Math 21a
Linear Approximation
Fall, 2015
We can use the first derivatives to estimate values of a function. This method is called linear
approximation (linearization).
The linearization of the function f (x, y) near the point (x0 , y0 ) is given by
f (x, y) ≈ L(x, y) = f (x0 , y0 ) + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ).
We will discuss the relation between the linearization and tangent spaces to the graph z = f (x, y)
next week!
1. (a) Suppose f (x, y) = 2x2 + y 2 . Find the linearization L(x, y) of f (x, y) at the point (1, 1)
and estimate the value of f (1.1, 0.9).
p
(b) Suppose f (x, y) = 1 + sin(x) + 2y 2 . Find the linearization L(x, y) of f (x, y) at the
point (0, 2) and estimate the value of f (0.01, 2.01).
√
2. Estimate the value of e0.01 3.9.
We can consider the linearization of functions of several variables. For example,
L(x, y, z) = g(x0 , y0 , z0 ) + gx (x0 , y0 , z0 )(x − x0 ) + gy (x0 , y0 , z0 )(y − y0 ) + gz (x0 , y0 , z0 )(z − z0 )
≈ g(x, y, z)
for (x, y, z) near (x0 , y0 , z0 ).
3. Use the linearization of functions of three or four variables and answer the following questions:
√
(a) Let f (x, y, z) = x + xyz. Use linear approximation to approximate the value of
f (1.1, 1.9, 3.1).
(b) Estimate 3(1.01)2
p
5 + (1.99)2 + (2.01) (0.99)2 .
Linear Approximation – Answers and Solutions
1. (a) It’s a bit annoying to evaluate f (1.1, 0.9). But evaluating f (x, y) at a closer point, i.e.
f (1, 1) is easy. So we will consider the linearization L(x, y) of f (x, y) at the point (1, 1)
to estimate the value of f (1.1, 0.9). We easily see that fx = 4x and fy = 2y. So
L(x, y) = f (1, 1) + fx (1, 1)(x − 1) + fy (1, 1)(y − 1) = 3 + 4(x − 1) + 2(y − 1).
This means that
f (x, y) = 2x2 + y 2 ≈ L(x, y) = 3 + 4(x − 1) + 2(y − 1)
for (x, y) near (1, 1).
At the point (x, y) = (1.1, 0.9), we get the approximation f (1.1, 0.9) ≈ 3 + 4(1.1 − 1) +
2(0.9 − 1) = 3.2. (Of course, we can fairly easily calculate f (1.1, 0.9) = 3.23, but this
isn’t always so simple.)
(b) As fx = √
2
cos(x)
1+sin(x)+y 2
and fy = √
2y
1+sin(x)+y 2
, we get
1
4
L(x, y) = f (0, 2) + fx (0, 2)(x − 0) + fy (0, 2)(y − 2) = 3 + x + (y − 2).
6
3
This means that
p
1
4
f (x, y) = 1 + sin(x) + y 2 ≈ L(x, y) = 3 + x + (y − 2)
6
3
for (x, y) near (0, 2).
At the point (x, y) = (0.01, 2.01), we get the approximation f (0.01, 2.01) ≈ 3 + 61 (0.01) +
4
(2.01 − 1) = 3.015. (The actual decimal value of f (0.01, 2.01) is 3.01499 . . ..)
3
√
2. Here a reasonable function is f (x, y) = ex y near the point (x0 , y0 ) = (0, 4). The key is that
we have to be able to easily evaluate f (x0 , y0 ), fx (x0 , y0 ) and fy (x0 , y0 ) (and, of course, that
(x0 , y0 ) has to be near the point at which we want to estimate f (x, y)).
√
Using our expression from above, we find f (0, 4) = 2, fx (x, y) = ex y so f (0, 4) = 2, and
x
fy (x, y) = 2e√y so fy (0, 4) = 41 . Thus the tangent plane approximation is
1
√
ex y ≈ L(x, y) = 2 + 2(x − 0) + (y − 4)
4
for (x, y) near (0, 4).
√
Thus e0.01 3.9 ≈ L(0.01, 3.9) = 2 + 2(0.01) + 14 (−0.1) = 1.995. (The actual decimal value of
√
e0.01 3.9 is 1.99469 . . ..)
3. (a) First, we should pick a point near (1.1, 1.9, 3.1) at which we can easily compute the
linear approximation. Let’s use (1, 2, 3). Then, we know that f (x, y, z) ≈ f (1, 2, 3) +
fx (1, 2, 3)(x − 1) + fy (1, 2, 3)(y − 2) + fz (1, 2, 3)(z − 3) for (x, y, z) near (1, 2, 3). Let’s
find the necessary partial derivatives:
fx = 21 x−1/2 + yz ⇒ fx (1, 2, 3) = 6.5
fy = xz
⇒ fy (1, 2, 3) = 3
fz = xy
⇒ fz (1, 2, 3) = 2
Also, f (1, 2, 3) = 7. So, for (x, y, z) near (1, 2, 3), f (x, y, z) ≈ 7 + 6.5(x − 1) + 3(y −
2) + 2(z − 3). Plugging in (x, y, z) = (1.1, 1.9, 3.1) gives f (1.1, 1.9, 3.1) ≈ 7 + 6.5(.1) +
3(−.1) + 2(.1) = 7.55. (The actual value of f (1.1, 1.9, 3.1) is about 7.5278.)
p
(b) We put f (x, y, z, w) = 3x2 5 + y 2 + zw2 , and evaluate f (1.01, 1.99, 2.01, 0.99) using
the linearization. We consider
the linearization L(x, y, z, w) of f (x, y, z, w) at the point
p
2
2
(1, 2, 2, 1). As fx = 6x 5 + y , fy = √3x y 2 , fz = w2 and fw = 2zw, we get
5+y
L(x, y, z, w) = f (1, 2, 2, 1) + fx (1, 2, 2, 1)(x − 1)fy (1, 2, 2, 1)(y − 2)
+ fz (1, 2, 2, 1)(z − 2) + fw (1, 2, 2, 1)(w − 1)
= 11 + 18(x − 1) + 2(y − 2) + (z − 2) + 4(w − 1).
So we get
f (1.01, 1.99, 2.01, 0.99) ≈ L(1.01, 1.99, 2.01, 0.99) = 11+0.18−0.02+0.01−0.04 = 11.13.
p
(The actual decimal value of 3(1.01)2 5 + (1.99)2 + (2.01) (0.99)2 is 11.13052 . . ..)