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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
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The graph of a function
z = f ( x, y ) is a surface in 3 (three dimensional space) and so we can now start thinking of the
plane that is “tangent” to the surface as a point
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e
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e
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Now, we know that f x ( x0 , y0 ) is

the slope of the tangent line to the trace C1 and f y ( x0 , y0 ) is the slope of the tangent line to the

trace C2
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The tangent plane will then be the plane that contains the two lines L1 and L2
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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
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Note that this gives us a point that is on the
plane
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( x0 , y0 , z0 ) = ( x0 , y0 , f ( x0 , y0 ) )

What we need to do now is determine the equation of the tangent plane
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Let’s rewrite this a
little
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Doing this
gives,

a
b
z − z0 = − ( x − x0 ) − ( y − y0 )
c
c
Now, let’s rename the constants to simplify up the notation a little
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Let’s first think about what happens if we hold y fixed, i
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if we assume that y = y0
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In addition, this line assumes that y = y0 (i
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fixed) and A is the slope of
this line
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In other words,

z − z0 = A ( x − x0 )

is the equation for L1 and we know that the slope of L1 is given by f x ( x0 , y0 )
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So,

B = f y ( x0 , y0 )

The equation of the tangent plane to the surface given by z = f ( x, y ) at ( x0 , y0 ) is then,

z − z0 = f x ( x0 , y0 )( x − x0 ) + f y ( x0 , y0 )( y − y0 )

Also, if we use the fact that z0 = f ( x0 , y0 ) we can rewrite the equation of the tangent plane as,

z − f ( x0 , y0 ) = f x ( x0 , y0 )( x − x0 ) + f y ( x0 , y0 )( y − y0 )

z = f ( x0 , y0 ) + f x ( x0 , y0 )( x − x0 ) + f y ( x0 , y0 )( y − y0 )
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
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Solution
There really isn’t too much to do here other than taking a couple of derivatives and doing some
quick evaluations
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As
long as we are near to the point ( x0 , y0 ) then the tangent plane should nearly approximate the
function at that point
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16 9
Solution
So, we’re really asking for the tangent plane so let’s find that
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