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Wize University Calculus 3 Textbook > Applications of Partial Derivatives

Tangent Planes & Linear Approximation

Introduction to Applications of Partial DerivativesPrevious Section
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Tangent Planes & Linear Approximation

Tangent Planes

The equation of the tangent plane to the surface z=f(x,y)z=f(x,y) at (x0,y0)(x_0,y_0) is:

z=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)\boxed{z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)}


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Linear Approximation

Linear approximation is one application of tangent planes and is given by:

L(x)≈f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)\boxed{L(x)\approx{}f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)}


As long as we are near (x0,y0)(x_0,y_0) then the tangent plane should nearly approximate the function at (x0,y0)(x_0,y_0).
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Example

Find the equation of tangent plane to f(x,y)=x2+y3−2xyf(x,y)=x^2+y^3-2xy at point x=1x=1, y=1y=1.

f(x,y)=x2+y3−2xyf(x,y)=x^2+y^3-2xy, (x0,y0)=(1,1)(x_0,y_0)=(1,1)
z=f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)

z0=f(1,1)=1+1−2=0z_0=f(1,1)=1+1-2=0
fx=2x−2y ⇒ fx(1,1)=2−2=0f_x=2x-2y\ \Rightarrow\ f_x(1,1)=2-2=0
fy=3y2−2x ⇒ fy(1,1)=3−2=1f_y=3y^2-2x\ \Rightarrow\ f_y(1,1)=3-2=1

∴z=0+0(x−1)+1(y−1)z=y−1                          \therefore z=0+0(x-1)+1(y-1)\newline z=y-1~~~~~~~~~~~~~~~~~~~~~~~~~~

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Example

Find the equation of the tangent plane to z=ln⁡(8x2+4y)  at  (−1, 2).z=\ln(8x^2+4y)~~\text{at}~~(-1,~2).

From the equation of the tangent plane, we need the z=f(−1, 2), fx(−1, 2), & fy(−1, 2)z=f(-1,~2),~f_x(-1,~2),~\&~f_y(-1,~2). So,


fx(x, y)=16x8x2+4yfy(x, y)=48x2+4yf(x, y)=ln(8x2+4y)fx(−1, 2)=16(−1)8(−1)2+4(2)fy(−1, 2)=48(−1)2+4(2)f(−1, 2)=ln(8(−1)2+4(2))fx(−1, 2)=−1fy(−1, 2)=14f(−1, 2)=ln16\begin{array}{c|c|c} f_x(x,~y)=\frac{16x}{8x^2+4y} & f_y(x,~y)=\frac{4}{8x2+4y} & f(x,~y)=ln(8x^2+4y)\\\\ f_x(-1,~2)=\frac{16(-1)}{8(-1)^2+4(2)} & f_y(-1,~2)=\frac{4}{8(-1)^2+4(2)} & f(-1,~2)=ln(8(-1)^2+4(2))\\\\ f_x(-1,~2)=-1 & f_y(-1,~2)=\frac{1}{4} & f(-1,~2)=ln16 \end{array}

Then, the equation of the tangent plane is:
z=−1(x+1)+14(y−2)+ln16z=-1(x+1)+\frac{1}{4}(y-2)+ln16

Practice

Find the linear approximation to z=3+x225+y249 at (1, 1)z=3+\frac{x^2}{25}+\frac{y^2}{49}~\text{at}~(1,~1).

Practice

Find the equation of the tangent plane to the surface z=f(x, y)=4cos⁡2(x)+4sin⁡2(y)z=f(x,~y)=4\cos^2(x)+4\sin^2(y) at the point (π, π3).\Big(\pi,~\dfrac{\pi}{3}\Big).