Problem 4
Question
Approximate the value of \(f\) at the given point. $$ f(x, y)=\sin \pi x y ;(-1.97,2.005) $$
Step-by-Step Solution
Verified Answer
The approximate value of \( f(-1.97, 2.005) \) is 0.
1Step 1: Identify the Function and Point
We are given the function \( f(x, y) = \sin(\pi x y) \) and the point \((-1.97, 2.005)\). Our goal is to approximate \( f(-1.97, 2.005) \).
2Step 2: Calculate \(\pi x y\) for the Point
Insert the given values into the expression \(\pi x y\): \[\pi (-1.97) \times 2.005 = -\pi \times (1.97 \times 2.005) = -3.95515\pi\]
3Step 3: Use Approximations for Constants
Use approximations for easier calculations. Assume \(\pi \approx 3.14\). Then,\[-3.95515 \times 3.14 \approx -12.4142\]
4Step 4: Use Sine Function for Approximation
Now, find \(\sin(-12.4142)\). Since this is a difficult calculation without a calculator, note that sine function values range between -1 and 1, and small perturbations do not greatly change their values, so approximate using periodic properties of sine function:\[\sin(-12.4142) \approx \sin(2\pi) = 0\]
5Step 5: Conclude with the Approximate Value
After all the calculations and approximations:\[ f(-1.97, 2.005) \approx 0 \]
Key Concepts
Trigonometric FunctionsApproximation MethodsPeriodic Properties
Trigonometric Functions
Trigonometric functions are a fundamental concept in mathematics, especially in topics like multivariable calculus. These functions, which include sine, cosine, and tangent, are used to relate the angles and sides of triangles. In the context of multivariable functions like \( f(x, y) = \sin(\pi x y) \), they help in understanding how values oscillate between -1 and 1 due to their cyclical nature.
- The sine function, \( \sin(x) \), is a periodic function with a period of \(2\pi\), which means it repeats its values every \(2\pi\) units.
- In multivariable calculus, trigonometric functions are often used to model periodic phenomena.
Approximation Methods
In calculus, and particularly multivariable calculus, approximation methods are vital tools. They help us estimate the values of functions at difficult points to calculate exactly. The exercise with the function \( f(x, y) = \sin(\pi x y) \) demonstrates such a technique.
- Approximating \(\pi\) as 3.14 is a common method for simplifying calculations.
- By approximating \(-3.95515\pi\) as \(-12.4142\), you simplify the calculation for the sine function.
- Since directly calculating \(\sin(-12.4142)\) is complex, using known values of the sine function at periodic intervals helps simplify this.
Periodic Properties
The periodic properties of functions like sine and cosine play a significant role in their application in multivariable calculus. Understanding periodicity simplifies calculations, especially when dealing with large inputs. In the given solution, recognizing that the sine function is periodic every \(2\pi\) allowed for an easier analysis of the problem.
- Even complex values, when examined through their periodic nature, often cycle back to simpler, known values.
- Sine has a period of \(2\pi\), so any multiple of \(2\pi\) results in the sine value cycling back to zero, e.g., \(\sin(2\pi) = 0\) or \(\sin(4\pi) = 0\).
- Recognizing the periodic interval can often reduce lengthy calculations, giving accurate approximations quickly.
Other exercises in this chapter
Problem 4
Find the extreme values of \(f\) subject to the given constraint. In each case assume that the extreme values exist. $$ f(x, y, z)=y^{3}+x z^{2} ; x^{2}+y^{2}+z
View solution Problem 4
Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ g(x, y)=x^{2}-x
View solution Problem 4
Find the gradient of the function. $$ f(x, y)=\frac{x y-1}{x^{2}+y^{2}} $$
View solution Problem 4
Compute \(d z / d t\). $$ z=\tan ^{-1}\left(y^{2}-x^{2}\right) ; x=\sin t, y=\cos t $$
View solution