IB Mathematics: Analysis and Approaches Quiz: Implicit And Parametric Differentiation
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Implicit And Parametric DifferentiationQuestion 1 of 20

A curve is described by the parametric equations x=4+2ln⁡(t)x = 4 + 2\ln(t) and y=t2+1y = t^2 + 1. Find the equation of the tangent line to the curve at the point where t=1t=1.

y=x−2y = x - 2
y=x+2y = x + 2
y=−x+6y = -x + 6
y=−x−6y = -x - 6
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Implicit And Parametric Differentiation

Practice Implicit And Parametric Differentiation in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Implicit And Parametric Differentiation, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A curve is described by the parametric equations x=4+2ln⁡(t)x = 4 + 2\ln(t) and y=t2+1y = t^2 + 1. Find the equation of the tangent line to the curve at the point where t=1t=1.

  1. y=x−2y = x - 2 (correct answer)
  2. y=x+2y = x + 2
  3. y=−x+6y = -x + 6
  4. y=−x−6y = -x - 6
Explanation: First, find the point on the curve when t=1t=1. x(1)=4+2ln⁡(1)=4+0=4x(1) = 4 + 2\ln(1) = 4 + 0 = 4. y(1)=12+1=2y(1) = 1^2 + 1 = 2. The point is (4,2)(4, 2). Next, find the slope of the tangent, dydx\frac{dy}{dx}. dxdt=2t\frac{dx}{dt} = \frac{2}{t} and dydt=2t\frac{dy}{dt} = 2t. So, dydx=dy/dtdx/dt=2t2/t=t2\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{2/t} = t^2. At t=1t=1, the slope is m=12=1m = 1^2 = 1. Now, use the point-slope form of a line: y−y1=m(x−x1)y - y_1 = m(x - x_1). y−2=1(x−4)y - 2 = 1(x - 4), which simplifies to y=x−2y = x - 2.

Question 2

What is the slope of the tangent line to the curve defined by the parametric equations x=t2+1,y=arctan⁡(t)x=t^2+1, y=\arctan(t) at the point (2,π/4)(2, \pi/4)?

  1. 1/4 (correct answer)
  2. 1/2
  3. 1
  4. 2
Explanation: First, find the value of the parameter tt corresponding to the point (2,π/4)(2, \pi/4). From x=t2+1x=t^2+1, we have 2=t2+1  ⟹  t2=1  ⟹  t=±12=t^2+1 \implies t^2=1 \implies t=\pm 1. From y=arctan⁡(t)y=\arctan(t), we have π/4=arctan⁡(t)  ⟹  t=tan⁡(π/4)=1\pi/4=\arctan(t) \implies t=\tan(\pi/4)=1. So we use t=1t=1. Next, find dydx\frac{dy}{dx}. dxdt=2t\frac{dx}{dt} = 2t and dydt=11+t2\frac{dy}{dt} = \frac{1}{1+t^2}. Then dydx=dy/dtdx/dt=1/(1+t2)2t=12t(1+t2)\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/(1+t^2)}{2t} = \frac{1}{2t(1+t^2)}. Evaluate the slope at t=1t=1: dydx=12(1)(1+12)=12(2)=14\frac{dy}{dx} = \frac{1}{2(1)(1+1^2)} = \frac{1}{2(2)} = \frac{1}{4}.

Question 3

Consider the curve defined by sin⁡(x)+cos⁡(y)=1\sin(x) + \cos(y) = 1. Which statement is true about the tangent to the curve at the point (π/6,π/3)(\pi/6, \pi/3)?

  1. The tangent line is horizontal.
  2. The tangent line is vertical.
  3. The tangent line has a slope of 1. (correct answer)
  4. The tangent line has a slope of -1.
Explanation: Differentiate implicitly with respect to xx: cos⁡(x)−sin⁡(y)dydx=0\cos(x) - \sin(y)\frac{dy}{dx} = 0. Solving for dydx\frac{dy}{dx} gives dydx=cos⁡(x)sin⁡(y)\frac{dy}{dx} = \frac{\cos(x)}{\sin(y)}. Now, evaluate the slope at the point (π/6,π/3)(\pi/6, \pi/3): dydx=cos⁡(π/6)sin⁡(π/3)=3/23/2=1\frac{dy}{dx} = \frac{\cos(\pi/6)}{\sin(\pi/3)} = \frac{\sqrt{3}/2}{\sqrt{3}/2} = 1. So, the tangent line has a slope of 1. A horizontal tangent would mean the slope is 0, which would require cos⁡(x)=0\cos(x)=0. A vertical tangent would mean the slope is undefined, which would require sin⁡(y)=0\sin(y)=0. A slope of -1 is a plausible distractor if a sign error is made or trig values are mixed up.

Question 4

A particle's position is given by x(t)=3t2−2x(t) = 3t^2 - 2 and y(t)=t3−3ty(t) = t^3 - 3t. For which interval(s) of tt is the particle moving to the right and downwards?

  1. t>1t > 1
  2. 0<t<10 < t < 1 (correct answer)
  3. t<−1t < -1
  4. −1<t<0-1 < t < 0
Explanation: The particle moves to the right when its horizontal velocity, dxdt\frac{dx}{dt}, is positive. dxdt=6t\frac{dx}{dt} = 6t. So, 6t>0  ⟹  t>06t > 0 \implies t > 0. The particle moves downwards when its vertical velocity, dydt\frac{dy}{dt}, is negative. dydt=3t2−3\frac{dy}{dt} = 3t^2 - 3. So, 3t2−3<0  ⟹  t2<1  ⟹  −1<t<13t^2 - 3 < 0 \implies t^2 < 1 \implies -1 < t < 1. For the particle to be moving both to the right and downwards, both conditions must be met. We need to find the intersection of the intervals t>0t > 0 and −1<t<1-1 < t < 1. The intersection is 0<t<10 < t < 1.

Question 5

What is the slope of the tangent line to the curve defined by the parametric equations x=t2+1,y=arctan⁡(t)x=t^2+1, y=\arctan(t) at the point (2,π/4)(2, \pi/4)?

  1. 1/4 (correct answer)
  2. 1/2
  3. 1
  4. 2
Explanation: First, find the value of the parameter tt corresponding to the point (2,π/4)(2, \pi/4). From x=t2+1x=t^2+1, we have 2=t2+1  ⟹  t2=1  ⟹  t=±12=t^2+1 \implies t^2=1 \implies t=\pm 1. From y=arctan⁡(t)y=\arctan(t), we have π/4=arctan⁡(t)  ⟹  t=tan⁡(π/4)=1\pi/4=\arctan(t) \implies t=\tan(\pi/4)=1. So we use t=1t=1. Next, find dydx\frac{dy}{dx}. dxdt=2t\frac{dx}{dt} = 2t and dydt=11+t2\frac{dy}{dt} = \frac{1}{1+t^2}. Then dydx=dy/dtdx/dt=1/(1+t2)2t=12t(1+t2)\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/(1+t^2)}{2t} = \frac{1}{2t(1+t^2)}. Evaluate the slope at t=1t=1: dydx=12(1)(1+12)=12(2)=14\frac{dy}{dx} = \frac{1}{2(1)(1+1^2)} = \frac{1}{2(2)} = \frac{1}{4}.

Question 6

A curve is defined by the parametric equations x=sec⁡(t)x = \sec(t) and y=tan⁡(t)y = \tan(t). Find dydx\frac{dy}{dx} in terms of tt.

  1. sin⁡(t)\sin(t)
  2. cos⁡(t)\cos(t)
  3. csc⁡(t)\csc(t) (correct answer)
  4. sec⁡(t)\sec(t)
Explanation: First, find the derivatives with respect to the parameter tt. dxdt=ddt(sec⁡(t))=sec⁡(t)tan⁡(t)\frac{dx}{dt} = \frac{d}{dt}(\sec(t)) = \sec(t)\tan(t). dydt=ddt(tan⁡(t))=sec⁡2(t)\frac{dy}{dt} = \frac{d}{dt}(\tan(t)) = \sec^2(t). The derivative dydx\frac{dy}{dx} is given by the ratio dy/dtdx/dt\frac{dy/dt}{dx/dt}. So, dydx=sec⁡2(t)sec⁡(t)tan⁡(t)=sec⁡(t)tan⁡(t)\frac{dy}{dx} = \frac{\sec^2(t)}{\sec(t)\tan(t)} = \frac{\sec(t)}{\tan(t)}. To simplify, we can write this in terms of sine and cosine: 1/cos⁡(t)sin⁡(t)/cos⁡(t)=1sin⁡(t)=csc⁡(t)\frac{1/\cos(t)}{\sin(t)/\cos(t)} = \frac{1}{\sin(t)} = \csc(t). Distractor B, cos⁡(t)\cos(t), is dx/dydx/dy. Distractors A and D arise from incorrect simplification of the trigonometric expression.

Question 7

A curve is defined by the parametric equations x=t3−3tx = t^3 - 3t and y=t2−2y = t^2 - 2. Find the coordinates of the points where the tangent to the curve is horizontal.

  1. (-2, -1) and (2, -1)
  2. (0, -2) (correct answer)
  3. (-2, -1)
  4. (0, -2) and (0, -1)
Explanation: A horizontal tangent occurs when dydx=0\frac{dy}{dx} = 0, which means dydt=0\frac{dy}{dt} = 0 and dxdt≠0\frac{dx}{dt} \neq 0. First, find the derivatives with respect to tt: dxdt=3t2−3\frac{dx}{dt} = 3t^2 - 3 and dydt=2t\frac{dy}{dt} = 2t. Set dydt=0\frac{dy}{dt} = 0, which gives 2t=0  ⟹  t=02t = 0 \implies t = 0. We must check that dxdt≠0\frac{dx}{dt} \neq 0 at t=0t=0. dxdt\frac{dx}{dt} at t=0t=0 is 3(0)2−3=−3≠03(0)^2 - 3 = -3 \neq 0. So there is a horizontal tangent at t=0t=0. The coordinates of this point are x=03−3(0)=0x = 0^3 - 3(0) = 0 and y=02−2=−2y = 0^2 - 2 = -2. The point is (0,−2)(0, -2). Distractor A corresponds to the points of vertical tangency, where dxdt=0  ⟹  3t2−3=0  ⟹  t=±1\frac{dx}{dt} = 0 \implies 3t^2 - 3 = 0 \implies t = \pm 1. At t=1t=1, (x,y)=(−2,−1)(x,y) = (-2,-1). At t=−1t=-1, (x,y)=(2,−1)(x,y) = (2,-1).

Question 8

Find the equation of the normal to the curve x2+3y2=13x^2 + 3y^2 = 13 at the point (2,3)(2, \sqrt{3}).

  1. y=−33x+533y = -\frac{\sqrt{3}}{3}x + \frac{5\sqrt{3}}{3}
  2. y=33x+33y = \frac{\sqrt{3}}{3}x + \frac{\sqrt{3}}{3}
  3. y=332x−23y = \frac{3\sqrt{3}}{2}x - 2\sqrt{3} (correct answer)
  4. y=−233x+733y = -\frac{2\sqrt{3}}{3}x + \frac{7\sqrt{3}}{3}
Explanation: First, find the slope of the tangent by implicit differentiation: 2x+6ydydx=02x + 6y\frac{dy}{dx} = 0, which gives dydx=−2x6y=−x3y\frac{dy}{dx} = -\frac{2x}{6y} = -\frac{x}{3y}. At the point (2,3)(2, \sqrt{3}), the slope of the tangent is mtan=−233m_{tan} = -\frac{2}{3\sqrt{3}}. The slope of the normal is the negative reciprocal of the tangent's slope: mnorm=−1mtan=332m_{norm} = -\frac{1}{m_{tan}} = \frac{3\sqrt{3}}{2}. Now, use the point-slope form for the equation of the normal line: y−y1=mnorm(x−x1)y - y_1 = m_{norm}(x - x_1), which is y−3=332(x−2)y - \sqrt{3} = \frac{3\sqrt{3}}{2}(x - 2). Simplifying gives y−3=332x−33y - \sqrt{3} = \frac{3\sqrt{3}}{2}x - 3\sqrt{3}, so y=332x−23y = \frac{3\sqrt{3}}{2}x - 2\sqrt{3}. Distractor A uses the tangent slope. Distractors B and D result from calculation errors with the slopes or the point-slope formula.

Question 9

The tangent to the curve y=ln⁡(x2+y2)y = \ln(x^2 + y^2) at the point (1,0)(1, 0) is horizontal. What can be concluded about dydx\frac{dy}{dx} at this point?

  1. The value of dydx\frac{dy}{dx} cannot be determined.
  2. The value of dydx\frac{dy}{dx} is undefined.
  3. The value of dydx\frac{dy}{dx} is 0.
  4. The statement is false; the tangent is not horizontal. (correct answer)
Explanation: Let's find the slope dydx\frac{dy}{dx} by implicit differentiation. dydx=1x2+y2(2x+2ydydx)\frac{dy}{dx} = \frac{1}{x^2+y^2}(2x+2y\frac{dy}{dx}). Rearranging to solve for dydx\frac{dy}{dx}: dydx(1−2yx2+y2)=2xx2+y2\frac{dy}{dx}(1 - \frac{2y}{x^2+y^2}) = \frac{2x}{x^2+y^2}. dydx(x2+y2−2yx2+y2)=2xx2+y2\frac{dy}{dx}(\frac{x^2+y^2-2y}{x^2+y^2}) = \frac{2x}{x^2+y^2}. dydx=2xx2+y2−2y\frac{dy}{dx} = \frac{2x}{x^2+y^2-2y}. Now evaluate at the point (1,0)(1, 0): dydx=2(1)12+02−2(0)=21=2\frac{dy}{dx} = \frac{2(1)}{1^2+0^2-2(0)} = \frac{2}{1} = 2. The slope is 2, not 0. Therefore, the tangent line is not horizontal, and the initial statement is false. This type of question tests whether students check the given information before proceeding.

Question 10

A curve is parametrically defined by x=sin⁡3(t)x = \sin^3(t) and y=cos⁡3(t)y = \cos^3(t). Find the slope of the tangent line at t=π/4t = \pi/4.

  1. -1 (correct answer)
  2. 0
  3. 1
  4. Undefined
Explanation: First, we find the derivatives with respect to tt using the chain rule. dxdt=3sin⁡2(t)⋅cos⁡(t)\frac{dx}{dt} = 3\sin^2(t) \cdot \cos(t). dydt=3cos⁡2(t)⋅(−sin⁡(t))=−3cos⁡2(t)sin⁡(t)\frac{dy}{dt} = 3\cos^2(t) \cdot (-\sin(t)) = -3\cos^2(t)\sin(t). The slope of the tangent line is dydx=dy/dtdx/dt=−3cos⁡2(t)sin⁡(t)3sin⁡2(t)cos⁡(t)\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3\cos^2(t)\sin(t)}{3\sin^2(t)\cos(t)}. For tt where sin⁡(t)\sin(t) and cos⁡(t)\cos(t) are non-zero, this simplifies to −cos⁡(t)sin⁡(t)=−cot⁡(t)-\frac{\cos(t)}{\sin(t)} = -\cot(t). Now, we evaluate the slope at t=π/4t = \pi/4: m=−cot⁡(π/4)=−1m = -\cot(\pi/4) = -1.

Question 11

A curve is defined by the parametric equations x=t3−3tx = t^3 - 3t and y=t2−2y = t^2 - 2. Find the coordinates of the points where the tangent to the curve is horizontal.

  1. (-2, -1) and (2, -1)
  2. (0, -2) (correct answer)
  3. (-2, -1)
  4. (0, -2) and (0, -1)
Explanation: A horizontal tangent occurs when dydx=0\frac{dy}{dx} = 0, which means dydt=0\frac{dy}{dt} = 0 and dxdt≠0\frac{dx}{dt} \neq 0. First, find the derivatives with respect to tt: dxdt=3t2−3\frac{dx}{dt} = 3t^2 - 3 and dydt=2t\frac{dy}{dt} = 2t. Set dydt=0\frac{dy}{dt} = 0, which gives 2t=0  ⟹  t=02t = 0 \implies t = 0. We must check that dxdt≠0\frac{dx}{dt} \neq 0 at t=0t=0. dxdt\frac{dx}{dt} at t=0t=0 is 3(0)2−3=−3≠03(0)^2 - 3 = -3 \neq 0. So there is a horizontal tangent at t=0t=0. The coordinates of this point are x=03−3(0)=0x = 0^3 - 3(0) = 0 and y=02−2=−2y = 0^2 - 2 = -2. The point is (0,−2)(0, -2). Distractor A corresponds to the points of vertical tangency, where dxdt=0  ⟹  3t2−3=0  ⟹  t=±1\frac{dx}{dt} = 0 \implies 3t^2 - 3 = 0 \implies t = \pm 1. At t=1t=1, (x,y)=(−2,−1)(x,y) = (-2,-1). At t=−1t=-1, (x,y)=(2,−1)(x,y) = (2,-1).

Question 12

The tangent to the curve y=ln⁡(x2+y2)y = \ln(x^2 + y^2) at the point (1,0)(1, 0) is horizontal. What can be concluded about dydx\frac{dy}{dx} at this point?

  1. The value of dydx\frac{dy}{dx} cannot be determined.
  2. The value of dydx\frac{dy}{dx} is undefined.
  3. The value of dydx\frac{dy}{dx} is 0.
  4. The statement is false; the tangent is not horizontal. (correct answer)
Explanation: Let's find the slope dydx\frac{dy}{dx} by implicit differentiation. dydx=1x2+y2(2x+2ydydx)\frac{dy}{dx} = \frac{1}{x^2+y^2}(2x+2y\frac{dy}{dx}). Rearranging to solve for dydx\frac{dy}{dx}: dydx(1−2yx2+y2)=2xx2+y2\frac{dy}{dx}(1 - \frac{2y}{x^2+y^2}) = \frac{2x}{x^2+y^2}. dydx(x2+y2−2yx2+y2)=2xx2+y2\frac{dy}{dx}(\frac{x^2+y^2-2y}{x^2+y^2}) = \frac{2x}{x^2+y^2}. dydx=2xx2+y2−2y\frac{dy}{dx} = \frac{2x}{x^2+y^2-2y}. Now evaluate at the point (1,0)(1, 0): dydx=2(1)12+02−2(0)=21=2\frac{dy}{dx} = \frac{2(1)}{1^2+0^2-2(0)} = \frac{2}{1} = 2. The slope is 2, not 0. Therefore, the tangent line is not horizontal, and the initial statement is false. This type of question tests whether students check the given information before proceeding.

Question 13

For the curve defined by exy=x−ye^{xy} = x - y, what is the value of dydx\frac{dy}{dx} at the point (1,0)(1, 0)?

  1. -1/2
  2. 0
  3. 1/2 (correct answer)
  4. 1
Explanation: We use implicit differentiation. Differentiating both sides with respect to xx, we apply the product rule to the term xyxy inside the exponential's chain rule: exy(y+xdydx)=1−dydxe^{xy}(y + x\frac{dy}{dx}) = 1 - \frac{dy}{dx}. Now, we rearrange the equation to solve for dydx\frac{dy}{dx}: yexy+xexydydx=1−dydxye^{xy} + xe^{xy}\frac{dy}{dx} = 1 - \frac{dy}{dx}. xexydydx+dydx=1−yexyxe^{xy}\frac{dy}{dx} + \frac{dy}{dx} = 1 - ye^{xy}. dydx(xexy+1)=1−yexy\frac{dy}{dx}(xe^{xy} + 1) = 1 - ye^{xy}. dydx=1−yexyxexy+1\frac{dy}{dx} = \frac{1 - ye^{xy}}{xe^{xy} + 1}. Substitute the point (1,0)(1, 0): dydx=1−0⋅e1⋅01⋅e1⋅0+1=1−01⋅1+1=12\frac{dy}{dx} = \frac{1 - 0 \cdot e^{1 \cdot 0}}{1 \cdot e^{1 \cdot 0} + 1} = \frac{1 - 0}{1 \cdot 1 + 1} = \frac{1}{2}. Distractor A results from a sign error. Distractor B results from ignoring the product rule inside the chain rule. Distractor D results from algebraic errors when isolating dydx\frac{dy}{dx}.

Question 14

A particle's position is given by x(t)=3t2−2x(t) = 3t^2 - 2 and y(t)=t3−3ty(t) = t^3 - 3t. For which interval(s) of tt is the particle moving to the right and downwards?

  1. t>1t > 1
  2. 0<t<10 < t < 1 (correct answer)
  3. t<−1t < -1
  4. −1<t<0-1 < t < 0
Explanation: The particle moves to the right when its horizontal velocity, dxdt\frac{dx}{dt}, is positive. dxdt=6t\frac{dx}{dt} = 6t. So, 6t>0  ⟹  t>06t > 0 \implies t > 0. The particle moves downwards when its vertical velocity, dydt\frac{dy}{dt}, is negative. dydt=3t2−3\frac{dy}{dt} = 3t^2 - 3. So, 3t2−3<0  ⟹  t2<1  ⟹  −1<t<13t^2 - 3 < 0 \implies t^2 < 1 \implies -1 < t < 1. For the particle to be moving both to the right and downwards, both conditions must be met. We need to find the intersection of the intervals t>0t > 0 and −1<t<1-1 < t < 1. The intersection is 0<t<10 < t < 1.

Question 15

The curve known as the lemniscate of Bernoulli is given by (x2+y2)2=2(x2−y2)(x^2 + y^2)^2 = 2(x^2 - y^2). Which of the following is an expression for dydx\frac{dy}{dx} in terms of xx and yy?

  1. x(1−x2−y2)y(1+x2+y2)\frac{x(1 - x^2 - y^2)}{y(1 + x^2 + y^2)} (correct answer)
  2. x(1+x2+y2)y(1−x2−y2)\frac{x(1 + x^2 + y^2)}{y(1 - x^2 - y^2)}
  3. x(2−x2−y2)y(2+x2+y2)\frac{x(2 - x^2 - y^2)}{y(2 + x^2 + y^2)}
  4. xy\frac{x}{y}
Explanation: Differentiate the equation implicitly with respect to xx. 2(x2+y2)(2x+2ydydx)=2(2x−2ydydx)2(x^2+y^2)(2x+2y\frac{dy}{dx}) = 2(2x - 2y\frac{dy}{dx}). Divide by 4: (x2+y2)(x+ydydx)=x−ydydx(x^2+y^2)(x+y\frac{dy}{dx}) = x - y\frac{dy}{dx}. Expand: x(x2+y2)+y(x2+y2)dydx=x−ydydxx(x^2+y^2) + y(x^2+y^2)\frac{dy}{dx} = x - y\frac{dy}{dx}. Group terms with dydx\frac{dy}{dx}: dydx(y(x2+y2)+y)=x−x(x2+y2)\frac{dy}{dx}(y(x^2+y^2) + y) = x - x(x^2+y^2). Factor out common terms: dydx⋅y(x2+y2+1)=x(1−(x2+y2))\frac{dy}{dx} \cdot y(x^2+y^2+1) = x(1 - (x^2+y^2)). Finally, solve for dydx\frac{dy}{dx}: dydx=x(1−x2−y2)y(1+x2+y2)\frac{dy}{dx} = \frac{x(1 - x^2 - y^2)}{y(1 + x^2 + y^2)}. The other options are distractors based on sign errors or incorrect factoring during the algebraic manipulation.

Question 16

Consider the curve defined by sin⁡(x)+cos⁡(y)=1\sin(x) + \cos(y) = 1. Which statement is true about the tangent to the curve at the point (π/6,π/3)(\pi/6, \pi/3)?

  1. The tangent line is horizontal.
  2. The tangent line is vertical.
  3. The tangent line has a slope of 1. (correct answer)
  4. The tangent line has a slope of -1.
Explanation: Differentiate implicitly with respect to xx: cos⁡(x)−sin⁡(y)dydx=0\cos(x) - \sin(y)\frac{dy}{dx} = 0. Solving for dydx\frac{dy}{dx} gives dydx=cos⁡(x)sin⁡(y)\frac{dy}{dx} = \frac{\cos(x)}{\sin(y)}. Now, evaluate the slope at the point (π/6,π/3)(\pi/6, \pi/3): dydx=cos⁡(π/6)sin⁡(π/3)=3/23/2=1\frac{dy}{dx} = \frac{\cos(\pi/6)}{\sin(\pi/3)} = \frac{\sqrt{3}/2}{\sqrt{3}/2} = 1. So, the tangent line has a slope of 1. A horizontal tangent would mean the slope is 0, which would require cos⁡(x)=0\cos(x)=0. A vertical tangent would mean the slope is undefined, which would require sin⁡(y)=0\sin(y)=0. A slope of -1 is a plausible distractor if a sign error is made or trig values are mixed up.

Question 17

A curve is given by x=a(t−sin⁡t)x = a(t - \sin t) and y=a(1−cos⁡t)y = a(1 - \cos t), where a>0a > 0. For what value of tt in (0,2π)(0, 2\pi) is the slope of the tangent line equal to 1?

  1. π/4\pi/4
  2. π/2\pi/2 (correct answer)
  3. 3π/43\pi/4
  4. π\pi
Explanation: First, we find dydx\frac{dy}{dx}. dxdt=a(1−cos⁡t)\frac{dx}{dt} = a(1 - \cos t) and dydt=asin⁡t\frac{dy}{dt} = a\sin t. The slope is dydx=asin⁡ta(1−cos⁡t)=sin⁡t1−cos⁡t\frac{dy}{dx} = \frac{a\sin t}{a(1 - \cos t)} = \frac{\sin t}{1 - \cos t}. We set the slope equal to 1: sin⁡t1−cos⁡t=1\frac{\sin t}{1 - \cos t} = 1, which means sin⁡t=1−cos⁡t\sin t = 1 - \cos t. Squaring both sides: sin⁡2t=(1−cos⁡t)2=1−2cos⁡t+cos⁡2t\sin^2 t = (1 - \cos t)^2 = 1 - 2\cos t + \cos^2 t. Since sin⁡2t+cos⁡2t=1\sin^2 t + \cos^2 t = 1, we have 1−cos⁡2t=1−2cos⁡t+cos⁡2t1 - \cos^2 t = 1 - 2\cos t + \cos^2 t, which simplifies to 2cos⁡2t−2cos⁡t=02\cos^2 t - 2\cos t = 0, or 2cos⁡t(cos⁡t−1)=02\cos t(\cos t - 1) = 0. This gives cos⁡t=0\cos t = 0 or cos⁡t=1\cos t = 1. For t∈(0,2π)t \in (0, 2\pi), cos⁡t=0\cos t = 0 gives t=π/2t = \pi/2 or t=3π/2t = 3\pi/2, and cos⁡t=1\cos t = 1 gives t=2πt = 2\pi (excluded). Checking: at t=π/2t = \pi/2, sin⁡t=1\sin t = 1 and 1−cos⁡t=11 - \cos t = 1, so the slope is 1. At t=3π/2t = 3\pi/2, sin⁡t=−1\sin t = -1 and 1−cos⁡t=11 - \cos t = 1, so the slope is -1. Therefore, t=π/2t = \pi/2.

Question 18

The curve known as the lemniscate of Bernoulli is given by (x2+y2)2=2(x2−y2)(x^2 + y^2)^2 = 2(x^2 - y^2). Which of the following is an expression for dydx\frac{dy}{dx} in terms of xx and yy?

  1. x(1−x2−y2)y(1+x2+y2)\frac{x(1 - x^2 - y^2)}{y(1 + x^2 + y^2)} (correct answer)
  2. x(1+x2+y2)y(1−x2−y2)\frac{x(1 + x^2 + y^2)}{y(1 - x^2 - y^2)}
  3. x(2−x2−y2)y(2+x2+y2)\frac{x(2 - x^2 - y^2)}{y(2 + x^2 + y^2)}
  4. xy\frac{x}{y}
Explanation: Differentiate the equation implicitly with respect to xx. 2(x2+y2)(2x+2ydydx)=2(2x−2ydydx)2(x^2+y^2)(2x+2y\frac{dy}{dx}) = 2(2x - 2y\frac{dy}{dx}). Divide by 4: (x2+y2)(x+ydydx)=x−ydydx(x^2+y^2)(x+y\frac{dy}{dx}) = x - y\frac{dy}{dx}. Expand: x(x2+y2)+y(x2+y2)dydx=x−ydydxx(x^2+y^2) + y(x^2+y^2)\frac{dy}{dx} = x - y\frac{dy}{dx}. Group terms with dydx\frac{dy}{dx}: dydx(y(x2+y2)+y)=x−x(x2+y2)\frac{dy}{dx}(y(x^2+y^2) + y) = x - x(x^2+y^2). Factor out common terms: dydx⋅y(x2+y2+1)=x(1−(x2+y2))\frac{dy}{dx} \cdot y(x^2+y^2+1) = x(1 - (x^2+y^2)). Finally, solve for dydx\frac{dy}{dx}: dydx=x(1−x2−y2)y(1+x2+y2)\frac{dy}{dx} = \frac{x(1 - x^2 - y^2)}{y(1 + x^2 + y^2)}. The other options are distractors based on sign errors or incorrect factoring during the algebraic manipulation.

Question 19

A curve is defined by the parametric equations x=sec⁡(t)x = \sec(t) and y=tan⁡(t)y = \tan(t). Find dydx\frac{dy}{dx} in terms of tt.

  1. sin⁡(t)\sin(t)
  2. cos⁡(t)\cos(t)
  3. csc⁡(t)\csc(t) (correct answer)
  4. sec⁡(t)\sec(t)
Explanation: First, find the derivatives with respect to the parameter tt. dxdt=ddt(sec⁡(t))=sec⁡(t)tan⁡(t)\frac{dx}{dt} = \frac{d}{dt}(\sec(t)) = \sec(t)\tan(t). dydt=ddt(tan⁡(t))=sec⁡2(t)\frac{dy}{dt} = \frac{d}{dt}(\tan(t)) = \sec^2(t). The derivative dydx\frac{dy}{dx} is given by the ratio dy/dtdx/dt\frac{dy/dt}{dx/dt}. So, dydx=sec⁡2(t)sec⁡(t)tan⁡(t)=sec⁡(t)tan⁡(t)\frac{dy}{dx} = \frac{\sec^2(t)}{\sec(t)\tan(t)} = \frac{\sec(t)}{\tan(t)}. To simplify, we can write this in terms of sine and cosine: 1/cos⁡(t)sin⁡(t)/cos⁡(t)=1sin⁡(t)=csc⁡(t)\frac{1/\cos(t)}{\sin(t)/\cos(t)} = \frac{1}{\sin(t)} = \csc(t). Distractor B, cos⁡(t)\cos(t), is dx/dydx/dy. Distractors A and D arise from incorrect simplification of the trigonometric expression.

Question 20

Find the equation of the normal to the curve x2+3y2=13x^2 + 3y^2 = 13 at the point (2,3)(2, \sqrt{3}).

  1. y=−33x+533y = -\frac{\sqrt{3}}{3}x + \frac{5\sqrt{3}}{3}
  2. y=33x+33y = \frac{\sqrt{3}}{3}x + \frac{\sqrt{3}}{3}
  3. y=332x−23y = \frac{3\sqrt{3}}{2}x - 2\sqrt{3} (correct answer)
  4. y=−233x+733y = -\frac{2\sqrt{3}}{3}x + \frac{7\sqrt{3}}{3}
Explanation: First, find the slope of the tangent by implicit differentiation: 2x+6ydydx=02x + 6y\frac{dy}{dx} = 0, which gives dydx=−2x6y=−x3y\frac{dy}{dx} = -\frac{2x}{6y} = -\frac{x}{3y}. At the point (2,3)(2, \sqrt{3}), the slope of the tangent is mtan=−233m_{tan} = -\frac{2}{3\sqrt{3}}. The slope of the normal is the negative reciprocal of the tangent's slope: mnorm=−1mtan=332m_{norm} = -\frac{1}{m_{tan}} = \frac{3\sqrt{3}}{2}. Now, use the point-slope form for the equation of the normal line: y−y1=mnorm(x−x1)y - y_1 = m_{norm}(x - x_1), which is y−3=332(x−2)y - \sqrt{3} = \frac{3\sqrt{3}}{2}(x - 2). Simplifying gives y−3=332x−33y - \sqrt{3} = \frac{3\sqrt{3}}{2}x - 3\sqrt{3}, so y=332x−23y = \frac{3\sqrt{3}}{2}x - 2\sqrt{3}. Distractor A uses the tangent slope. Distractors B and D result from calculation errors with the slopes or the point-slope formula.