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JEECUP (UP Polytechnic) Application of Derivatives — practice questions — page 2

Explore 69 free Application of Derivatives questions in JEECUP (UP Polytechnic) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Application of Derivatives questions and exercises with answers

Showing questions 31–60 of 69.

Rate of change of volume V = (4/3)πr³ of a sphere with radius:Read the answer and worked solution →Linear approximation of f(x) near x = a:Read the answer and worked solution →A spherical balloon's volume increases at 30 cm³/s. At r = 5 cm, how fast is the radius increasing?Read the answer and worked solution →Find equation of tangent to y = x² - 1 at point (2, 3):Read the answer and worked solution →Maximum area of a rectangle inscribed in a semicircle of radius R (base on diameter):Read the answer and worked solution →Use L'Hôpital's rule: lim_(x→0) (sin x - x)/x³ =Read the answer and worked solution →For f(x) = x² ln x on (0, ∞), find x where f has local minimum:Read the answer and worked solution →A 5 m ladder slides down a wall. Bottom moves outward at 1 m/s. Speed of top when bottom is 3 m from wall:Read the answer and worked solution →For y = e^x − x, find values where tangent is horizontal:Read the answer and worked solution →For f(x) = x³ - 3x² + 4, the function is increasing on intervals:Read the answer and worked solution →Find minimum value of f(x) = (sin x)² + (cos x)² + 2 (sin x)(cos x):Read the answer and worked solution →Volume of cylinder inscribed in sphere of radius R, expressed using h (height of cylinder):Read the answer and worked solution →Concavity test: f(x) = e^x is:Read the answer and worked solution →Find the rate at which the surface area of a balloon (sphere) is changing when radius is 10 cm and dr/dt = 0.1 cm/s:Read the answer and worked solution →What does the sign of f ′(x) tell you?Read the answer and worked solution →Complete the definition of a critical pointRead the answer and worked solution →Order the first derivative testRead the answer and worked solution →Match each sign change to its verdictRead the answer and worked solution →The second derivative testRead the answer and worked solution →True or false: maxima and minimaRead the answer and worked solution →Order the working rule for absolute maxima and minimaRead the answer and worked solution →Tap every point you must test for an absolute maximumRead the answer and worked solution →Local or absolute?Read the answer and worked solution →Complete the rate of change statementRead the answer and worked solution →The chain rule for related ratesRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →True or false: increasing and decreasingRead the answer and worked solution →

Worked practice questions

Rate of change of volume V = (4/3)πr³ of a sphere with radius:

  1. 3πr²
  2. 4πr³
  3. πr²
  4. 4πr² (= surface area of sphere)
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Correct answer: D. 4πr² (= surface area of sphere)

dV/dr = 4πr². Equals surface area of sphere (the shell-area interpretation — V is integral of A from 0 to r).

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Linear approximation of f(x) near x = a:

  1. f'(x) × a
  2. f(a) × x
  3. f(a) + f'(a)(x - a)
  4. f(a)²
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Correct answer: C. f(a) + f'(a)(x - a)

**Linearization:** L(x) = f(a) + f'(a)(x - a). The tangent-line approximation. Valid for x near a (small Δx). Foundation of Taylor series (1st-order term).

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A spherical balloon's volume increases at 30 cm³/s. At r = 5 cm, how fast is the radius increasing?

  1. 1 cm/s
  2. 3/10π cm/s ≈ 0.095 cm/s
  3. 30 cm/s
  4. 2 cm/s
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Correct answer: B. 3/10π cm/s ≈ 0.095 cm/s

V = (4/3)πr³. dV/dt = 4πr² dr/dt. We have **So dr/dt = (dV/dt)/(4πr²). That is **So dr/dt = 30/(4π × 25). That is **So dr/dt = 30/(100π). So **So dr/dt = 3/(10π) ≈ 0.0955 cm/s.**.

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