Approximating with the gradient

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We use the gradient to approximate values for functions of several variables.

We’ve studied differentials in our previous courses: If $f$ is differentiable, then: $\d f=f'(x)\d x$ Here $\d f$ and $\d x$ are two new variables that have been “cooked-up” to ensure that $\dd [f]{x} = f'(x).$ It is worthwhile to compare and contrast $\Delta x, \ \d x \quad {and}\quad \Delta f,\ \d f.$ The values $\Delta x \quad \text {and}\quad \Delta f$ are the change $x$ and change in $f$ when $x$ and $f$ are related. On the other hand, if $f$ is a function of $x$ , $\d x = \Delta x$ but $\d f$ is not necessarily equal to $\Delta f$ . Instead $\d f$ is the value that satisfies this equation: $\d f=f'(x)\d x$ When $\d x$ is small, $\d f\approx \Delta f$ , the change in $f$ resulting from the change in $x$ . The key idea is that as $\d x\to 0$ $(\Delta f - \d f) \to 0$ Another way of stating this is: as $\d x$ goes to $0$ , the error in approximating $\Delta f$ with $\d f$ goes to $0$ .

Let’s extend this idea to functions of two variables. Consider $F(x,y)$ , and let $\Delta x = \d x$ and $\Delta y=\d y$ represent changes in $x$ and $y$ , respectively. Now $\d F = F(x+\d x,y+\d y) - F(x,y)$ is the change in $F$ over the change in $x$ and $y$ . Recalling that $F^{(1,0)}$ and $F^{(0,1)}$ give the instantaneous rates of change of $F$ in the $x$ and $y$ -directions respectively, we can approximate $\d F$ as $\d F = F^{(1,0)}\d x+F^{(0,1)}\d y.$ In words, this says:

The total change in $F$ is approximately the change caused by changing $x$ plus the change caused by changing $y$ .

Setting $\vec {x}=\vector {x,y}$ , we can rewrite this in terms of the dot product: $\d F = \grad F \dotp \d \vec {x}$ This leads us to our next definition.

Let $\vec {x} = \vector {x_1,x_2,\dots ,x_n}$ and $F:\R ^n\to \R$ be continuous on an open set $S$ . Let $\d \vec {x}$ represent changes in $x_1,x_2,\dots ,x_n$ . Where $\grad F$ exists, the total differential is $\d F = \grad F \dotp \d \vec {x}$

Let $F(x,y) = x^4e^{3y}$ . Find $\d F$ . $\d F = \answer {4x^3e^{3y}}\d x+\answer {3x^4e^{3y}}\d y.$

We can approximate $\Delta F$ with $\d F$ , but as with all approximations, there is error involved. Approximating $\Delta F$ with $\d F$ relies on the fact that if a function is differentiable, then we can “zoom-in” on our surface until at some point, the surface looks like a plane. Of course as we’ve learned, this property is called differentiability, meaning quite literally can we use differentials to describe this surface.

If we believe that discrete data has been gathered from a function that is differentiable, it makes sense to estimate values of the function using differentials.

Let $F:\R ^2\to \R$ be a differentiable function described by the following table of values:

Estimate the total differential $\d F$ in terms of $\d x$ and $\d y$ at the point $(2,6)$ .

We estimate $\grad F(2,6)$ as $\vector {\answer [given]{-7},\answer [given]{2}}$ . Hence $\d F \approx \answer [given]{-7}\d x + \answer [given]{2}\d y.$

Approximating with the total differential

Suppose you want to approximate $F(x,y)=\sqrt {x}\sin (y)$ at the point $(4.1,0.8)$ . Without knowledge of calculus, your approximation might go like this:

We try to find numbers near $4.1$ and $0.8$ where $F(x,y)=\sqrt {x}\sin (y)$ is easy to evaluate. For example, we know that $\sqrt {4}= 2$ , so instead of looking at $x=4.1$ , we’ll use $x=4$ . Also, we know $\sin (\pi /4)= \sqrt {2}/2$ since $\pi /4\approx 0.8$ , we’ll use it in our approximation. Hence

$\begin{align*} F(4.1,0.8) &\approx F(4,\pi/4) \\ &= \sqrt{4}\sin(\pi/4)\\ &= 2\left(\frac{\sqrt{2}}2\right)\\ &= \sqrt{2}. \end{align*}$

Without calculus, (or some other insight) this is the best approximation we could reasonably come up with. The total differential gives us a way of adjusting this initial approximation to hopefully get a more accurate answer.

Let $F(x,y)=\sqrt {x}\sin (y)$ . Approximate $F(4.1,0.8)$ .

By the definition, when $F$ is differentiable $\d F$ is a good approximation for $\Delta F$ when $\d x$ and $\d y$ are largesmall .

As before we will attempt to compute $F(4.1,0.8)$ using $F(\answer [given]{4},\answer [given]{\pi /4})$ . Recall from directly above $F(4,\pi /4) = \answer [given]{\sqrt {2}}$ Now we will use the total differential to acquire a better approximation. Consider $\Delta F = F(4.1,0.8) - F(4,\pi /4).$ Since $\Delta F\approx \d F$ , we write

$\begin{align*} \d F &\approx F(4.1,0.8) - F(4,\pi/4)\\ F(4.1,0.8) &\approx F(4,\pi/4)+ \d F \end{align*}$

To find $\d F$ , we need $\grad F$ , as

$\begin{align*} \d F &= \grad F \dotp \d\vec{x}\\ &=\vector{\answer[given]{\frac{\sin (y)}{2\sqrt{x}}},\answer[given]{\sqrt{x}\cos (y)}}\dotp\vector{\d x,\d y} \end{align*}$

Evaluating at $(4,\pi /4)$ we find

$\begin{align*} \d F &= \grad F \dotp \d\vec{x}\\ &=\vector{\answer[given]{\frac{\sin (\pi/4)}{2\sqrt{4}}}, \answer[given]{\sqrt{4}\cos (\pi/4)}}\dotp\vector{\d x,\d y }\\ &=\vector{\frac{\sqrt{2}}{8}, \sqrt{2}}\dotp\vector{\d x,\d y } \end{align*}$

Now we need to know the change in $x$ and $y$ :

$\begin{align*} \d x &= 4.1-4= \answer[given]{0.1}\\ \d y &= 0.8 -\answer[given]{\pi/4} \approx 0.015.\\ \end{align*}$

Thus

$\begin{align*} \d F &= \vector{\answer[given]{\frac{\sqrt{2}}{8}}, \answer[given]{\sqrt{2}}}\dotp\vector{\answer[given]{0.1},\answer[given]{0.015}}\\ &= \frac{\sqrt{2}}8(0.1) + \sqrt{2}(0.015)\\ \end{align*}$

So finally,

$\begin{align*} F(4.1,0.8) &\approx F(4,\pi/4) + \d F\\ F(4.1,0.8) &\approx \answer[given]{\sqrt{2} + \frac{\sqrt{2}}{8}(0.1) + \sqrt{2}(0.015)} \end{align*}$

If we compare our estimates of $F(x,y) = \sqrt {x}\sin (y)$ at $(4.1,0.8)$ we see that without knowledge of calculus we found $F(4.1,0.8) \approx \sqrt {2},$ but with knowledge of calculus we found $F(4.1,0.8) \approx \sqrt {2} + \frac {\sqrt {2}}8(0.1) + \sqrt {2}(0.015)$ We can compute the actual value of $F(4.1,0.8)$ with a calculator. We find the value $1.45254$ , accurate to $5$ places after the decimal. We see that calculus has served us well.

The point of the previous example was not to develop an approximation method for known functions. After all, we can very easily compute $F(4.1,0.8)$ using readily available technology. Rather, it serves to illustrate how well this method of approximation works, and to reinforce the following concept:

New position = old position $+$ amount of change,
New position $\approx$ old position + approximate amount of change.

In the previous example, we could easily compute $F(4,\pi /4)$ and could approximate the change in $F$ when computing $F(4.1,0.8)$ , letting us approximate the new value for $F$ .

It may be surprising to learn that it is not uncommon to know the values of $F$ and $\grad F$ at a particular point without actually knowing a formula for $F$ . The total differential gives a good method of approximating $F$ by looking at nearby points.

Given that $F(2,-3) = 6$ , $\grad F(2,-3) = \vector {1.3,-0.6}$ , approximate $F(2.1,-3.03)$ .

The total differential approximates how much $F$ changes from the point $(2,-3)$ to the point $(2.1,-3.03)$ . With $\d x = \answer [given]{0.1}$ and $\d y = \answer [given]{-0.03}$ , we have $\begin{align*} \d F &= \grad F \dotp \d \vec{x}\\ &= (1.3)(\answer[given]{0.1}) + (-0.6)(\answer[given]{-0.03}) \\ &= 0.148. \end{align*}$

The change in $F$ is approximately $0.148$ , so we approximate $F(2.1,-3.03)\approx \answer [given]{6.148}$ .

Error analysis

The total differential gives an approximation of the change in $F$ given small changes in $x$ and $y$ . We can use this to approximate error propagation; that is, if the input is a little off from what it should be, how far from correct will the output be? We demonstrate this in an example.

A cylindrical steel storage tank is to be built that is $10\unit {ft}$ tall and $4\unit {ft}$ across in diameter. It is known that the steel will expand/contract with temperature changes; is the overall volume of the tank more sensitive to changes in the diameter or in the height of the tank?

A cylindrical solid with height $h$ and radius $r$ has volume $V = \pi r^2h.$ We can view $V$ as a function of two variables, $r$ and $h$ (in that order). We can compute $\grad V$ $\grad V = \vector {\answer [given]{2\pi r h},\answer [given]{\pi r^2}}.$ Setting $\vec {x}=\vector {r,h}$ , the total differential is $\begin{align*} \d V &=\grad V \dotp \d \vec{x}\\ &=\vector{\answer[given]{2\pi rh},\answer[given]{ \pi r^2}}\dotp\vector{\d r,\d h} \end{align*}$

When $h = 10$ and $r = 2$ , we have $\d V = 40\pi \d r + 4\pi \d h.$ Note that the coefficient of $\d r$ is $40\pi \approx 125.7$ ; the coefficient of $\d h$ is , approximately $12.57$ . A small change in radius will be multiplied by $125.7$ , whereas a small change in height will be multiplied by $12.57$ . Thus the volume of the tank is more sensitive to changes in radius than in height.

The previous example showed that the volume of a particular tank was more sensitive to changes in radius than in height. Keep in mind that this analysis only applies to a tank of the dimensions given in the problem. A tank with a height of $1$ ft and radius of $5$ ft would be more sensitive to changes in height than in radius.

One could make a chart of small changes in radius and height and find exact changes in volume given specific changes. While this provides exact numbers, it does not give as much insight as the error analysis using the total differential.