📝 Differentials in calculus (16 MCQs)
📖 From Calculus • 4. Topics in Differentiation • 16 questions available
What is Differentials in calculus?
Definition:
Differentials represent infinitesimal changes in variables. If , the differential is defined as , where is an arbitrary small change in . Unlike , which is the actual change, is the change along the tangent line. Differentials provide a way to estimate changes in the function value corresponding to small changes in the input variable.
Example:
For , find when and . Since , we have . The actual change , showing is a close approximation.
Reason:
Differentials form the basis of integral calculus and are essential for understanding sensitivity analysis. They allow engineers and scientists to estimate how small errors or variations in measurements propagate through calculations, providing a tool for precision and error bounding.
📝 All Differentials in calculus MCQs
Q1. In differential notation, how is defined in terms of the derivative of at a point and the increment ?
📖 Explanation: The definition of the differential follows directly from the derivative: multiplying the derivative f'(x) by the independent increment gives the linear change . This relationship treats as an actual ratio, not just a symbol.
Q2. If is a small increment and the derivative f'(x_0)=-3, what can be said about the sign of the differential at ?
📖 Explanation: Since dy = f'(x_0)dx and is positive, the sign of is determined by the sign of the derivative. With f'(x_0) = -3, the product yields a negative result, so must be negative.
Q3. Using the local linear approximation at for , what is the approximate value of \\\sqrt{1.21}\?
📖 Explanation: The linear approximation at is \\\sqrt{x}\\approx 1+\\tfrac12(x-1)\. Substituting gives \1+\\tfrac12(0.21)=1+0.105=1.105\, which rounds to 1.10. This estimate is close to the true value \\\sqrt{1.21}=1.1\.
Q4. Which of the following statements correctly describes the relative size of the linear approximation error for \f(x)=\\sin x\ near \x=0\ versus near \x=\\pi/2\ for the same small \|dx|\?
📖 Explanation: The error of a linear approximation depends on the second derivative. At \x=0\, \f''(0)=0\, while at \x=\\pi/2\, \f''(\\pi/2)=-1\. Because the magnitude of the second derivative is larger at \\\pi/2\, the approximation error is greater there for the same \dx\.
Q5. If a function \f\ is differentiable at \x_0\, which of the following must be true about its graph near \(x_0,f(x_0))\?
📖 Explanation: Differentiability guarantees that the function behaves locally like its tangent line: as the interval around \x_0\ shrinks, the graph approaches the line \y=f(x_0)+f'(x_0)(x-x_0)\. This means a sufficiently magnified view looks linear, though the curve need not be exactly the line.
Q6. For \f(x)=e^x\ at \x_0=0\ with \dx=0.1\, what is the differential \dy\ and what does it approximate?
📖 Explanation: Using the definition \dy = f'(x_0)dx\ gives \dy = e^{0}\\cdot0.1 = 0.1\. This linear estimate represents the change in the function value over the small interval and is close to the exact change \e^{0.1}-1\\approx0.1052\, illustrating how differentials provide a quick approximation.
Q7. Using differentials, approximate \\\ln(1.05)\ by linearizing \f(x)=\\ln x\ at \x_0=1\ with \dx=0.05\. Which value best matches the approximation?
📖 Explanation: The derivative of \\\ln x\ at \x=1\ is \1\. Multiplying by \dx=0.05\ gives \dy = 0.05\. Hence the linear approximation predicts \\\ln(1.05)\\approx0.05\. The true value is \0.04879\, showing the differential gives a slightly high estimate.
Q8. Which function generally yields a smaller linear approximation error for the same small \|dx|\: \f(x)=x^2\ or \g(x)=\\sqrt[3]{x}\ at a point where both have derivative magnitude 2?
📖 Explanation: Error of a linear approximation is proportional to the second derivative. At the point where both derivatives equal 2, \f''(x)=2\ for the quadratic, whereas the cube‑root function has a second derivative \g''(x) = -\\tfrac{2}{9}x^{-5/3}\, whose magnitude is less than 2. Therefore, \g\ typically yields a smaller error.
Q9. If the radius of a circle changes by a small amount \dr\, which expression using differentials gives the approximate change in area \dA\?
📖 Explanation: The area of a circle is \A=\\pi r^2\. Differentiating yields \dA = 2\\pi r\\,dr\. This linear relation shows that a tiny increase in radius produces an area increase proportional to the current circumference, providing a quick estimate for small changes.
Q10. For \f(x)=x^3\ at \x_0=2\ with \dx=0.01\, use differentials to approximate \(2.01)^3\. Which result is correct?
📖 Explanation: The derivative is \f'(x)=3x^2\; at \x_0=2\ this equals 12. Multiplying by \dx=0.01\ gives \dy=0.12\. Adding to \f(2)=8\ yields the approximation \8+0.12=8.12\. The true value is \8.1206\, confirming the differential estimate.
Q11. Given \\\frac{dy}{dx}=3x^2\, \dx=-0.02\ at \x=4\, what is the approximate change in \y\?
📖 Explanation: Compute \dy = 3x^2\\,dx = 3\\cdot16\\cdot(-0.02) = 48(-0.02) = -0.96\. The negative sign indicates a decrease. Thus the differential predicts that \y\ will drop by about 0.96 when \x\ is reduced by 0.02 at \x=4\.
Q12. Can differentials \dx\ and \dy\ be defined at a point where a function is not differentiable?
📖 Explanation: The definition of a differential relies on the existence of the derivative at the point: \dy = f'(x)dx\. If \f'(x)\ does not exist, the product cannot be formed, so differentials are not defined at points of non‑differentiability.
Q13. Using differentials, approximate \(1.02)^5\ by linearizing \f(x)=x^5\ at \x_0=1\ with \dx=0.02\. Which value is closest to the approximation?
📖 Explanation: The derivative at \x=1\ is \f'(1)=5\. Multiplying by \dx=0.02\ gives \dy=0.1\. Adding to \f(1)=1\ yields the estimate \1+0.1=1.10\. The exact value \(1.02)^5\\approx1.10408\, showing the differential provides a close approximation.
Q14. Which statement best captures why the error of the local linear approximation for a differentiable function \f\ at \x_0\ is of order \o(dx)\?
📖 Explanation: Taylor's theorem expresses \f(x_0+dx)=f(x_0)+f'(x_0)dx+\\frac{f''(c)}{2}(dx)^2\ for some \c\. The remainder term contains \(dx)^2\, which becomes insignificant relative to \dx\ as \dx\ approaches zero. Hence the approximation error is of smaller order than \dx\, denoted \o(dx)\.
Q15. Estimate the error when using the linear approximation for \f(x)=\\tan x\ at \x_0=0.1\ to compute \\\tan 0.12\. Choose the closest bound.
📖 Explanation: The second derivative of \\\tan x\ is \2\\sec^2 x\\tan x\. Near \0.1\, \\\sec^2 0.1\\approx1.01\ and \\\tan 0.1\\approx0.1003\, giving \|f''|\\approx0.202\. Using the error bound \\\frac{M}{2}(dx)^2\ with \dx=0.02\ yields \\\frac{0.202}{2}(0.02)^2\\approx4.0\\times10^{-5}\.
Q16. When solving \f(x)=x^{3}-2x+1=0\ near the root \x\\approx1\, one iteration of Newton's method starting from \x_{0}=1\ gives \x_{1}=1-\\frac{f(1)}{f'(1)}\. Using differentials, the linear approximation predicts the root at \x\\approx1-\\frac{f(1)}{f'(1)}\ as well. Which statement correctly compares the two approaches?
📖 Explanation: Both methods start with the same linearization: \x\\approx x_0-\\frac{f(x_0)}{f'(x_0)}\. Newton's method then repeats this process, using the updated point to improve accuracy, while the differential approach stops after the first linear step. Consequently, Newton's method typically yields a more precise approximation after one iteration.