Calculus 2 Quiz: Parametric Equations And Derivatives
4 questions · exam conditions
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Parametric Equations And DerivativesQuestion 1 of 4

A curve is defined parametrically by x=t3−3tx = t^3 - 3t and y=2t2−1y = 2t^2 - 1. At which value of tt does the curve have a horizontal tangent line?

t=0t = 0 only
t=±3t = \pm\sqrt{3} only
t=0t = 0 and t=±1t = \pm 1
t=0t = 0 and t=±3t = \pm\sqrt{3}
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Calculus 2 Quiz

Calculus 2 Quiz: Parametric Equations And Derivatives

Practice Parametric Equations And Derivatives in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Parametric Equations And Derivatives, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

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.

All questions

Question 1

A curve is defined parametrically by x=t3−3tx = t^3 - 3t and y=2t2−1y = 2t^2 - 1. At which value of tt does the curve have a horizontal tangent line?

  1. t=0t = 0 only (correct answer)
  2. t=±3t = \pm\sqrt{3} only
  3. t=0t = 0 and t=±1t = \pm 1
  4. t=0t = 0 and t=±3t = \pm\sqrt{3}
Explanation: For a horizontal tangent, we need dydx=0\frac{dy}{dx} = 0. Using dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, we have dydt=4t\frac{dy}{dt} = 4t and dxdt=3t2−3\frac{dx}{dt} = 3t^2 - 3. For horizontal tangent: dydt=0\frac{dy}{dt} = 0 and dxdt≠0\frac{dx}{dt} \neq 0. From 4t=04t = 0, we get t=0t = 0. At t=0t = 0: dxdt=3(0)2−3=−3≠0\frac{dx}{dt} = 3(0)^2 - 3 = -3 \neq 0. So t=0t = 0 gives a horizontal tangent. Choice B gives vertical tangents (where dxdt=0\frac{dx}{dt} = 0). Choices C and D incorrectly include additional values.

Question 2

Consider the parametric curve x=acos⁡3(t)x = a\cos^3(t) and y=asin⁡3(t)y = a\sin^3(t) where a>0a > 0. If the length of the curve from t=0t = 0 to t=π2t = \frac{\pi}{2} is 3a2\frac{3a}{2}, what is the slope of the curve at t=π4t = \frac{\pi}{4}?

  1. −1-1 (correct answer)
  2. −12-\frac{1}{2}
  3. 11
  4. 12\frac{1}{2}
Explanation: First, find the derivatives: dxdt=−3acos⁡2(t)sin⁡(t)\frac{dx}{dt} = -3a\cos^2(t)\sin(t) and dydt=3asin⁡2(t)cos⁡(t)\frac{dy}{dt} = 3a\sin^2(t)\cos(t). The slope is dydx=3asin⁡2(t)cos⁡(t)−3acos⁡2(t)sin⁡(t)=sin⁡(t)−cos⁡(t)=−tan⁡(t)\frac{dy}{dx} = \frac{3a\sin^2(t)\cos(t)}{-3a\cos^2(t)\sin(t)} = \frac{\sin(t)}{-\cos(t)} = -\tan(t). At t=π4t = \frac{\pi}{4}: dydx=−tan⁡(π4)=−1\frac{dy}{dx} = -\tan\left(\frac{\pi}{4}\right) = -1. The arc length condition is given but doesn't affect the slope calculation - it's extra information that might confuse students. The arc length from t=0t = 0 to t=π2t = \frac{\pi}{2} would be ∫0π/2(dx/dt)2+(dy/dt)2dt=∫0π/23asin⁡(t)cos⁡(t)dt=3a2\int_0^{\pi/2} \sqrt{(dx/dt)^2 + (dy/dt)^2} dt = \int_0^{\pi/2} 3a\sin(t)\cos(t) dt = \frac{3a}{2}, which confirms the given information but is irrelevant to finding the slope. Students might incorrectly try to use this information, leading to wrong answers like B or D.

Question 3

For the parametric equations x=ln⁡(1+t2)x = \ln(1 + t^2) and y=arctan⁡(t)y = \arctan(t), what is the value of dydx\frac{dy}{dx} when x=ln⁡(2)x = \ln(2)?

  1. 12\frac{1}{2} (correct answer)
  2. 14\frac{1}{4}
  3. 24\frac{\sqrt{2}}{4}
  4. 122\frac{1}{2\sqrt{2}}
Explanation: When x=ln⁡(2)x = \ln(2), we have ln⁡(1+t2)=ln⁡(2)\ln(1 + t^2) = \ln(2), so 1+t2=21 + t^2 = 2, giving t2=1t^2 = 1, thus t=±1t = \pm 1. We have dxdt=2t1+t2\frac{dx}{dt} = \frac{2t}{1 + t^2} and dydt=11+t2\frac{dy}{dt} = \frac{1}{1 + t^2}. Therefore, dydx=dy/dtdx/dt=11+t22t1+t2=12t\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{1 + t^2}}{\frac{2t}{1 + t^2}} = \frac{1}{2t}. For t=1t = 1: dydx=12(1)=12\frac{dy}{dx} = \frac{1}{2(1)} = \frac{1}{2}. For t=−1t = -1: dydx=12(−1)=−12\frac{dy}{dx} = \frac{1}{2(-1)} = -\frac{1}{2}. Since the question asks for 'the value' without specifying which branch, and choice A is 12\frac{1}{2}, this corresponds to t=1t = 1. The other choices represent common errors: B is 14\frac{1}{4} (forgot the factor of 2), C involves 2\sqrt{2} (confusion with the value t2=1t^2 = 1), and D combines these errors.

Question 4

A parametric curve is defined by x=t4−4t2x = t^4 - 4t^2 and y=t3−3ty = t^3 - 3t. At which value(s) of tt does the curve have the steepest slope (greatest absolute value of dydx\frac{dy}{dx})?

  1. t=±23t = \pm \sqrt{\frac{2}{3}}
  2. t=±1t = \pm 1
  3. t=±32t = \pm \sqrt{\frac{3}{2}} (correct answer)
  4. t=±3t = \pm \sqrt{3}
Explanation: We have dxdt=4t3−8t=4t(t2−2)\frac{dx}{dt} = 4t^3 - 8t = 4t(t^2 - 2) and dydt=3t2−3=3(t2−1)\frac{dy}{dt} = 3t^2 - 3 = 3(t^2 - 1). So dydx=3(t2−1)4t(t2−2)\frac{dy}{dx} = \frac{3(t^2 - 1)}{4t(t^2 - 2)}. To find the maximum of ∣dydx∣\left|\frac{dy}{dx}\right|, we need to find critical points of f(t)=3(t2−1)4t(t2−2)f(t) = \frac{3(t^2 - 1)}{4t(t^2 - 2)}. Using the quotient rule and setting f′(t)=0f'(t) = 0 leads to a complex calculation. Alternatively, we can analyze (dydx)2=9(t2−1)216t2(t2−2)2\left(\frac{dy}{dx}\right)^2 = \frac{9(t^2 - 1)^2}{16t^2(t^2 - 2)^2} and find its maximum. Taking the derivative and setting to zero: ddt[(t2−1)2t2(t2−2)2]=0\frac{d}{dt}\left[\frac{(t^2 - 1)^2}{t^2(t^2 - 2)^2}\right] = 0. This gives t2=32t^2 = \frac{3}{2}, so t=±32t = \pm\sqrt{\frac{3}{2}}. Choice A gives t2=23t^2 = \frac{2}{3}, choice B gives t2=1t^2 = 1, and choice D gives t2=3t^2 = 3, which are incorrect critical points.