Suppose
f
(
x
,
t
)
=
e
−
2
t
sin
⁡
(
x
+
5
t
)
. Which of the following is a good approximation of the value of
f
(
2.02
,
−
0.03
)
? (Hint: first find the differential
d
f
at the point
(
2
,
0
)
.)

Question

Suppose 
f
(
x
,
t
)
=
e
−
2
t
sin
⁡
(
x
+
5
t
)
. Which of the following is a good approximation of the value of 
f
(
2.02
,
−
0.03
)
? (Hint: first find the differential 
d
f
 at the point 
(
2
,
0
)
.)

BlackTom AI · Accepted Answer

We start by restating the given problem and the available option to set the stage for step-by-step evaluation.
Question: Suppose f(x,t) = e^{-2t} sin(x + 5t). Which of the following is a good approximation of f(2.02, -0.03) using a differential at the point (2, 0)? The provided option is: 1.06 sin(2) − 0.13 cos(2).

First, identify the differential at (2, 0).
- Write f(x,t) = e^{-2t} sin(u) with u = x + 5t.
- Partial derivatives at (x,t):
  • f_x = ∂f/∂x = e^{-2t} cos(u).
  • f_t = ∂f/∂t = e^{-2t} [−2 sin(u) + 5 cos(u)].
- Evaluate at (x,t) = (2,0): u = 2, e^{-2t} = 1, so
  • f_x(2,0) = cos(2),
  • f_t(2,0) = −2 sin(2) + 5 cos(2).

Next, relate the differential to the target values.
- The shifts are dx = 2.02 − 2 = 0.02 and dt = −0.03.
- Linear approximation (differential): df ≈ f_x dx + f_t dt = cos(2)·0.02 + [−2 sin(2) + 5 cos(2)]·(−0.03).
- Compute term by term using approximate numeric values: sin(2) ≈ 0.909297, cos(2) ≈ −0.416147.
  • f_x dx = cos(2)·0.02 ≈ (−0.416147)·0.02 ≈ −0.0083229.
  • f_t dt = [−2 sin(2) + 5 cos(2)]·(−0.03) = [−2·0.909297 + 5·(−0.416147)]·(−0.03)
    = [−1.818594 − 2.080735]·(−0.03) ≈ (−3.899329)·(−0.03) ≈ 0.1169799.
- Sum: df ≈ −0.0083229 + 0.1169799 ≈ 0.108657.

Thus, the differential approximation gives f(2.02, −0.03) ≈ f(2,0) + df. Since f(2,0) = e^0 sin(2) = sin(2) ≈ 0.909297, the approximation would be
  f(2.02, −0.03) ≈ 0.909297 + 0.108657 ≈ 1.017954.

Now examine the provided option and compare with the computed result.
- The option given is: 1.06 sin(2) − 0.13 cos(2).
- Evaluate this expression numerically using sin(2) ≈ 0.909297 and cos(2) ≈ −0.416147:
  1.06 sin(2) − 0.13 cos(2) ≈ 1.06·0.909297 − 0.13·(−0.416147)
  ≈ 0.963658 + 0.054099 ≈ 1.017757.

Why this option is not a good approximation:
- Although the numeric value 1.017757 is close to the approximately computed 1.017954, the expression 1.06 sin(2) − 0.13 cos(2) is not the same as the derived differential form df ≈ cos(2)·0.02 + [−2 sin(2) + 5 cos(2)]·(−0.03). It happens to yield a similar magnitude but is not the direct result of the differential calculation, and using a coefficient combination like 1.06 and −0.13 lacks direct justification from the linearization process.
- The core uncertainty here is that the option expresses the result in a mixed trigonometric form with particular coefficients, whereas the principled differential approximation yields a linear combination with coefficients determined strictly by the partial derivatives and the given dx, dt. Unless those coefficients match the exact linearization expression, the option cannot be guaranteed to be a correct or faithful differential-based approximation.
- Moreover, if one were to equate the option to the differential form, the coefficients would need to satisfy: df ≈ (dx) cos(2) + (dt)[−2 sin(2) + 5 cos(2)]. Substituting dx = 0.02 and dt = −0.03 gives the exact numeric combination above, which does not neatly simplify to 1.06 sin(2) − 0.13 cos(2) in a general, justified way.

In summary, the option evaluates to a numeric value near 1.0178, matching the rough differential-based estimate, but the form 1.06 sin(2) − 0.13 cos(2) is not the direct, correct representation of the differential expression f_x dx + f_t dt. Therefore, this option does not provide a properly justified differential approximation when expressed in terms of the original partial derivatives and increments.

Suppose f ( x , t ) = e − 2 t sin ⁡ ( x + 5 t ) . Which of the following is a good approximation of the value of f ( 2.02 , − 0.03 ) ? (Hint: first find the differential d f at the point ( 2 , 0 ) .) 单项选择题

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