Differential Calculus Questions and Answers

Answer: B

The derivatives run in a cycle of four:

\[ \sin x \;\to\; \cos x \;\to\; -\sin x \;\to\; -\cos x \;\to\; \sin x. \]

So the fourth derivative returns the original function, and in general \( y^{(n)} = \sin(x + n\pi/2) \). Any derivative whose order is a multiple of four reproduces the function unchanged

Answer: A

Curvature \( \kappa \) measures how fast the tangent direction turns per unit arc length, and the radius of curvature is \( \rho = 1/\kappa \). It is the radius of the circle that matches the curve in position, tangent and second derivative at the point, so a straight line has zero curvature and infinite radius

Answer: D

The standard Cartesian formula is

\[ \rho \;=\; \frac{\left(1 + y'^{2}\right)^{3/2}}{|y''|}. \]

The inverted form is the curvature itself. The exponent must be \( 3/2 \), not \( 1/2 \): the arc length contributes one power of \( (1+y'^{2})^{1/2} \) and the turning of the tangent the remaining one

Answer: C

At the origin \( y' = 2x = 0 \) and \( y'' = 2 \), so the formula collapses to the reciprocal of the second derivative:

\[ \rho \;=\; \frac{(1 + 0)^{3/2}}{|2|} \;=\; \frac{1}{2}. \]

The parabola is therefore locally the circle \( x^{2} + (y - 1/2)^{2} = 1/4 \), whose centre sits on the axis at the height of the focus

Answer: B

A circle turns at a constant rate, and its curvature is the reciprocal of its radius:

\[ \kappa \;=\; \frac{1}{\rho} \;=\; \frac{1}{5} \;=\; 0.2. \]

This is the one curve whose curvature is the same at every point, which is what makes it the natural yardstick: any curve's curvature at a point is that of the circle which best fits it there

Answer: A

Each point of a curve has a centre of curvature, lying on the normal at a distance \( \rho \) on the concave side. As the point moves along the curve the centre traces the evolute, and the original curve is then one involute of it. Inflexion points, where the curvature vanishes, send the centre off to infinity

Answer: D

The denominator vanishes at \( x = 3 \), giving a vertical asymptote. For the horizontal one, divide through by \( x \):

\[ \lim_{x \to \pm\infty} \frac{2x+1}{x-3} \;=\; \lim_{x \to \pm\infty} \frac{2 + 1/x}{1 - 3/x} \;=\; 2. \]

So \( y = 2 \) is approached at both ends, the ratio of the leading coefficients

Answer: C

Dividing out leaves a linear quotient plus a remainder that dies away, so the curve approaches a slant line. Equal degrees give a horizontal asymptote at the ratio of leading coefficients, a smaller numerator gives \( y = 0 \), and an excess of two or more gives a parabolic or higher approach with no straight asymptote at all

Answer: B

The envelope is tangent to each curve of the family at some point, and is found by eliminating the parameter between \( F(x,y,c) = 0 \) and \( \partial F/\partial c = 0 \). The orthogonal-crossing curve is a member of the orthogonal family instead, and the locus of centres of curvature is the evolute

Answer: A

Differentiating with respect to \( x \) treats \( y \) as a constant, so the factor \( y^{2} \) comes along unchanged:

\[ f_{x} \;=\; \frac{\partial}{\partial x}\left(x^{3} y^{2}\right) \;=\; 3x^{2} y^{2}. \]

The form \( 2x^{3}y \) is \( f_{y} \) instead, and \( 6x^{2}y \) is the mixed derivative \( f_{xy} \), got by differentiating once with respect to each variable