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Chapter 6 · Application of Derivatives

Exercise 6.1 · Q17 (MCQ): Rate of change of a circle's area at r=6r = 6 cm

Exercise 6.1Rate of change of quantities · Section 6.2
The question

Choose the correct answer.

The rate of change of the area of a circle with respect to its radius rr at r=6r = 6 cm is

(A) 10π10\pi    (B) 12π12\pi    (C) 8π8\pi    (D) 11π11\pi

Exam answer

A=πr2⇒dAdr=2πrA = \pi r^2 \Rightarrow \frac{dA}{dr} = 2\pi r⁠.

At r=6r = 6⁠, dAdr=2π(6)=12π\frac{dA}{dr} = 2\pi(6) = 12\pi⁠.

Answer: (B) 12π12\pi

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

"Rate of change of the area with respect to the radius" is the derivative dAdr\frac{dA}{dr}⁠. For A=πr2A = \pi r^2 this is 2πr2\pi r⁠, which is the circumference: widening the radius by a little adds a thin ring whose area is about (circumference) ×\times (width). At r=6r = 6⁠, 2π(6)=12π2\pi(6) = 12\pi⁠, option (B).

Given, to find, and the plan
Given
A circle of radius rr⁠; its area is A=πr2A = \pi r^2⁠.
To find
dAdr\frac{dA}{dr} at r=6r = 6 cm, and which option it matches.
Plan
(1) Write the area formula. (2) Differentiate with respect to rr⁠. (3) Substitute r=6r = 6⁠. (4) Examine every option.

Basics you need

1Rate of change with respect to a variable

The rate of change of AA with respect to rr is dAdr\frac{dA}{dr}⁠, and its value at r=r0r = r_0 is dAdr∣r=r0\left.\frac{dA}{dr}\right|_{r = r_0} (Section 6.2). Basics: rate of change as a derivative

2Area of a circle

A=πr2A = \pi r^2⁠; circumference 2πr2\pi r⁠. Basics: area and volume formulas

3Power rule

ddr(r2)=2r\frac{d}{dr}\left(r^2\right) = 2r⁠, and a constant factor such as π\pi stays in front.

Solution

  1. Differentiate the area with respect to the radius

    A=πr2⟹dAdr=π⋅2r=2πr.A = \pi r^2 \quad\Longrightarrow\quad \frac{dA}{dr} = \pi\cdot 2r = 2\pi r.
    The question asks for a rate with respect to the radius (not time), so we differentiate AA directly with respect to rr⁠. No chain rule in time is needed. This is exactly Example 1 of Section 6.2, which found dAdr=2πr\frac{dA}{dr} = 2\pi r⁠.
    ddr(r2)=2r\frac{d}{dr}\left(r^2\right) = 2r (Table 5.3); constant multiple π\pi stays in front. Basics: rate of change as a derivative

  2. Substitute r=6r = 6

    dAdr∣r=6=2π(6)=12π cm2/cm.\left.\frac{dA}{dr}\right|_{r=6} = 2\pi(6) = 12\pi \ \text{cm}^2/\text{cm}.

    That is about 37.7 cm2^2 of area per cm of radius.

    Differentiate first, then substitute: the derivative 2πr2\pi r is valid for every rr⁠, and we want its value at one radius. The unit is cm2^2 (area) per cm (radius); the options give just the number 12π12\pi⁠.

  3. Examine every option

    OptionValueEqual to 2π(6)=12π2\pi(6) = 12\pi⁠?Where the number comes from
    (A)10π10\piNo2πr2\pi r at r=5r = 5 (Example 1), not 6
    (B)12π12\piYes2πr2\pi r at r=6r = 6
    (C)8π8\piNo2πr2\pi r at r=4r = 4 (Q1(b)), not 6
    (D)11π11\piNo2πr2\pi r at r=5.5r = 5.5⁠, not 6

    Only (B) equals 12π12\pi⁠. Note that the area itself, π(6)2=36π\pi(6)^2 = 36\pi⁠, is not offered: the question is about the rate, not the area.

    Each wrong option is the correct formula 2πr2\pi r evaluated at a different radius, so they all fail for the same reason: 2πr2\pi r is strictly increasing in rr⁠, and gives 12π12\pi only at r=6r = 6⁠.
    Rate of change at a point: evaluate the derivative at that point. Lesson: rate of change of quantities
2 4 6 8 50 100 150 200 r A (6, 36π) slope 12π ≈ 37.7 A = πr²
The area A=πr2A = \pi r^2 as the radius grows. At r=6r = 6 the tangent (red) has slope dAdr=12π≈37.7\frac{dA}{dr} = 12\pi \approx 37.7⁠: near that radius, each extra cm of radius adds about 12π12\pi cm2^2 of area.
Result

dAdr=2πr=12π\frac{dA}{dr} = 2\pi r = 12\pi at r=6r = 6 cm. The answer is (B) 12π12\pi. (Textbook answer: B.)

Check it with numbers

π(6.001)2≈113.1350\pi(6.001)^2 \approx 113.1350 and π(5.999)2≈113.0596\pi(5.999)^2 \approx 113.0596⁠. Their difference is about 0.0754 for a change of 0.002 cm in radius, so the rate is about 0.07540.002≈37.7\frac{0.0754}{0.002} \approx 37.7⁠, and 12π≈37.69912\pi \approx 37.699⁠. (The area itself at r=6r = 6 is 36π≈113.1036\pi \approx 113.10 cm2^2⁠.)

Common mistakes
  1. Giving the area instead of the rate. πr2=36π\pi r^2 = 36\pi at r=6r = 6 is the area, not its rate of change.
  2. Using πr\pi r or πr2/2\pi r^2/2⁠. The power rule gives ddr(πr2)=2πr\frac{d}{dr}(\pi r^2) = 2\pi r⁠.
  3. Using the wrong radius. Options (A), (C) and (D) are 2πr2\pi r at r=5r = 5⁠, 4 and 5.5.
What to take away
  1. ddr(πr2)=2πr\frac{d}{dr}\left(\pi r^2\right) = 2\pi r⁠: the rate of change of a circle's area with its radius is its circumference.
  2. Find the derivative as a formula first, then evaluate it at the given point.
  3. In an MCQ, explain why each wrong option fails; here each is the right formula at the wrong radius.

Every fact used, and where it comes from

FactUsed inSource
Rate of change of AA with respect to rr is dAdr\frac{dA}{dr}Steps 1, 2Section 6.2; Example 1 (Lesson: rate of change of quantities)
Area of a circle πr2\pi r^2Step 1Mensuration (Class X, Areas Related to Circles)
ddr(r2)=2r\frac{d}{dr}\left(r^2\right) = 2rStep 1Table 5.3, Section 5.3
10π10\pi at r=5r = 5⁠, 8π8\pi at r=4r = 4Step 3Example 1, Section 6.2; Exercise 6.1 Q1(b)
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