IB Mathematics: Analysis and Approaches Quiz: Derivative Definition
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Derivative DefinitionQuestion 1 of 20

The gradient of the tangent to the curve y=f(x)y=f(x) at x=1x=1 is given by the limit lim⁡h→03(1+h)2−3h\lim_{h \to 0} \frac{3(1+h)^2 - 3}{h}. Given that f(1)=3f(1)=3, find the equation of the normal line to the curve at x=1x=1.

y=6x−3y = 6x - 3
y=−6x+9y = -6x + 9
y=−16x+196y = -\frac{1}{6}x + \frac{19}{6}
y=16x+176y = \frac{1}{6}x + \frac{17}{6}
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Derivative Definition

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Question 1

The gradient of the tangent to the curve y=f(x)y=f(x) at x=1x=1 is given by the limit lim⁡h→03(1+h)2−3h\lim_{h \to 0} \frac{3(1+h)^2 - 3}{h}. Given that f(1)=3f(1)=3, find the equation of the normal line to the curve at x=1x=1.

  1. y=6x−3y = 6x - 3
  2. y=−6x+9y = -6x + 9
  3. y=−16x+196y = -\frac{1}{6}x + \frac{19}{6} (correct answer)
  4. y=16x+176y = \frac{1}{6}x + \frac{17}{6}
Explanation: Step 1: Find the gradient of the tangent, mtm_t, by evaluating the limit. mt=lim⁡h→03(1+2h+h2)−3h=lim⁡h→03+6h+3h2−3h=lim⁡h→06h+3h2h=lim⁡h→0(6+3h)=6m_t = \lim_{h \to 0} \frac{3(1+2h+h^2) - 3}{h} = \lim_{h \to 0} \frac{3+6h+3h^2 - 3}{h} = \lim_{h \to 0} \frac{6h+3h^2}{h} = \lim_{h \to 0} (6+3h) = 6. Step 2: Find the gradient of the normal, mnm_n. The normal is perpendicular to the tangent, so mn=−1mt=−16m_n = -\frac{1}{m_t} = -\frac{1}{6}. Step 3: Find the equation of the normal line. The line passes through the point (1,f(1))=(1,3)(1, f(1)) = (1, 3) and has gradient −16-\frac{1}{6}. Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1): y−3=−16(x−1)y - 3 = -\frac{1}{6}(x - 1). Rearranging to slope-intercept form: y=−16x+16+3=−16x+16+186=−16x+196y = -\frac{1}{6}x + \frac{1}{6} + 3 = -\frac{1}{6}x + \frac{1}{6} + \frac{18}{6} = -\frac{1}{6}x + \frac{19}{6}.

Question 2

The line y=8x−5y=8x-5 is tangent to the curve y=kx2y=kx^2 at some point. The gradient of the tangent at a point xx on the curve is given by lim⁡h→0k(x+h)2−kx2h\lim_{h \to 0} \frac{k(x+h)^2 - kx^2}{h}. Find the value of kk.

  1. 22
  2. 5/45/4
  3. 16/516/5 (correct answer)
  4. 44
Explanation: The gradient of the tangent line y=8x−5y=8x-5 is 8. The derivative of y=kx2y=kx^2, which is given by the limit, must be equal to 8 at the point of tangency. The derivative is y′=2kxy' = 2kx. So, 2kx=82kx=8, which simplifies to kx=4kx=4. At the point of tangency, the coordinates must be the same for the line and the curve, so kx2=8x−5kx^2 = 8x-5. We can substitute kx=4kx=4 into this equation: (kx)x=8x−5  ⟹  4x=8x−5(kx)x = 8x-5 \implies 4x = 8x-5. Solving for xx gives 4x=54x=5, so x=5/4x=5/4. Now we can find kk using kx=4kx=4: k(5/4)=4  ⟹  k=16/5k(5/4)=4 \implies k = 16/5.

Question 3

The limit expression lim⁡h→0(3+h)2−9h\lim_{h \to 0} \frac{(3+h)^2 - 9}{h} represents the derivative of a function f(x)f(x) at a point x=ax=a. What are f(x)f(x) and aa?

  1. f(x)=x2−9f(x) = x^2 - 9 and a=3a=3
  2. f(x)=x2f(x) = x^2 and a=3a=3 (correct answer)
  3. f(x)=(x+3)2f(x) = (x+3)^2 and a=0a=0
  4. f(x)=2xf(x) = 2x and a=3a=3
Explanation: The definition of the derivative of a function f(x)f(x) at a point x=ax=a is f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. Comparing this to the given limit, lim⁡h→0(3+h)2−9h\lim_{h \to 0} \frac{(3+h)^2 - 9}{h}, we can identify a=3a=3. This means f(a+h)=f(3+h)=(3+h)2f(a+h) = f(3+h) = (3+h)^2 and f(a)=f(3)=9f(a) = f(3) = 9. From these, we can deduce that the function is f(x)=x2f(x) = x^2, since f(3)=32=9f(3) = 3^2 = 9.

Question 4

The line y=8x−5y=8x-5 is tangent to the curve y=kx2y=kx^2 at some point. The gradient of the tangent at a point xx on the curve is given by lim⁡h→0k(x+h)2−kx2h\lim_{h \to 0} \frac{k(x+h)^2 - kx^2}{h}. Find the value of kk.

  1. 22
  2. 5/45/4
  3. 16/516/5 (correct answer)
  4. 44
Explanation: The gradient of the tangent line y=8x−5y=8x-5 is 8. The derivative of y=kx2y=kx^2, which is given by the limit, must be equal to 8 at the point of tangency. The derivative is y′=2kxy' = 2kx. So, 2kx=82kx=8, which simplifies to kx=4kx=4. At the point of tangency, the coordinates must be the same for the line and the curve, so kx2=8x−5kx^2 = 8x-5. We can substitute kx=4kx=4 into this equation: (kx)x=8x−5  ⟹  4x=8x−5(kx)x = 8x-5 \implies 4x = 8x-5. Solving for xx gives 4x=54x=5, so x=5/4x=5/4. Now we can find kk using kx=4kx=4: k(5/4)=4  ⟹  k=16/5k(5/4)=4 \implies k = 16/5.

Question 5

The height HH in metres of a ball thrown vertically upwards is modelled by H(t)=20t−5t2H(t) = 20t - 5t^2, where tt is time in seconds. The instantaneous velocity of the ball at t=1t=1 is defined by lim⁡h→0H(1+h)−H(1)h\lim_{h \to 0} \frac{H(1+h) - H(1)}{h}. Calculate this velocity.

  1. 55 m/s
  2. 1010 m/s (correct answer)
  3. 1515 m/s
  4. 2020 m/s
Explanation: We need to evaluate the limit for H(t)=20t−5t2H(t) = 20t - 5t^2 at t=1t=1. First, find H(1)=20(1)−5(1)2=15H(1) = 20(1) - 5(1)^2 = 15. Next, find H(1+h)=20(1+h)−5(1+h)2=20+20h−5(1+2h+h2)=20+20h−5−10h−5h2=15+10h−5h2H(1+h) = 20(1+h) - 5(1+h)^2 = 20 + 20h - 5(1 + 2h + h^2) = 20 + 20h - 5 - 10h - 5h^2 = 15 + 10h - 5h^2. Now, set up the difference quotient: H(1+h)−H(1)h=(15+10h−5h2)−15h=10h−5h2h\frac{H(1+h) - H(1)}{h} = \frac{(15 + 10h - 5h^2) - 15}{h} = \frac{10h - 5h^2}{h}. Factor out hh: h(10−5h)h=10−5h\frac{h(10 - 5h)}{h} = 10 - 5h. Finally, take the limit: lim⁡h→0(10−5h)=10\lim_{h \to 0} (10 - 5h) = 10. The instantaneous velocity at t=1t=1 is 10 m/s.

Question 6

The height HH in metres of a ball thrown vertically upwards is modelled by H(t)=20t−5t2H(t) = 20t - 5t^2, where tt is time in seconds. The instantaneous velocity of the ball at t=1t=1 is defined by lim⁡h→0H(1+h)−H(1)h\lim_{h \to 0} \frac{H(1+h) - H(1)}{h}. Calculate this velocity.

  1. 55 m/s
  2. 1010 m/s (correct answer)
  3. 1515 m/s
  4. 2020 m/s
Explanation: We need to evaluate the limit for H(t)=20t−5t2H(t) = 20t - 5t^2 at t=1t=1. First, find H(1)=20(1)−5(1)2=15H(1) = 20(1) - 5(1)^2 = 15. Next, find H(1+h)=20(1+h)−5(1+h)2=20+20h−5(1+2h+h2)=20+20h−5−10h−5h2=15+10h−5h2H(1+h) = 20(1+h) - 5(1+h)^2 = 20 + 20h - 5(1 + 2h + h^2) = 20 + 20h - 5 - 10h - 5h^2 = 15 + 10h - 5h^2. Now, set up the difference quotient: H(1+h)−H(1)h=(15+10h−5h2)−15h=10h−5h2h\frac{H(1+h) - H(1)}{h} = \frac{(15 + 10h - 5h^2) - 15}{h} = \frac{10h - 5h^2}{h}. Factor out hh: h(10−5h)h=10−5h\frac{h(10 - 5h)}{h} = 10 - 5h. Finally, take the limit: lim⁡h→0(10−5h)=10\lim_{h \to 0} (10 - 5h) = 10. The instantaneous velocity at t=1t=1 is 10 m/s.

Question 7

For the function f(x)=x3f(x) = x^3, let RavgR_{avg} be the average rate of change on the interval [1,3][1, 3] and RinstR_{inst} be the instantaneous rate of change at x=1x=1, defined by lim⁡h→0(1+h)3−1h\lim_{h \to 0} \frac{(1+h)^3-1}{h}. Find the value of Ravg−RinstR_{avg} - R_{inst}.

  1. 33
  2. 1010 (correct answer)
  3. 1313
  4. 1616
Explanation: First, calculate the average rate of change: Ravg=f(3)−f(1)3−1=33−132=27−12=262=13R_{avg} = \frac{f(3) - f(1)}{3-1} = \frac{3^3 - 1^3}{2} = \frac{27 - 1}{2} = \frac{26}{2} = 13. Next, calculate the instantaneous rate of change at x=1x=1 by finding the derivative f′(x)=3x2f'(x) = 3x^2 and evaluating it at x=1x=1, which gives Rinst=f′(1)=3(1)2=3R_{inst} = f'(1) = 3(1)^2 = 3. The question asks for Ravg−Rinst=13−3=10R_{avg} - R_{inst} = 13 - 3 = 10.

Question 8

The derivative of f(x)=x2−6xf(x) = x^2 - 6x is defined by f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. Find the value of cc such that the instantaneous rate of change of ff at x=cx=c is equal to 4.

  1. 11
  2. 22
  3. 44
  4. 55 (correct answer)
Explanation: The instantaneous rate of change of ff at x=cx=c is f′(c)f'(c). We need to find the value of cc for which f′(c)=4f'(c) = 4. First, find the derivative of f(x)=x2−6xf(x) = x^2 - 6x. Using the power rule, f′(x)=2x−6f'(x) = 2x - 6. Now, set f′(c)=4f'(c) = 4: 2c−6=42c - 6 = 4. Solving for cc, we get 2c=102c = 10, which means c=5c = 5.

Question 9

Let g(x)g(x) be an even, differentiable function. Find the value of lim⁡h→0g(h)−g(−h)h\lim_{h \to 0} \frac{g(h) - g(-h)}{h}.

  1. 00 (correct answer)
  2. g′(0)g'(0)
  3. 2g′(0)2g'(0)
  4. Does not exist
Explanation: A function g(x)g(x) is even if g(x)=g(−x)g(x) = g(-x) for all xx in its domain. Therefore, for the numerator of the given expression, we have g(h)−g(−h)=g(h)−g(h)=0g(h) - g(-h) = g(h) - g(h) = 0. The expression inside the limit is 0h\frac{0}{h}, which is 0 for all h≠0h \neq 0. The limit of 0 as h→0h \to 0 is 0.

Question 10

For the function f(x)=x3f(x) = x^3, let RavgR_{avg} be the average rate of change on the interval [1,3][1, 3] and RinstR_{inst} be the instantaneous rate of change at x=1x=1, defined by lim⁡h→0(1+h)3−1h\lim_{h \to 0} \frac{(1+h)^3-1}{h}. Find the value of Ravg−RinstR_{avg} - R_{inst}.

  1. 33
  2. 1010 (correct answer)
  3. 1313
  4. 1616
Explanation: First, calculate the average rate of change: Ravg=f(3)−f(1)3−1=33−132=27−12=262=13R_{avg} = \frac{f(3) - f(1)}{3-1} = \frac{3^3 - 1^3}{2} = \frac{27 - 1}{2} = \frac{26}{2} = 13. Next, calculate the instantaneous rate of change at x=1x=1 by finding the derivative f′(x)=3x2f'(x) = 3x^2 and evaluating it at x=1x=1, which gives Rinst=f′(1)=3(1)2=3R_{inst} = f'(1) = 3(1)^2 = 3. The question asks for Ravg−Rinst=13−3=10R_{avg} - R_{inst} = 13 - 3 = 10.

Question 11

The expression lim⁡h→0cos⁡(π3+h)−12h\lim_{h \to 0} \frac{\cos(\frac{\pi}{3}+h) - \frac{1}{2}}{h} represents the value of f′(a)f'(a) for some function ff and point aa. Find this value.

  1. −32-\frac{\sqrt{3}}{2} (correct answer)
  2. −12-\frac{1}{2}
  3. 12\frac{1}{2}
  4. 32\frac{\sqrt{3}}{2}
Explanation: The given expression matches the definition of the derivative, f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. By comparing the terms, we can identify the function as f(x)=cos⁡(x)f(x) = \cos(x) and the point as a=π3a = \frac{\pi}{3}, since cos⁡(π3)=12\cos(\frac{\pi}{3}) = \frac{1}{2}. The limit therefore represents the value of the derivative of cos⁡(x)\cos(x) at x=π3x = \frac{\pi}{3}. The derivative of f(x)=cos⁡(x)f(x) = \cos(x) is f′(x)=−sin⁡(x)f'(x) = -\sin(x). Evaluating at a=π3a = \frac{\pi}{3}, we get f′(π3)=−sin⁡(π3)=−32f'(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}.

Question 12

The gradient of the tangent to the curve y=f(x)y=f(x) at x=1x=1 is given by the limit lim⁡h→03(1+h)2−3h\lim_{h \to 0} \frac{3(1+h)^2 - 3}{h}. Given that f(1)=3f(1)=3, find the equation of the normal line to the curve at x=1x=1.

  1. y=6x−3y = 6x - 3
  2. y=−6x+9y = -6x + 9
  3. y=−16x+196y = -\frac{1}{6}x + \frac{19}{6} (correct answer)
  4. y=16x+176y = \frac{1}{6}x + \frac{17}{6}
Explanation: Step 1: Find the gradient of the tangent, mtm_t, by evaluating the limit. mt=lim⁡h→03(1+2h+h2)−3h=lim⁡h→03+6h+3h2−3h=lim⁡h→06h+3h2h=lim⁡h→0(6+3h)=6m_t = \lim_{h \to 0} \frac{3(1+2h+h^2) - 3}{h} = \lim_{h \to 0} \frac{3+6h+3h^2 - 3}{h} = \lim_{h \to 0} \frac{6h+3h^2}{h} = \lim_{h \to 0} (6+3h) = 6. Step 2: Find the gradient of the normal, mnm_n. The normal is perpendicular to the tangent, so mn=−1mt=−16m_n = -\frac{1}{m_t} = -\frac{1}{6}. Step 3: Find the equation of the normal line. The line passes through the point (1,f(1))=(1,3)(1, f(1)) = (1, 3) and has gradient −16-\frac{1}{6}. Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1): y−3=−16(x−1)y - 3 = -\frac{1}{6}(x - 1). Rearranging to slope-intercept form: y=−16x+16+3=−16x+16+186=−16x+196y = -\frac{1}{6}x + \frac{1}{6} + 3 = -\frac{1}{6}x + \frac{1}{6} + \frac{18}{6} = -\frac{1}{6}x + \frac{19}{6}.

Question 13

The derivative of f(x)=x2−6xf(x) = x^2 - 6x is defined by f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. Find the value of cc such that the instantaneous rate of change of ff at x=cx=c is equal to 4.

  1. 11
  2. 22
  3. 44
  4. 55 (correct answer)
Explanation: The instantaneous rate of change of ff at x=cx=c is f′(c)f'(c). We need to find the value of cc for which f′(c)=4f'(c) = 4. First, find the derivative of f(x)=x2−6xf(x) = x^2 - 6x. Using the power rule, f′(x)=2x−6f'(x) = 2x - 6. Now, set f′(c)=4f'(c) = 4: 2c−6=42c - 6 = 4. Solving for cc, we get 2c=102c = 10, which means c=5c = 5.

Question 14

Find the gradient of the tangent to the curve y=xy=\sqrt{x} at x=9x=9 by evaluating the limit lim⁡h→09+h−3h\lim_{h \to 0} \frac{\sqrt{9+h}-3}{h}.

  1. 33
  2. 1/31/3
  3. 1/61/6 (correct answer)
  4. 66
Explanation: To evaluate this limit, we multiply the numerator and denominator by the conjugate of the numerator, which is 9+h+3\sqrt{9+h}+3. This gives: lim⁡h→09+h−3h⋅9+h+39+h+3=lim⁡h→0(9+h)−9h(9+h+3)\lim_{h \to 0} \frac{\sqrt{9+h}-3}{h} \cdot \frac{\sqrt{9+h}+3}{\sqrt{9+h}+3} = \lim_{h \to 0} \frac{(9+h) - 9}{h(\sqrt{9+h}+3)}. Simplifying the numerator gives lim⁡h→0hh(9+h+3)\lim_{h \to 0} \frac{h}{h(\sqrt{9+h}+3)}. We can cancel hh (since h≠0h \neq 0), leaving lim⁡h→019+h+3\lim_{h \to 0} \frac{1}{\sqrt{9+h}+3}. Now, we can substitute h=0h=0 to find the limit: 19+0+3=13+3=16\frac{1}{\sqrt{9+0}+3} = \frac{1}{3+3} = \frac{1}{6}.

Question 15

Find the value of the limit lim⁡k→0(a+3k)2−a2k\lim_{k \to 0} \frac{(a+3k)^2 - a^2}{k}.

  1. 2a2a
  2. 3a3a
  3. 6a6a (correct answer)
  4. 9a29a^2
Explanation: We can evaluate this limit algebraically. First, expand the numerator: (a+3k)2−a2=(a2+6ak+9k2)−a2=6ak+9k2(a+3k)^2 - a^2 = (a^2 + 6ak + 9k^2) - a^2 = 6ak + 9k^2. Now, substitute this back into the limit expression: lim⁡k→06ak+9k2k\lim_{k \to 0} \frac{6ak + 9k^2}{k}. We can factor out kk from the numerator: lim⁡k→0k(6a+9k)k\lim_{k \to 0} \frac{k(6a + 9k)}{k}. For k≠0k \neq 0, we can cancel kk: lim⁡k→0(6a+9k)\lim_{k \to 0} (6a + 9k). Now, we can substitute k=0k=0 to evaluate the limit, which gives 6a+9(0)=6a6a + 9(0) = 6a. Alternatively, this limit is a variation of the derivative definition for f(x)=x2f(x)=x^2 at x=ax=a. It represents 3f′(a)=3(2a)=6a3f'(a) = 3(2a) = 6a.

Question 16

The limit expression lim⁡h→0(3+h)2−9h\lim_{h \to 0} \frac{(3+h)^2 - 9}{h} represents the derivative of a function f(x)f(x) at a point x=ax=a. What are f(x)f(x) and aa?

  1. f(x)=x2−9f(x) = x^2 - 9 and a=3a=3
  2. f(x)=x2f(x) = x^2 and a=3a=3 (correct answer)
  3. f(x)=(x+3)2f(x) = (x+3)^2 and a=0a=0
  4. f(x)=2xf(x) = 2x and a=3a=3
Explanation: The definition of the derivative of a function f(x)f(x) at a point x=ax=a is f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. Comparing this to the given limit, lim⁡h→0(3+h)2−9h\lim_{h \to 0} \frac{(3+h)^2 - 9}{h}, we can identify a=3a=3. This means f(a+h)=f(3+h)=(3+h)2f(a+h) = f(3+h) = (3+h)^2 and f(a)=f(3)=9f(a) = f(3) = 9. From these, we can deduce that the function is f(x)=x2f(x) = x^2, since f(3)=32=9f(3) = 3^2 = 9.

Question 17

The expression lim⁡h→0cos⁡(π3+h)−12h\lim_{h \to 0} \frac{\cos(\frac{\pi}{3}+h) - \frac{1}{2}}{h} represents the value of f′(a)f'(a) for some function ff and point aa. Find this value.

  1. −32-\frac{\sqrt{3}}{2} (correct answer)
  2. −12-\frac{1}{2}
  3. 12\frac{1}{2}
  4. 32\frac{\sqrt{3}}{2}
Explanation: The given expression matches the definition of the derivative, f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. By comparing the terms, we can identify the function as f(x)=cos⁡(x)f(x) = \cos(x) and the point as a=π3a = \frac{\pi}{3}, since cos⁡(π3)=12\cos(\frac{\pi}{3}) = \frac{1}{2}. The limit therefore represents the value of the derivative of cos⁡(x)\cos(x) at x=π3x = \frac{\pi}{3}. The derivative of f(x)=cos⁡(x)f(x) = \cos(x) is f′(x)=−sin⁡(x)f'(x) = -\sin(x). Evaluating at a=π3a = \frac{\pi}{3}, we get f′(π3)=−sin⁡(π3)=−32f'(\frac{\pi}{3}) = -\sin(\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}.

Question 18

Given a function f(x)f(x) such that f′(a)f'(a) exists, which of the following limits is also equal to f′(a)f'(a)?

  1. lim⁡h→0f(a+2h)−f(a)h\lim_{h \to 0} \frac{f(a+2h) - f(a)}{h}
  2. lim⁡h→0f(a+h)−f(a)2h\lim_{h \to 0} \frac{f(a+h) - f(a)}{2h}
  3. lim⁡h→0f(a)−f(a−h)h\lim_{h \to 0} \frac{f(a) - f(a-h)}{h} (correct answer)
  4. lim⁡h→0f(a+h)−f(a−h)h\lim_{h \to 0} \frac{f(a+h) - f(a-h)}{h}
Explanation: The standard definition is f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. Let's analyze the options. A: lim⁡h→02⋅f(a+2h)−f(a)2h=2f′(a)\lim_{h \to 0} 2 \cdot \frac{f(a+2h) - f(a)}{2h} = 2f'(a). B: 12lim⁡h→0f(a+h)−f(a)h=12f′(a)\frac{1}{2} \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \frac{1}{2}f'(a). C: Let k=−hk = -h. As h→0h \to 0, k→0k \to 0. The expression becomes lim⁡k→0f(a)−f(a+k)−k=lim⁡k→0−(f(a+k)−f(a))−k=lim⁡k→0f(a+k)−f(a)k=f′(a)\lim_{k \to 0} \frac{f(a) - f(a+k)}{-k} = \lim_{k \to 0} \frac{-(f(a+k)-f(a))}{-k} = \lim_{k \to 0} \frac{f(a+k)-f(a)}{k} = f'(a). This is correct. D: This limit equals 2f′(a)2f'(a).

Question 19

The quantity lim⁡h→0A(t+h)−A(t)h\lim_{h \to 0} \frac{A(t+h) - A(t)}{h}, where A(t)A(t) is the area of a circular oil slick in m2m^2 at time tt in seconds, represents:

  1. the average rate at which the area is changing over the interval [t,t+h][t, t+h].
  2. the instantaneous rate of change of the area at time tt. (correct answer)
  3. the total change in the area of the oil slick after time tt.
  4. the acceleration of the area's growth at time tt.
Explanation: The expression A(t+h)−A(t)h\frac{A(t+h) - A(t)}{h} represents the average rate of change of the area AA over a time interval of length hh. The limit of this expression as h→0h \to 0 is the definition of the derivative, A′(t)A'(t), which represents the instantaneous rate of change of the area with respect to time at the specific moment tt.

Question 20

Find the gradient of the tangent to the curve y=xy=\sqrt{x} at x=9x=9 by evaluating the limit lim⁡h→09+h−3h\lim_{h \to 0} \frac{\sqrt{9+h}-3}{h}.

  1. 33
  2. 1/31/3
  3. 1/61/6 (correct answer)
  4. 66
Explanation: To evaluate this limit, we multiply the numerator and denominator by the conjugate of the numerator, which is 9+h+3\sqrt{9+h}+3. This gives: lim⁡h→09+h−3h⋅9+h+39+h+3=lim⁡h→0(9+h)−9h(9+h+3)\lim_{h \to 0} \frac{\sqrt{9+h}-3}{h} \cdot \frac{\sqrt{9+h}+3}{\sqrt{9+h}+3} = \lim_{h \to 0} \frac{(9+h) - 9}{h(\sqrt{9+h}+3)}. Simplifying the numerator gives lim⁡h→0hh(9+h+3)\lim_{h \to 0} \frac{h}{h(\sqrt{9+h}+3)}. We can cancel hh (since h≠0h \neq 0), leaving lim⁡h→019+h+3\lim_{h \to 0} \frac{1}{\sqrt{9+h}+3}. Now, we can substitute h=0h=0 to find the limit: 19+0+3=13+3=16\frac{1}{\sqrt{9+0}+3} = \frac{1}{3+3} = \frac{1}{6}.