Suppose
f
(
x
,
t
)
=
e
−
2
t
sin
⁡
(
x
+
3
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
+
3
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

Let's unpack the problem by identifying the function and the differential approach we should use.
The function given is f(x,t) = e^{-2t} sin(x + 3t).
We are asked to approximate f(2.02, -0.03) by first evaluating near the point (2, 0) and using the differential df.
First, compute the partial derivatives at (2,0):
- f_x(x,t) = e^{-2t} cos(x + 3t). At (2,0), this becomes f_x(2,0) = cos(2).
- f_t(x,t) = e^{-2t}[3 cos(x + 3t) − 2 sin(x + 3t)]. At (2,0), this becomes f_t(2,0) = 3 cos(2) − 2 sin(2).
Also, f(2,0) = e^0 sin(2) = sin(2).
We then use the differential df = f_x dx + f_t dt with the changes dx = 2.02 − 2 = 0.02 and dt = −0.03.
Plugging in: df ≈ (cos(2))*(0.02) + (3 cos(2) − 2 sin(2))*(−0.03).
Compute the terms: 0.02 cos(2) − 0.09 cos(2) + 0.06 sin(2) = −0.07 cos(2) + 0.06 sin(2).
Therefore the linear approximation is f(2.02, −0.03) ≈ f(2,0) + df = sin(2) + (−0.07 cos(2) + 0.06 sin(2))
= (1.06) sin(2) − 0.07 cos(2).
Now let's examine each answer option and see how it compares to this expression.

Option 1: 1.06 sin(2) + 0.11 cos(2)
This option uses the same coefficient for sin(2) as our approximation, namely 1.06, which matches the sin(2) term after adding df. However, the cosine term has a positive 0.11 cos(2), whereas our differential calculation gives −0.07 cos(2). The sign is crucial; the extra 0.11 is also not present in the linear approximation. This option incorrectly changes the sign and magnitude of the cos(2) contribution.

Option 2: 0.94 sin(2) + 0.11 cos(2)
Here the sin(2) term has a smaller coefficient (0.94 instead of 1.06), which would correspond to subtracting 0.12 sin(2) somewhere, not allowed by our df calculation. The cosine term is positive 0.11 cos(2), which again conflicts with the correct negative contribution found in df. Overall, both the magnitude and the sign of both components do not align with the differential approximation.

Option 3: 1.06 sin(2) − 0.07 cos(2)
This option matches exactly the expression we derived from the differential approximation: the sin(2) term has coefficient 1.06, and the cos(2) term has coefficient −0.07. The signs and coefficients align with df = (0.02) f_x(2,0) + (−0.03) f_t(2,0) leading to f(2,0) + df = sin(2) + (−0.07 cos(2) + 0.06 sin(2)) = 1.06 sin(2) − 0.07 cos(2).

Option 4: 0.94 sin(2) − 0.07 cos(2)
This option keeps the correct cos(2) term sign, but the sin(2) coefficient is 0.94 instead of 1.06. That would require an incorrect adjustment to the sin(2) component, implying a miscalculation in the dx contribution or f(2,0) value. Since our derivation yields 1.06 for the sin(2) term, this option does not match the differential result.

In summary, evaluating the differential df at (2,0) and applying dx = 0.02 and dt = −0.03 leads to the approximation f(2.02, −0.03) ≈ 1.06 sin(2) − 0.07 cos(2). Among the given options, only the third option has this exact form, with the correct coefficients and signs for both sine and cosine terms.

MATH1062_MATH1005_MATH1023 MATH1062/1023 Calculus Quiz 8

Suppose f ( x , t ) = e − 2 t sin ⁡ ( x + 3 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 ) .)

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