A robot arm has two rigid links, each of length \(1\) , joined at an elbow. The shoulder is fixed at the origin. Motors set two angles: \(\theta_1\) , the angle of the first link above the \(x\) -axis, and \(\theta_2\) , the angle by which the second link is bent from the direction of the first. The hand is then at

An engineer wants to know two things. If both motors are slightly off, where does the hand end up? And is there a posture in which some small movements of the hand cannot be produced by any small movement of the motors?

Both questions are about a function with two inputs and two outputs. Its derivative cannot be a single number or even a single vector. This chapter shows that it is a matrix, or better, the linear map that the matrix represents. Once the derivative is understood as a linear map, the chain rule becomes the statement that the derivative of a composition is the composition of the derivatives, and every coordinate formula of the subject follows by multiplying matrices.

3.1 The derivative of a map is a linear map

Definition 2.3 called a real-valued function differentiable when a linear function of the displacement predicts its change with relatively small error. Nothing in that sentence requires the values to be numbers.

The matrix acts on displacements: it takes a small change \(\mathbf h\in\mathbb R^n\) of the input and returns the predicted change \(A\mathbf h\in\mathbb R^m\) of the output. A matrix used in this way is a linear map in the sense of linear algebra (math.linear-algebra.matrix-as-map), and it is the map, more than the array of numbers, that deserves the name derivative. Its shape is forced: \(n\) columns because the input has \(n\) coordinates, \(m\) rows because the output has \(m\) .

Together with Theorem 2.6 the proposition gives a practical test: if all the partial derivatives \(\partial F_i/\partial x_j\) are continuous on an open set, then \(F\) is differentiable there and its derivative is the matrix of those partial derivatives. Three special cases are worth naming. When \(m=1\) the matrix is a single row, the gradient written horizontally. When \(n=1\) it is a single column, the velocity vector of Chapter 1. And when \(F(\mathbf x)=M\mathbf x\) is itself linear, the error in Definition 3.1 is identically zero with \(A=M\) , so a linear map is its own derivative at every point.

Example 3.1: How the robot’s hand responds to its motors

For the arm of the opening paragraph, differentiate each coordinate of \(F\) with respect to each angle:

\[ DF(\theta_1,\theta_2)=\begin{pmatrix}-\sin\theta_1-\sin(\theta_1+\theta_2)&-\sin(\theta_1+\theta_2)\\ \cos\theta_1+\cos(\theta_1+\theta_2)&\cos(\theta_1+\theta_2)\end{pmatrix}. \]

Take the posture \(\theta_1=0\) , \(\theta_2=\pi/2\) : the first link lies along the \(x\) -axis, the second points straight up, and the hand is at \(F(0,\pi/2)=(1,1)\) . There

\[ DF(0,\pi/2)=\begin{pmatrix}-1&-1\\ 1&0\end{pmatrix}. \]

The first column, \((-1,1)\) , is the velocity of the hand per unit rate of the shoulder motor: turning the shoulder swings the whole arm about the origin. The second column, \((-1,0)\) , is the response to the elbow motor: the hand moves left.

Suppose the motors are off by \(\mathbf h=(0.1,-0.05)\) radians. The linear prediction for the displacement of the hand is \(DF\,\mathbf h=(-0.1+0.05,\ 0.1)=(-0.05,0.1)\) , so the hand should be near \((0.95,\,1.10)\) . Direct evaluation of \(F(0.1,\pi/2-0.05)\) gives \((0.9450,\,1.0986)\) to four decimal places. The prediction misses by a distance of about \(0.0052\) , while the step \(\mathbf h\) has length about \(0.112\) : the error is less than a twentieth of the step, and Definition 3.1 says the ratio goes to zero as the step shrinks.

For the engineer’s second question, compute the determinant. Expanding and using the subtraction formula for the sine,

\[ \det DF=\sin\theta_2 . \]

When \(\theta_2=0\) the arm is straight, the determinant vanishes, and

\[ DF(\theta_1,0)=\begin{pmatrix}-2\sin\theta_1&-\sin\theta_1\\ 2\cos\theta_1&\cos\theta_1\end{pmatrix}. \]

Both columns are multiples of \((-\sin\theta_1,\cos\theta_1)\) , the direction perpendicular to the arm. To first order the hand can move only sideways; no small motion of the motors moves it along the arm. That is the posture the engineer was worried about.

Figure 3.1 shows what the matrix means. In each panel the motors are perturbed by every pair of angle errors on a circle of radius \(0.3\) . The solid curve is where the hand really goes; the dashed curve is the image of the same circle under the linear map \(\mathbf h\mapsto F(\mathbf a)+DF(\mathbf a)\mathbf h\) . A linear map sends a circle to an ellipse. In the bent posture the ellipse has positive area and the true curve follows it closely. In the straight posture the ellipse has collapsed to a segment, and the true curve is a thin sliver that departs from it only at second order.

3.2 The chain rule

If the hand of the robot carries a sensor, and the reading depends on where the hand is, then the reading depends on the motor angles through the position. One map is fed into another. The rule for differentiating such a composition is the central computational theorem of differential calculus, and with derivatives understood as matrices it has the only form it could have.

The proof needs one estimate, which says that a matrix cannot stretch a vector by more than a fixed factor.

The proof is the one-variable proof with absolute values replaced by norms and one extra ingredient, Lemma 3.3, to control what a matrix does to an error term. Notice the order of the factors. The matrix of the map applied first stands on the right, next to the vector it acts on, exactly as in \(F(G(\mathbf x))\) . Since matrix multiplication is not commutative the order matters, and the shapes usually enforce it: \(p\times m\) times \(m\times n\) is defined, and the reverse product generally is not.

3.3 The chain rule in coordinates

Written out entry by entry, the matrix product in Theorem 3.4 says the following. Suppose \(z\) is a differentiable function of \(x_1,\dots,x_m\) and each \(x_k\) is a differentiable function of \(s_1,\dots,s_n\) . Then

In words: to find how \(z\) responds to \(s_j\) , follow every route from \(s_j\) to \(z\) through an intermediate variable, multiply the rates along each route, and add the routes. Theorem 2.10 was the case of a single independent variable. Many people draw the dependencies as a tree, with \(z\) at the top, the intermediate variables below it, and the independent variables at the bottom, and read the sum off the branches.

3.4 Curves and surfaces defined implicitly

An equation \(F(x,y)=c\) usually defines a curve, but not as the graph of an explicit function. When can it be solved, at least locally, for \(y\) in terms of \(x\) , and what is the slope of the resulting graph?

As with Theorem 3.5, the existence statement is an analysis theorem, proved from the same fixed-point argument. The formula, once existence is granted, is a one-line application of the chain rule. Differentiate the identity \(F(x,\varphi(x))=c\) with respect to \(x\) using Theorem 2.10 along the curve \(x\mapsto(x,\varphi(x))\) , whose velocity is \((1,\varphi'(x))\) :

and solve for \(\varphi'\) . The hypothesis \(F_y\ne 0\) is exactly what the division needs. Geometrically, it says that the gradient \((F_x,F_y)\) is not horizontal, so the level curve, which is perpendicular to the gradient by Corollary 2.11, is not vertical.

3.5 A mean value theorem, and its limits

The Mean Value Theorem of one-variable calculus says that the average rate of change over an interval is attained as an instantaneous rate somewhere inside. It extends to real-valued functions of several variables along a line segment.

The hypothesis on \(U\) cannot be dropped: a function equal to \(0\) on one disc and to \(1\) on a second, disjoint disc has zero gradient on their union. A second consequence of Theorem 3.7 is an inequality. If \(\lVert\nabla f\rVert\le M\) along the segment, then by Cauchy–Schwarz \(\lvert f(\mathbf b)-f(\mathbf a)\rvert\le M\lVert\mathbf b-\mathbf a\rVert\) : a bound on the gradient is a bound on how fast the function can change.

Theorem 3.7 is a statement about real-valued functions, and it does not survive the passage to maps. For \(F(t)=(\cos t,\sin t)\) on \([0,2\pi]\) we have \(F(2\pi)-F(0)=\mathbf 0\) , yet \(F'(t)=(-\sin t,\cos t)\) is never zero, so no single \(t_0\) satisfies \(F(2\pi)-F(0)=2\pi F'(t_0)\) . Each coordinate function has its own mean-value point, and the two points differ. What does survive for maps is the inequality, in the form \(\lVert F(\mathbf b)-F(\mathbf a)\rVert\le M\lVert\mathbf b-\mathbf a\rVert\) when \(\lVert DF\rVert\le M\) on the segment; its proof and its many uses are part of analysis.

With the derivative of a map in hand, the first-order theory is complete: a differentiable map is, near each point, a linear map plus a small error, and compositions, inverses, and implicit functions are all governed by the algebra of the linear parts. The next chapter pushes the approximation one order further for real-valued functions, replacing the tangent plane by the best-fitting quadratic surface, because at a critical point the linear part vanishes and the quadratic part decides everything.

Exercises

1. State Definition 3.1, give the shape of \(DF(\mathbf a)\) for a map from \(\mathbb R^n\) to \(\mathbb R^m\) , and describe what the matrix reduces to when \(m=1\) and when \(n=1\) .

2. Compute the derivative matrix of each map and state its shape: (a) \(F(x,y)=(x^{2}-y^{2},\ 2xy)\) ; (b) \(G(x,y,z)=(xyz,\ x+y+z)\) ; (c) \(H(t)=(t,t^{2},t^{3})\) . For (a) also compute the determinant.

   Answer

   (a) \(DF=\begin{pmatrix}2x&-2y\\ 2y&2x\end{pmatrix}\) , of shape \(2\times 2\) , with determinant \(4(x^{2}+y^{2})\) . (b) \(DG=\begin{pmatrix}yz&xz&xy\\ 1&1&1\end{pmatrix}\) , of shape \(2\times 3\) . (c) \(DH=\begin{pmatrix}1\\ 2t\\ 3t^{2}\end{pmatrix}\) , of shape \(3\times 1\) , the velocity as a column.

3. For \(F(x,y)=(x^{2}-y^{2},\ 2xy)\) , use the derivative at \((2,1)\) to estimate \(F(2.1,0.9)\) , compare with the exact value, and explain why one coordinate of the estimate happens to be exact.

   Solution

   From Exercise 2(a), \(DF(2,1)=\begin{pmatrix}4&-2\\ 2&4\end{pmatrix}\) and \(F(2,1)=(3,4)\) . The displacement is \(\mathbf h=(0.1,-0.1)\) , and \(DF(2,1)\mathbf h=(0.4+0.2,\ 0.2-0.4)=(0.6,-0.2)\) . The estimate is \((3.6,3.8)\) .

   The exact value is \(F(2.1,0.9)=(4.41-0.81,\ 2\cdot 2.1\cdot 0.9)\) , which is \((3.6,3.78)\) . The second coordinate is off by \(0.02\) ; the first is exact.

   To see why, expand exactly. For a step \((h,k)\) from \((x,y)\) the first coordinate changes by \((x+h)^{2}-(y+k)^{2}-x^{2}+y^{2}=2xh-2yk+(h^{2}-k^{2})\) . The linear part \(2xh-2yk\) is the prediction, and the error is \(h^{2}-k^{2}\) , which vanishes whenever \(\lvert h\rvert\) equals \(\lvert k\rvert\) , as here. The second coordinate changes by \(2(x+h)(y+k)-2xy=2yh+2xk+2hk\) , with error \(2hk=-0.02\) . In both coordinates the error is quadratic in the step, as differentiability requires; in the first it happens to be zero along the two diagonal directions.

4. Let \(G(u,v)=(u+v,\ uv,\ u^{2})\) and \(F(x,y,z)=xy+z^{2}\) . Compute \(D(F\circ G)(1,2)\) as a product of two matrices, stating the shape of each factor, and verify the result by composing first.

   Solution

   The inner map \(G\) goes from \(\mathbb R^2\) to \(\mathbb R^3\) , so \(DG\) is \(3\times 2\) ; the outer function \(F\) goes from \(\mathbb R^3\) to \(\mathbb R\) , so \(DF\) is \(1\times 3\) ; the product \(DF\,DG\) is \(1\times 2\) , as the derivative of a real-valued function of \((u,v)\) should be.

   Compute \(DG(u,v)=\begin{pmatrix}1&1\\ v&u\\ 2u&0\end{pmatrix}\) , so \(DG(1,2)=\begin{pmatrix}1&1\\ 2&1\\ 2&0\end{pmatrix}\) , and \(G(1,2)=(3,2,1)\) . Next \(DF(x,y,z)=\begin{pmatrix}y&x&2z\end{pmatrix}\) , which at \((3,2,1)\) is \(\begin{pmatrix}2&3&2\end{pmatrix}\) . Then

   \[ D(F\circ G)(1,2)=\begin{pmatrix}2&3&2\end{pmatrix}\begin{pmatrix}1&1\\ 2&1\\ 2&0\end{pmatrix}=\begin{pmatrix}12&5\end{pmatrix}. \]

   To verify, compose: \(F(G(u,v))=(u+v)uv+u^{4}=u^{2}v+uv^{2}+u^{4}\) . Its partial derivative in \(u\) is \(2uv+v^{2}+4u^{3}=4+4+4=12\) at \((1,2)\) , and in \(v\) it is \(u^{2}+2uv=1+4=5\) . The two methods agree.

5. Let \(w=f(x,y)\) with \(x=s^{2}-t^{2}\) and \(y=2st\) , where \(f\) is differentiable. Express \(w_s\) and \(w_t\) in terms of \(f_x\) and \(f_y\) , and show that \(w_s^{2}+w_t^{2}=4(s^{2}+t^{2})(f_x^{2}+f_y^{2})\) .

   Solution

   By the coordinate form of the chain rule, with \(x_s=2s\) , \(y_s=2t\) , \(x_t=-2t\) , \(y_t=2s\) ,

   \[ w_s=2s\,f_x+2t\,f_y,\qquad w_t=-2t\,f_x+2s\,f_y , \]

   where \(f_x\) and \(f_y\) are evaluated at \((s^{2}-t^{2},2st)\) . Squaring,

   \[ w_s^{2}=4s^{2}f_x^{2}+8st\,f_xf_y+4t^{2}f_y^{2},\qquad w_t^{2}=4t^{2}f_x^{2}-8st\,f_xf_y+4s^{2}f_y^{2}. \]

   Adding, the cross terms cancel and the rest collects as \(4(s^{2}+t^{2})f_x^{2}+4(s^{2}+t^{2})f_y^{2}=4(s^{2}+t^{2})(f_x^{2}+f_y^{2})\) .

   The reason for the clean answer is visible in the derivative matrix of the substitution, \(\begin{pmatrix}2s&-2t\\ 2t&2s\end{pmatrix}\) : its columns are orthogonal and both have length \(2\sqrt{s^{2}+t^{2}}\) , so it is a rotation followed by a scaling. Such a map multiplies the length of every gradient by the same factor.

6. A differentiable function \(f\) on \(\mathbb R^n\) minus the origin is homogeneous of degree \(k\) if \(f(t\mathbf x)=t^{k}f(\mathbf x)\) for all \(t>0\) . Prove that then \(\mathbf x\cdot\nabla f(\mathbf x)=kf(\mathbf x)\) , and verify the identity for \(f(x,y)=x^{3}+xy^{2}\) .

   Solution

   Fix \(\mathbf x\ne\mathbf 0\) and consider the function \(\varphi(t)=f(t\mathbf x)\) for \(t>0\) . It is \(f\) composed with the curve \(t\mapsto t\mathbf x\) , whose velocity is \(\mathbf x\) , so by Theorem 2.10

   \[ \varphi'(t)=\nabla f(t\mathbf x)\cdot\mathbf x . \]

   On the other hand homogeneity says \(\varphi(t)=t^{k}f(\mathbf x)\) , and differentiating this expression in \(t\) gives \(\varphi'(t)=kt^{k-1}f(\mathbf x)\) . Equating the two expressions and setting \(t=1\) yields \(\nabla f(\mathbf x)\cdot\mathbf x=kf(\mathbf x)\) .

   For \(f(x,y)=x^{3}+xy^{2}\) , which is homogeneous of degree \(3\) because every term has total degree \(3\) , we have \(f_x=3x^{2}+y^{2}\) and \(f_y=2xy\) . Then \(xf_x+yf_y=3x^{3}+xy^{2}+2xy^{2}=3x^{3}+3xy^{2}=3f\) , as the identity predicts.

7. The radius of a cone is \(3\) and increasing at the rate \(0.2\) per second while its height is \(4\) and decreasing at \(0.1\) per second. Use the chain rule to find the rate of change of the volume \(V=\tfrac13\pi r^{2}h\) .

   Answer

   \(\dfrac{dV}{dt}=\dfrac{\pi}{3}\bigl(2rh\,r'+r^{2}h'\bigr)=\dfrac{\pi}{3}\bigl(2\cdot 3\cdot 4\cdot 0.2+9\cdot(-0.1)\bigr)=1.3\pi\approx 4.08\) cubic units per second. The volume is increasing even though the height is falling.

8. For the curve \(x^{2}+xy+y^{2}=7\) , find the slope and the tangent line at \((1,2)\) by implicit differentiation, checking the hypothesis of Theorem 3.6, and find all points of the curve where the tangent is horizontal.

   Solution

   Let \(F(x,y)=x^{2}+xy+y^{2}\) , a polynomial and hence of class \(C^1\) . At \((1,2)\) we have \(F=1+2+4=7\) , so the point is on the curve, and \(F_x=2x+y=4\) , \(F_y=x+2y=5\) . Since \(F_y(1,2)\ne 0\) , Theorem 3.6 applies: near \((1,2)\) the curve is a graph \(y=\varphi(x)\) with slope \(-F_x/F_y\) , which is \(-\tfrac45\) . The tangent line is \(y-2=-\tfrac45(x-1)\) , that is, \(4x+5y=14\) .

   The tangent is horizontal where the slope \(-F_x/F_y\) is zero, which requires \(F_x=0\) and \(F_y\ne 0\) . The condition \(2x+y=0\) gives \(y=-2x\) ; substituting into the curve, \(x^{2}-2x^{2}+4x^{2}=3x^{2}=7\) , so \(x=\pm\sqrt{7/3}\) . The two points are \(\bigl(\sqrt{7/3},-2\sqrt{7/3}\bigr)\) and \(\bigl(-\sqrt{7/3},2\sqrt{7/3}\bigr)\) . At both, \(F_y=x+2y=-3x\ne 0\) , so the theorem applies and the tangent is indeed horizontal. These are the lowest and highest points of the curve, which is an ellipse.

9. The equation \(z^{3}+xz-y=0\) holds at \((x,y,z)=(1,2,1)\) . Show that it defines \(z\) as a \(C^1\) function of \((x,y)\) near that point, compute \(z_x\) and \(z_y\) there, and use them to estimate the value of \(z\) at \((1.1,1.9)\) .

   Solution

   Let \(F(x,y,z)=z^{3}+xz-y\) , a polynomial. At \((1,2,1)\) , \(F=1+1-2=0\) . The partial derivative with respect to the variable we want to solve for is \(F_z=3z^{2}+x\) , which equals \(4\) at the point and so is not zero. By the general form of Theorem 3.6, near \((1,2,1)\) the level set \(F=0\) is the graph of a unique \(C^1\) function \(z=\varphi(x,y)\) with \(\varphi(1,2)=1\) .

   Its partial derivatives are \(z_x=-F_x/F_z\) and \(z_y=-F_y/F_z\) . Here \(F_x=z=1\) and \(F_y=-1\) , so \(z_x=-\tfrac14\) and \(z_y=\tfrac14\) at the point.

   The linear estimate for a displacement \((0.1,-0.1)\) is \(z\approx 1-\tfrac14(0.1)+\tfrac14(-0.1)\) , that is, \(0.95\) . Solving \(z^{3}+1.1z-1.9=0\) numerically gives \(0.9494\) to four places, so the estimate is off by less than \(0.001\) . Notice that we differentiated a function for which we have no formula.

10. A student reasons: “Since \(y=r\sin\theta\) , we have \(\partial y/\partial\theta=r\cos\theta\) , and therefore \(\partial\theta/\partial y=1/(r\cos\theta)\) .” Find the correct value of \(\partial\theta/\partial y\) , identify precisely what is wrong with the reasoning, and state the correct relation between the two sets of partial derivatives.

    Solution

    The correct value comes from the inverse of the derivative matrix of the polar map, as in Example 3.4: \(\partial\theta/\partial y=\cos\theta/r\) . It can be confirmed from \(\theta=\arctan(y/x)\) , whose partial derivative in \(y\) is \(x/(x^{2}+y^{2})=r\cos\theta/r^{2}\) . The student’s answer \(1/(r\cos\theta)\) differs from this by the factor \(\cos^{2}\theta\) .

    The flaw is that the two partial derivatives hold different variables fixed. In \(\partial y/\partial\theta=r\cos\theta\) the variable \(r\) is constant: the point moves along a circle. In \(\partial\theta/\partial y\) the variable \(x\) is constant: the point moves along a vertical line. A reciprocal relation holds for a function of one variable and its inverse, where there is only one way to move. With two variables there are many, and the two derivatives describe motions along different curves.

    The correct statement is that the matrices of partial derivatives are inverse to each other, \(D(T^{-1})=(DT)^{-1}\) , which follows from the chain rule applied to \(T^{-1}\circ T\) . An individual entry of the inverse matrix is not the reciprocal of the corresponding entry of the original.

11. For \(F(x,y)=(x^{2}-y^{2},\ 2xy)\) , determine where Theorem 3.5 applies, show that \(F\) is nevertheless not one-to-one on the plane with the origin removed, and compute the derivative of a local inverse at the point \(F(2,1)=(3,4)\) .

    Hint

    From Exercise 2, \(\det DF=4(x^{2}+y^{2})\) , which is nonzero except at the origin, so Theorem 3.5 applies at every other point. For the failure of injectivity compare \(F(x,y)\) with \(F(-x,-y)\) . For the last part invert the matrix \(DF(2,1)\) of Exercise 3; you should find \(\begin{pmatrix}0.2&0.1\\ -0.1&0.2\end{pmatrix}\) .

12. Generalize Example 3.1 to an arm with link lengths \(\ell_1\) and \(\ell_2\) : write down the hand position, compute \(\det DF\) , and describe the postures in which the hand loses a direction of motion and which direction is lost.

    Hint

    The hand is at \((\ell_1\cos\theta_1+\ell_2\cos(\theta_1+\theta_2),\ \ell_1\sin\theta_1+\ell_2\sin(\theta_1+\theta_2))\) . Expanding the determinant as in Example 3.1 gives \(\ell_1\ell_2\sin\theta_2\) . It vanishes when \(\theta_2\) is \(0\) or \(\pi\) , the arm fully extended or folded back on itself, and in both cases the columns of \(DF\) are perpendicular to the arm, so motion along the arm is lost.

13. Let \(f\) be differentiable on a convex open set \(U\) with \(\lVert\nabla f(\mathbf x)\rVert\le M\) for all \(\mathbf x\in U\) . Prove that \(\lvert f(\mathbf b)-f(\mathbf a)\rvert\le M\lVert\mathbf b-\mathbf a\rVert\) for all \(\mathbf a,\mathbf b\in U\) , and find such an \(M\) for \(f(x,y)=\sin(x+2y)\) on the plane.

    Solution

    A set is convex when it contains the segment joining any two of its points, so for \(\mathbf a,\mathbf b\in U\) the whole segment lies in \(U\) and Theorem 3.7 applies: there is a point \(\mathbf c\) on the segment with \(f(\mathbf b)-f(\mathbf a)=\nabla f(\mathbf c)\cdot(\mathbf b-\mathbf a)\) . By the Cauchy–Schwarz inequality and the hypothesis at the point \(\mathbf c\) ,

    \[ \lvert f(\mathbf b)-f(\mathbf a)\rvert\le\lVert\nabla f(\mathbf c)\rVert\,\lVert\mathbf b-\mathbf a\rVert\le M\lVert\mathbf b-\mathbf a\rVert . \]

    For \(f(x,y)=\sin(x+2y)\) the gradient is \(\cos(x+2y)\,(1,2)\) , of length \(\lvert\cos(x+2y)\rvert\sqrt5\le\sqrt5\) . So \(M=\sqrt5\) works on the whole plane, which is convex, and \(\lvert\sin(x_1+2y_1)-\sin(x_2+2y_2)\rvert\le\sqrt5\,\lVert(x_1,y_1)-(x_2,y_2)\rVert\) for all pairs of points. No smaller constant works, because the bound is attained in the limit of nearby points along the direction \((1,2)\) where the cosine is \(1\) .

14. Show by direct computation that for \(F(t)=(t^{2},t^{3})\) on \([0,1]\) there is no \(t_0\) with \(F(1)-F(0)=F'(t_0)\) , and explain which step in the proof of Theorem 3.7 has no counterpart for vector-valued functions.

    Solution

    Here \(F(1)-F(0)=(1,1)\) and \(F'(t)=(2t,3t^{2})\) . The equation \(F'(t_0)=(1,1)\) requires \(2t_0=1\) and \(3t_0^{2}=1\) simultaneously, that is, \(t_0=\tfrac12\) and \(t_0=1/\sqrt3\) . These are different numbers, so no such \(t_0\) exists.

    In the proof of Theorem 3.7 the one-variable Mean Value Theorem is applied to the single real-valued function \(g\) . For a map with two coordinates one can apply it to each coordinate function separately: the first coordinate \(t^{2}\) has its mean-value point at \(\tfrac12\) and the second coordinate \(t^{3}\) has its own at \(1/\sqrt3\) . The step that has no counterpart is the production of one point that serves all coordinates at once. The Mean Value Theorem is an existence statement about a real-valued function, and existence statements for several functions need not be satisfied by a common point.

15. Let \(f\) and \(g\) be twice differentiable functions of one variable and \(c\) a constant. Use the chain rule to show that \(u(x,t)=f(x-ct)+g(x+ct)\) satisfies \(u_{tt}=c^{2}u_{xx}\) .

    Hint

    Put \(p=x-ct\) and \(q=x+ct\) . Then \(u_x=f'(p)+g'(q)\) and \(u_t=-cf'(p)+cg'(q)\) by the chain rule. Differentiate once more in the same way and compare \(u_{tt}\) with \(u_{xx}\) ; each differentiation in \(t\) brings out a factor \(\mp c\) .

16. The period of a pendulum is \(T=2\pi\sqrt{L/g}\) . Use the linearization of \(T\) as a function of \((L,g)\) to show that the relative error in \(T\) is approximately half the relative error in \(L\) minus half the relative error in \(g\) , and bound it when \(L\) is known to within \(1\) percent and \(g\) to within \(0.5\) percent.

    Answer

    \(T_L=\pi/\sqrt{Lg}\) and \(T_g=-\pi\sqrt L/g^{3/2}\) , so \(\Delta T\approx T_L\,\Delta L+T_g\,\Delta g\) and, dividing by \(T\) , \(\Delta T/T\approx\tfrac12\Delta L/L-\tfrac12\Delta g/g\) . In the worst case the two errors have opposite signs, giving a relative error in \(T\) of at most about \(0.5+0.25=0.75\) percent.

17. A camera maps a point \((x,y,z)\) with \(z>0\) to the image point \(P(x,y,z)=(x/z,\ y/z)\) . A particle moves along the curve \(\mathbf r(t)=(t,t^{2},1+t)\) from Chapter 1. Find the velocity of its image at \(t=1\) as the product \(DP(\mathbf r(1))\,\mathbf r'(1)\) , and check by differentiating the image curve directly.

    Solution

    At \(t=1\) the particle is at \(\mathbf r(1)=(1,1,2)\) with velocity \(\mathbf r'(1)=(1,2t,1)=(1,2,1)\) . The derivative of the camera map is

    \[ DP(x,y,z)=\begin{pmatrix}1/z&0&-x/z^{2}\\ 0&1/z&-y/z^{2}\end{pmatrix},\qquad DP(1,1,2)=\begin{pmatrix}\tfrac12&0&-\tfrac14\\ 0&\tfrac12&-\tfrac14\end{pmatrix}. \]

    By the chain rule the velocity of the image point is the product

    \[ DP(\mathbf r(1))\,\mathbf r'(1)=\bigl(\tfrac12-\tfrac14,\ 1-\tfrac14\bigr), \]

    which is \((\tfrac14,\tfrac34)\) .

    Directly, the image curve is \(\bigl(t/(1+t),\ t^{2}/(1+t)\bigr)\) . The quotient rule gives the derivatives \(1/(1+t)^{2}\) and \((2t(1+t)-t^{2})/(1+t)^{2}\) , which at \(t=1\) are \(\tfrac14\) and \(\tfrac34\) . The third column of \(DP\) records that moving away from the camera pulls the image toward the centre of the picture.

18. Prove directly from Definition 3.1 that the derivative matrix is unique: if matrices \(A\) and \(B\) both satisfy the definition at \(\mathbf a\) , then \(A=B\) .

    Solution

    Suppose \(A\) and \(B\) both satisfy Definition 3.1 at \(\mathbf a\) . For any \(\mathbf h\ne\mathbf 0\) ,

    \[ (A-B)\mathbf h=\bigl[F(\mathbf a+\mathbf h)-F(\mathbf a)-B\mathbf h\bigr]-\bigl[F(\mathbf a+\mathbf h)-F(\mathbf a)-A\mathbf h\bigr], \]

    so by the triangle inequality \(\lVert(A-B)\mathbf h\rVert/\lVert\mathbf h\rVert\) is at most the sum of the two quotients in the definition, and therefore tends to \(0\) as \(\mathbf h\to\mathbf 0\) .

    Now fix any nonzero vector \(\mathbf v\) and take \(\mathbf h=t\mathbf v\) with \(t>0\) . By linearity \((A-B)(t\mathbf v)=t(A-B)\mathbf v\) , so

    \[ \frac{\lVert(A-B)\mathbf h\rVert}{\lVert\mathbf h\rVert}=\frac{\lVert(A-B)\mathbf v\rVert}{\lVert\mathbf v\rVert}, \]

    which does not depend on \(t\) . A constant that tends to \(0\) as \(t\to 0\) is \(0\) . Hence \((A-B)\mathbf v=\mathbf 0\) for every \(\mathbf v\) , and taking \(\mathbf v\) to be each standard basis vector shows that every column of \(A-B\) is zero. So \(A=B\) . The argument uses exactly one feature of linear maps: their relative size along a ray does not change with scale, so a linear map that is relatively small near the origin must vanish.