📝 Local linear approximation linearization (16 MCQs)
📖 From Calculus • 4. Topics in Differentiation • 16 questions available
What is Local linear approximation linearization?
Definition:
Local linear approximation, or linearization, uses the tangent line to a curve at a specific point to approximate the function's values near that point. The formula is . This method relies on the fact that differentiable functions look nearly straight when viewed at a sufficiently small scale, providing a simple linear model for complex nonlinear behavior.
Example:
Approximate using linearization of at . , . So, . For , .
Reason:
Linearization simplifies calculations for functions that are difficult to evaluate directly. It is widely used in engineering and physics for error analysis and in numerical methods to provide initial estimates for iterative algorithms like Newton's method.
📝 All Local linear approximation linearization MCQs
Q1. In the context of local linear approximation, how is the differential \dy\ defined in terms of \dx\ and the derivative of \f\ at \x_0\?
📖 Explanation: The differential \dy\ captures the infinitesimal change in \f\ caused by a small change \dx\. By definition, \dy = f'(x_0)\\,dx\, which directly uses the derivative evaluated at \x_0\. This relationship lets us treat the derivative as a ratio of two infinitesimals, not just a limit.
Q2. What is the linear approximation \L(x)\ of a differentiable function \f\ near a point \a\?
📖 Explanation: The linear approximation, also called the tangent line approximation, uses the function value at \a\ and the slope \f'(a)\. It is given by \L(x)=f(a)+f'(a)(x-a)\. This formula reproduces the function’s value and first‑order behavior near \a\, providing a simple linear model.
Q3. Suppose \f(x)=\\sqrt{x}\. Using the linear approximation at \a=4\, which of the following estimates for \f(4.1)\ is closest to the true value? (Assume \dx=0.1\)
📖 Explanation: The derivative of \\\sqrt{x}\ is \\\frac{1}{2\\sqrt{x}}\. At \x=4\ this equals \\\frac{1}{4}=0.25\. The linear estimate is \2+0.25\\times0.1=2.025\, which matches option A. The other options use incorrect slopes, leading to larger deviations from the true value \\\sqrt{4.1}\.
Q4. Given \f(x)=e^{x}\ and its linear approximation at \a=0\, which expression correctly represents the error term \E(x)=f(x)-L(x)\ for small \x\?
📖 Explanation: Expanding \e^{x}=1+x+\\frac{x^{2}}{2}+\\dots\, the linear part \L(x)=1+x\. Subtracting gives the error \E(x)=\\frac{x^{2}}{2}+O(x^{3})\. Option D repeats this expression, indicating the leading quadratic term dominates the error for small \x\.
Q5. Why does the tangent line provide a good approximation to a differentiable function near the point of tangency?
📖 Explanation: A differentiable function and its tangent line coincide in both value and slope at the point of tangency. This matching of the zeroth‑ and first‑order terms ensures that, in a small neighborhood, the linear model captures the dominant behavior of the function, making the approximation accurate.
Q6. Let \f(x)=\\ln(x)\. Using linear approximation at \a=1\, estimate \f(1.05)\. Which of the following is the correct estimate?
📖 Explanation: The derivative of \\\ln x\ at \x=1\ is \1\. With \dx=0.05\, the linear estimate is \0+1\\times0.05=0.05\. This matches option B. The other numbers result from using incorrect slopes or higher‑order corrections, which are not part of the first‑order linear approximation.
Q7. Consider \f(x)=\\sin x\. The linear approximation at \a=\\pi/6\ is used to estimate \\\sin(\\pi/6+0.1)\. Which statement correctly compares the approximation error to the bound given by the remainder term of the Taylor series?
📖 Explanation: For a first‑order Taylor approximation, the remainder satisfies \|R_2|\\le \\frac{M}{2}(\\Delta x)^2\ with \M\\le1\. With \\\Delta x=0.1\ this gives a bound of \0.005\. The true error is strictly smaller than the bound, so option D correctly states the relationship.
Q8. How does the concept of differentials \dx\ and \dy\ help interpret the derivative \\\frac{dy}{dx}\ as a ratio rather than a single limit object?
📖 Explanation: By defining \dy = f'(x)\\,dx\, the derivative becomes the quotient \\\frac{dy}{dx}=f'(x)\. This viewpoint lets us manipulate \dy\ and \dx\ like ordinary quantities, facilitating chain‑rule calculations and related‑rates problems, while preserving the rigorous limit definition underneath.
Q9. A particle moves along a curve described by \y = x^3\. At \x=2\, the differential \dy\ for a small change \dx = 0.01\ is approximated using the derivative. What is the approximate change in \y\?
📖 Explanation: The derivative \dy/dx = 3x^2\ gives \3(2)^2 = 12\ at \x=2\. Multiplying by the small increment \dx=0.01\ yields \dy \\approx 12\\times0.01 = 0.12\. This linear estimate captures the immediate slope of the cubic curve near the point.
Q10. For the function \f(x)=\\frac{1}{x}\, the linear approximation at \a=2\ is used to estimate \f(2.1)\. Which statement about the sign of the approximation error is true?
📖 Explanation: Because \f''(x)=2/x^3>0\ on \x>0\, the function is convex. The tangent line at \a=2\ therefore lies below the curve, making the linear estimate slightly smaller than the true value. Hence the error (true minus estimate) is positive, which corresponds to option D.
Q11. When approximating a function \f\ near \x=a\ using its tangent line, which condition guarantees that the approximation error tends to zero faster than \|x-a|\ as \x\\to a\?
📖 Explanation: If \f\ possesses a continuous second derivative, the Taylor remainder is of order \(x-a)^2\. Consequently, the error shrinks quadratically, i.e., faster than the linear term \|x-a|\. Differentiability alone only ensures a first‑order error term, not the faster decay.
Q12. If \f(x)=\\tan^{-1}x\ and we use the linear approximation at \a=0\ to estimate \f(0.2)\, which statement best describes the relationship between the estimated value and the true value?
📖 Explanation: The derivative of \\\tan^{-1}x\ at 0 is 1, so the linear estimate is \0+1\\times0.2=0.2\. The actual value \\\tan^{-1}(0.2)\\approx0.1974\ is a bit smaller, making the linear approximation an overestimate, as expressed in option B.
Q13. Consider \f(x)=\\ln(1+x)\. Using the linear approximation at \a=0\, derive an expression for the relative error \\\frac{f(x)-L(x)}{f(x)}\ for small positive \x\. Which expression correctly represents the leading term of this relative error?
📖 Explanation: Expanding \\\ln(1+x)=x-\\frac{x^{2}}{2}+\\dots\ and noting \L(x)=x\, the absolute error is \-\\frac{x^{2}}{2}+\\dots\. Dividing by \f(x)\\approx x\ gives a leading relative error of \-\\frac{x}{2}\. Option D presents this term correctly.
Q14. Explain why the linear approximation fails to provide accurate predictions for functions with vertical tangents, such as \f(x)=\\sqrt[3]{x}\ at \x=0\.
📖 Explanation: A vertical tangent implies an undefined or infinite slope, so the notion of a single linear function with a constant finite slope breaks down. The linear approximation, which relies on a finite derivative to construct the tangent line, cannot be formed, leading to inaccurate or meaningless predictions.
Q15. A particle moves along a curve described by \y = x^3\. At \x=2\, the differential \dy\ for a small change \dx = 0.01\ is approximated using the derivative. What is the approximate change in \y\?
📖 Explanation: The derivative \dy/dx = 3x^2\ gives \3(2)^2 = 12\ at \x=2\. Multiplying by the small increment \dx=0.01\ yields \dy \\approx 12\\times0.01 = 0.12\. This linear estimate captures the immediate slope of the cubic curve near the point.
Q16. A temperature sensor follows \T(t)=20+5\\sin(\\pi t/12)\ degrees Celsius. Using linear approximation at \t=6\ hours, estimate the temperature change after \\\Delta t = 0.5\ hour. Which statement about the sign of the actual temperature change compared to the estimate is correct?
📖 Explanation: At \t=6\ the derivative \T'(t) = (5\\pi/12)\\cos(\\pi t/12)\ equals zero, so the linear model predicts no temperature change. Because the sine curve peaks at \t=6\ and then declines, the true temperature actually drops slightly, matching option C.