>\r\n

We can extend this idea to the linear approximation of a function of two variables at the point[latex](x_0, y_0)[\/latex] to create the formula for the total differential<\/strong> for a function of two variables.<\/p>\r\n\r\n

\r\n

total differential<\/h3>\r\nLet [latex]z=f(x,y)[\/latex] be a function of two variables with [latex](x_0, y_0)[\/latex] in the domain of [latex]f[\/latex] and let [latex]\\Delta x[\/latex] and [latex]\\Delta y[\/latex] be chosen so that [latex](x_0+\\Delta x, y_0+\\Delta y)[\/latex] is also in the domain of [latex]f[\/latex]. If [latex]f[\/latex] is differentiable at the point [latex](x_0, y_0)[\/latex],\u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>then the differentials [latex]dx[\/latex] and [latex]dy[\/latex] are defined as:<\/span>\r\n

[latex]dx= \\Delta x[\/latex] and [latex]dy= \\Delta y[\/latex]<\/p>\r\nThe differential [latex]dz[\/latex] also called the\u00a0total differential<\/span><\/strong>\u00a0of [latex]z=f(x,y)[\/latex] at [latex](x_0,y_0)[\/latex],\u00a0<\/span>is defined as:<\/span>\r\n

[latex]dz=f_x(x_0,y_0)dx+f_y(x_0,y_0)dy[\/latex].<\/p>\r\n\r\n<\/section>

Notice that we use [latex]d[\/latex] (not the partial derivative symbol [latex]\\partial[\/latex]) for the total differential.<\/section>\r\n

Now, let's define [latex]\\Delta z=f(x+\\Delta x, y+\\Delta y) - f(x,y)[\/latex]. We use [latex]dz[\/latex] to approximate [latex]\\Delta z[\/latex]:<\/p>\r\n

[latex]\\Delta z\\approx dz=f_x(x_0,y_0)dx+f_y(x_0,y_0)dy[\/latex]<\/p>\r\n

The differential approximates the change in [latex]z=f(x_0, y_0)[\/latex] at point [latex](x_0, y_0)[\/latex] for given values of [latex]\\Delta x[\/latex] and [latex]\\Delta y[\/latex]. Since [latex]\\Delta z=f(x+ \\Delta x, y+ \\Delta y)-f(x, y)[\/latex], we can use this to approximate [latex]f(x+ \\Delta x, y+ \\Delta y)[\/latex]:<\/p>\r\n[latex]\\hspace{8cm}\\begin{alignat}{2} f(x+\\Delta{x},y+\\Delta{y})&=f(x,y)+\\Delta{z}\\\\\r\n\r\n&\\approx{f}(x,y)+f_x(x_0,y_0)\\Delta{x}+f_y(x_0,y_0)\\Delta{y}.\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n

See the following figure.<\/p>\r\n\r\n[caption id=\"attachment_1242\" align=\"aligncenter\" width=\"435\"]\"A Figure 2. The linear approximation is calculated via the formula\u00a0[latex]\\small{f(x+\\Delta{x},y+\\Delta{y}) \\approx f(x,y)+f_{x}(x_{0},y_{0})\\Delta{x}+f_{y}(x_{0},y_{0})\\Delta{y}}[\/latex].[\/caption]

\r\n

Applications to Error Propagation<\/strong><\/p>\r\n

One practical application of differentials is determining error propagation. For example, if you're manufacturing a gadget and your measurements are off by a small amount, you can use the differential to estimate the error in the total volume of the gadget.<\/p>\r\n\r\n<\/section>

Find the differential [latex]dz[\/latex] of the function [latex]f(x,y)=3x^{2}-2xy+y^{2}[\/latex] and use it to approximate [latex]\\Delta z[\/latex] at point [latex](2, -3)[\/latex]. Use [latex]\\Delta x=0.1[\/latex] and [latex]\\Delta y=-0.05[\/latex] What is the exact value of [latex]\\Delta z[\/latex]?<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>[reveal-answer q=\"875201369\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"875201369\"]First, we must calculate [latex]f(x_0, y_0)[\/latex], [latex]f_x(x_0, y_0)[\/latex], and [latex]f_y(x_0, y_0)[\/latex] using [latex]x_0=2[\/latex] and [latex]y_0=-3[\/latex]:[latex]\\begin{alignat}{2} \\hspace{6cm}f(x_0, y_0)&=f(2,-3)=3(2)^2-2(2)(-3)+(-3)^2=12+12+9=33\\\\\r\n\r\nf_x(x,y)&=6x-2y\\\\\r\n\r\nf_y(x,y)&=-2x+2y\\\\\r\n\r\nf_x(x_0, y_0)&=f_x(2,-3)=6(2)-2(-3)=12+6=18\\\\\r\n\r\nf_y(x_0, y_0)&=f_y(2,-3)=-2(2)+2(-3)=-4-6=-10.\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\nThen, we substitute these quantities into\u00a0the Total Differential Equation:\r\n\r\n[latex]\\begin{alignat}{2} \\hspace{7.2cm}dz&=f_x(x_0, y_0)dx+f_y(x_0,y_0)dy\\\\\r\n\r\ndz&=18(0.1)-10(-0.05)=1.8+0.5=2.3.\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\nThis is the approximation to [latex]\\Delta{z}=f(x_0+\\Delta{x},y_0+\\Delta{y})-f(x_0,y_0)[\/latex]\u00a0The exact value of [latex]\\Delta z[\/latex] is given by\r\n\r\n[latex]\\begin{alignat}{2} \\hspace{7.2cm}\\Delta{z}&=f(x_0+\\Delta{x},y_0+\\Delta{y})-f(x_0,y_0)\\\\\r\n\r\n&=f(2+0.1,-3-0.05)-f(2,-3)\\\\\r\n\r\n&=f(2.1,-3.05)-f(2,-3)\\\\\r\n\r\n&=2.3425.\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
[ohm_question hide_question_numbers=1]250904[\/ohm_question]<\/section>\r\n

Differentiability of a Function of Three Variables<\/h3>\r\nAll of the results for differentiability of functions of two variables generalize naturally to functions of three variables.\r\n\r\n
\r\n

differentiable function of three variables<\/h3>\r\nA function [latex]f(x, y, z)[\/latex] is differentiable at a point [latex]P(x_0, y_0, z_0)[\/latex] if for all points [latex](x, y, z)[\/latex] in a [latex]\\delta[\/latex]\u00a0<\/span>disk around [latex]P[\/latex] we can write:\r\n\r\n[latex]\\begin{alignat}{2}\\hspace{6cm}f(x, y, z)&=f_x(x_0, y_0, z_0)(x-x_0)+f_y(x_0, y_0, z_0)(y-y_0)\\\\\r\n\r\n& \\;\\; +f_z(x_0, y_0, z_0)(z-z_0)+E(x,y,z),\\\\\r\n\r\n\\end{alignat}[\/latex]\r\n

where the error term [latex]E[\/latex] satisfies:<\/p>\r\n

[latex]\\displaystyle\\lim_{(x,y,z)\\to(x_0,y_0,z_0)}=\\dfrac{E(x,y,x)}{\\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}}=0.[\/latex]<\/p>\r\n\r\n<\/section>\r\n

\r\n
\r\n

If a function of three variables is differentiable at [latex](x_0, y_0, z_0)[\/latex], then it is continuous there. Furthermore, continuity of first partial derivatives at that point guarantees differentiability.<\/p>\r\n\r\n<\/div>\r\n<\/div>","rendered":"

Differentials<\/h2>\n

You first encountered differentials when studying linear approximations in single-variable calculus. The differential of [latex]y[\/latex], written [latex]dy[\/latex], is defined as [latex]f'(x)dx[\/latex]. We use differentials to approximate [latex]\\Delta y = f(x+\\Delta x)-f(x)[\/latex], where [latex]\\Delta x=dx[\/latex].<\/p>\n

We can extend this idea to the linear approximation of a function of two variables at the point[latex](x_0, y_0)[\/latex] to create the formula for the total differential<\/strong> for a function of two variables.<\/p>\n

\n

total differential<\/h3>\n

Let [latex]z=f(x,y)[\/latex] be a function of two variables with [latex](x_0, y_0)[\/latex] in the domain of [latex]f[\/latex] and let [latex]\\Delta x[\/latex] and [latex]\\Delta y[\/latex] be chosen so that [latex](x_0+\\Delta x, y_0+\\Delta y)[\/latex] is also in the domain of [latex]f[\/latex]. If [latex]f[\/latex] is differentiable at the point [latex](x_0, y_0)[\/latex],\u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>then the differentials [latex]dx[\/latex] and [latex]dy[\/latex] are defined as:<\/span><\/p>\n

[latex]dx= \\Delta x[\/latex] and [latex]dy= \\Delta y[\/latex]<\/p>\n

The differential [latex]dz[\/latex] also called the\u00a0total differential<\/span><\/strong>\u00a0of [latex]z=f(x,y)[\/latex] at [latex](x_0,y_0)[\/latex],\u00a0<\/span>is defined as:<\/span><\/p>\n

[latex]dz=f_x(x_0,y_0)dx+f_y(x_0,y_0)dy[\/latex].<\/p>\n<\/section>\n

Notice that we use [latex]d[\/latex] (not the partial derivative symbol [latex]\\partial[\/latex]) for the total differential.<\/section>\n

Now, let’s define [latex]\\Delta z=f(x+\\Delta x, y+\\Delta y) - f(x,y)[\/latex]. We use [latex]dz[\/latex] to approximate [latex]\\Delta z[\/latex]:<\/p>\n

[latex]\\Delta z\\approx dz=f_x(x_0,y_0)dx+f_y(x_0,y_0)dy[\/latex]<\/p>\n

The differential approximates the change in [latex]z=f(x_0, y_0)[\/latex] at point [latex](x_0, y_0)[\/latex] for given values of [latex]\\Delta x[\/latex] and [latex]\\Delta y[\/latex]. Since [latex]\\Delta z=f(x+ \\Delta x, y+ \\Delta y)-f(x, y)[\/latex], we can use this to approximate [latex]f(x+ \\Delta x, y+ \\Delta y)[\/latex]:<\/p>\n

[latex]\\hspace{8cm}\\begin{alignat}{2} f(x+\\Delta{x},y+\\Delta{y})&=f(x,y)+\\Delta{z}\\\\ &\\approx{f}(x,y)+f_x(x_0,y_0)\\Delta{x}+f_y(x_0,y_0)\\Delta{y}.\\\\ \\end{alignat}[\/latex]<\/p>\n

See the following figure.<\/p>\n

\"A
Figure 2. The linear approximation is calculated via the formula\u00a0[latex]\\small{f(x+\\Delta{x},y+\\Delta{y}) \\approx f(x,y)+f_{x}(x_{0},y_{0})\\Delta{x}+f_{y}(x_{0},y_{0})\\Delta{y}}[\/latex].<\/figcaption><\/figure>\n
\n

Applications to Error Propagation<\/strong><\/p>\n

One practical application of differentials is determining error propagation. For example, if you’re manufacturing a gadget and your measurements are off by a small amount, you can use the differential to estimate the error in the total volume of the gadget.<\/p>\n<\/section>\n

Find the differential [latex]dz[\/latex] of the function [latex]f(x,y)=3x^{2}-2xy+y^{2}[\/latex] and use it to approximate [latex]\\Delta z[\/latex] at point [latex](2, -3)[\/latex]. Use [latex]\\Delta x=0.1[\/latex] and [latex]\\Delta y=-0.05[\/latex] What is the exact value of [latex]\\Delta z[\/latex]?<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n