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CalcVR Supplemental Materials

Jeremy Becnel, Nicholas Long

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Section 6.5 Partial Derivatives

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Figure 6.5.1. A screenshot from the lesson on partial derivatives of multivariable functions.
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The CalcVR lesson defines partial derivatives and motivates this definition by examining the slope of tangent lines along level curves or traces in one of the independent variables. Several different surfaces are given as examples where the curves generated by holding one variable constant are highlighted. We have also examined the concept of derivatives of paths (or parametric equations) in Section 5.2. Both of these concepts involve a rate of change and thus should be connected.
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Subsection 6.5.1 Connection to Vector Valued Functions

Example 6.5.2.

Consider the example of the multivariable function
(6.5.1)f(x,y)=2x2+y2+1.
First let’s find the partial derivative at (1,0). The partial derivative of f with respect to x is given by
fx(x,y)=−4x(x2+y2+1)2
and at the point (1,0) we have
fx(1,0)=−4(12+02+1)2=−44=−1.
Let’s take another approach. If we consider the partial derivative at (1,0), we know that y is fixed at 0. Referring back to Subsection 6.4.1 we should be able to find this path on the function. This can be accomplished by allowing
x(t)=ty(t)=0z(t)=f(t,0)=2t2+02+1=2t2+1.
in r(t)=⟨x(t),y(t),z(t)⟩. This parametric equation follows the function along y=0 and passes through (1,0,1) when t=1.
This parametric equation has a derivative, which we learned about in Section 5.2. The derivative at of this path is given by r\'(t)=⟨1,0,−4t(t2+1)2⟩ At t=1 (the value at which the path passing through (1,0,1)), r\'(1)=⟨1,0,−4(12+1)2⟩=⟨1,0,−1⟩.
Notice we have that
fx(1,0)=−1andr\'(1)=⟨1,0,−1⟩.
Let’s rewrite this,
fx(1,0)=−11andr\'(1)=⟨1,0,−1⟩.
It is no coincidence that these are similar. The partial fx(1,0)=−11 can be interpreted using the typical riserun for slope. This means at the point (x=1,y=0) the rate of elevation change is −1 (down 1) for a one unit change in the positive x direction. For the rate of change r\'(1)=⟨1,0,−1⟩ we can interpret the 0 as we are not changing in the y direction This corresponds to y being fixed when we consider partials with respect to x. The other values are interpreted the same as before. That is, a one unit change in x corresponds to an elevation change of −1.
You can explore the function and the partial with respect to x using the tool below.
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Activity 6.5.1.

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In this activity, we want to repeat the discussion above and work through a similar calculation for fy at (1,0) using the function f in (6.5.1).
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(a)
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Find the partial with respect to y at (1,0).
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(b)
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Fix x=1 and find a parametric equation r for the path that follows the function in the y direction.
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(c)
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At what value of t does the function r pass through the point (1,0,1)?
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(d)
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Find the derivative of r at the value from the previous part.
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(e)
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Interpret both the partial derivative with respect to y at (1,0) and the value of r\' from the previous part. Explain the relationship between the two.
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(f)
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Find an equation for the tangent line to r at the point (1,0,1).
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(g)
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Use the tool below to graph the function, path, and tangent line on the same graph in order to confirm your answer is correct.
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Activity 6.5.2.

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The graph of f(x,y)=sin⁡(x(y−π2))+1 is shown below.
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(a)
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We defined the tangent line as a limit of secant lines. On the surface above, draw the tangent line to the graph at the given point in the y direction. Find the slope of this tangent line (algebraically, not geometrically).
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(b)
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On the above surface, draw the line in the x direction that is tangent to the graph at (3,1.5,f(3,1.5)) (highlighted on the graph). Find the slope of this tangent line.
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(c)
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The two tangent lines drawn in parts a and b are part of the tangent plane to the graph of the function at the point (3,1.5,f(3,1.5)) (highlighted on the graph). Find the equation of the tangent plane (You know a point and two slopes).
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