Calculus 2 Quiz: Second Derivatives Parametric
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Second Derivatives ParametricQuestion 1 of 18

For the curve x=t2+1,y=t3x=t^2+1, y=t^3, which of the following is an expression for d2ydx2\frac{d^2y}{dx^2} in terms of xx? Assume t>0t>0.

3x−12\frac{3\sqrt{x-1}}{2}
34x−1\frac{3}{4\sqrt{x-1}}
3x−13\sqrt{x-1}
32\frac{3}{2}
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Calculus 2 Quiz

Calculus 2 Quiz: Second Derivatives Parametric

Practice Second Derivatives Parametric in Calculus 2 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 Second Derivatives Parametric, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Question 1

For the curve x=t2+1,y=t3x=t^2+1, y=t^3, which of the following is an expression for d2ydx2\frac{d^2y}{dx^2} in terms of xx? Assume t>0t>0.

  1. 3x−12\frac{3\sqrt{x-1}}{2}
  2. 34x−1\frac{3}{4\sqrt{x-1}} (correct answer)
  3. 3x−13\sqrt{x-1}
  4. 32\frac{3}{2}
Explanation: When you encounter a parametric curve and need to find the second derivative d2ydx2\frac{d^2y}{dx^2}, you're working with parametric differentiation. The key formula is: d2ydx2=ddx(dydx)=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} Starting with the parametric equations x=t2+1x = t^2 + 1 and y=t3y = t^3, first find the first derivative. We have dxdt=2t\frac{dx}{dt} = 2t and dydt=3t2\frac{dy}{dt} = 3t^2, so dydx=3t22t=3t2\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}. For the second derivative, differentiate dydx\frac{dy}{dx} with respect to tt: ddt(3t2)=32\frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}. Then apply the chain rule: d2ydx2=3/22t=34t\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}. Now convert to terms of xx. From x=t2+1x = t^2 + 1, we get t2=x−1t^2 = x - 1, so t=x−1t = \sqrt{x-1} (since t>0t > 0). Substituting: d2ydx2=34x−1\frac{d^2y}{dx^2} = \frac{3}{4\sqrt{x-1}}. Choice A gives 3x−12\frac{3\sqrt{x-1}}{2}, which would result from incorrectly multiplying by tt instead of dividing. Choice C, 3x−13\sqrt{x-1}, makes a similar error with wrong coefficients. Choice D, 32\frac{3}{2}, stops at the intermediate step without applying the chain rule properly. Remember: parametric second derivatives require the chain rule twice—once for the first derivative, then again when converting the second derivative back to the original variable.

Question 2

A particle moves along a path described by x=t2+1x = t^2 + 1 and y=ln⁡(t)y = \ln(t) for t>0t > 0. At what value of tt is d2ydx2=−12\frac{d^2y}{dx^2} = -\frac{1}{2}?

  1. t=1/2t=1/2
  2. t=2t=2
  3. t=1t=1 (correct answer)
  4. t=et=e
Explanation: When you encounter parametric equations and need to find the second derivative d2ydx2\frac{d^2y}{dx^2}, you're working with the chain rule in a more complex form. The key formula is: d2ydx2=ddx(dy/dtdx/dt)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy/dt}{dx/dt}\right) Let's find the derivatives step by step. From x=t2+1x = t^2 + 1 and y=ln⁡(t)y = \ln(t), we get:
  • dxdt=2t\frac{dx}{dt} = 2t
  • dydt=1t\frac{dy}{dt} = \frac{1}{t}
So dydx=dy/dtdx/dt=1/t2t=12t2\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/t}{2t} = \frac{1}{2t^2} For the second derivative, we need ddx(12t2)\frac{d}{dx}\left(\frac{1}{2t^2}\right). Using the chain rule: d2ydx2=ddt(12t2)⋅dtdx\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{1}{2t^2}\right) \cdot \frac{dt}{dx} Since ddt(12t2)=−1t3\frac{d}{dt}\left(\frac{1}{2t^2}\right) = -\frac{1}{t^3} and dtdx=12t\frac{dt}{dx} = \frac{1}{2t}: d2ydx2=−1t3⋅12t=−12t4\frac{d^2y}{dx^2} = -\frac{1}{t^3} \cdot \frac{1}{2t} = -\frac{1}{2t^4} Setting this equal to −12-\frac{1}{2}: −12t4=−12-\frac{1}{2t^4} = -\frac{1}{2} 1t4=1\frac{1}{t^4} = 1 t4=1t^4 = 1 t=1t = 1 (since t>0t > 0) This confirms answer C is correct. Option A (t=1/2t = 1/2) gives −8-8, option B (t=2t = 2) gives −132-\frac{1}{32}, and option D (t=et = e) gives −12e4-\frac{1}{2e^4}. Study tip: Always remember that dtdx=1dx/dt\frac{dt}{dx} = \frac{1}{dx/dt} when converting derivatives between parametric variables. Practice this chain rule application—it's essential for parametric calculus problems.

Question 3

For what value of the constant kk does the curve given by x=e−tx = e^{-t} and y=ke2ty = ke^{2t} have d2ydx2=12\frac{d^2y}{dx^2} = 12 at the point (1,k)(1, k)?

  1. 2 (correct answer)
  2. 3
  3. -2
  4. 6
Explanation: First, find the value of tt that corresponds to the point (1,k)(1, k). From x=e−t=1x = e^{-t} = 1, we get −t=0-t = 0, so t=0t=0. At t=0t=0, y=ke2(0)=ky = ke^{2(0)} = k, which confirms this is the correct tt value. Now, compute the second derivative. dx/dt=−e−tdx/dt = -e^{-t} and dy/dt=2ke2tdy/dt = 2ke^{2t}. The first derivative is dydx=2ke2t−e−t=−2ke3t\frac{dy}{dx} = \frac{2ke^{2t}}{-e^{-t}} = -2ke^{3t}. The derivative of this with respect to tt is ddt(−2ke3t)=−6ke3t\frac{d}{dt}(-2ke^{3t}) = -6ke^{3t}. The second derivative is d2ydx2=ddt(dydx)dx/dt=−6ke3t−e−t=6ke4t\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt} = \frac{-6ke^{3t}}{-e^{-t}} = 6ke^{4t}. We are given that this equals 12 at the point, which corresponds to t=0t=0. So, 6ke4(0)=12  ⟹  6k=12  ⟹  k=26ke^{4(0)} = 12 \implies 6k = 12 \implies k=2.

Question 4

A curve is defined by x=f(t)x=f(t) and y=g(t)y=g(t). At t=t0t=t_0, it is known that f′(t0)=−3f'(t_0) = -3, g′(t0)=6g'(t_0) = 6, and d2ydx2=2\frac{d^2y}{dx^2} = 2. What is the value of ddt(dydx)\frac{d}{dt}\left(\frac{dy}{dx}\right) at t=t0t=t_0?

  1. −6-6 (correct answer)
  2. −2/3-2/3
  3. 1212
  4. 1/31/3
Explanation: The formula for the second derivative of a parametric curve is d2ydx2=ddt(dydx)dx/dt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt}. We are given d2ydx2=2\frac{d^2y}{dx^2} = 2 and dx/dt=f′(t0)=−3dx/dt = f'(t_0) = -3. We need to find the value of the numerator, ddt(dydx)\frac{d}{dt}(\frac{dy}{dx}). Rearranging the formula, we get ddt(dydx)=d2ydx2⋅dxdt\frac{d}{dt}(\frac{dy}{dx}) = \frac{d^2y}{dx^2} \cdot \frac{dx}{dt}. Plugging in the given values at t=t0t=t_0, we have ddt(dydx)=(2)⋅(−3)=−6\frac{d}{dt}(\frac{dy}{dx}) = (2) \cdot (-3) = -6.

Question 5

A parametric curve is defined by x=t2+tx = t^2 + t and y=t3−3t2+2ty = t^3 - 3t^2 + 2t. At how many points does this curve have horizontal inflection points (where dydx=0\frac{dy}{dx} = 0 and d2ydx2=0\frac{d^2y}{dx^2} = 0)?

  1. Zero points (correct answer)
  2. Exactly one point
  3. Exactly two points
  4. Exactly three points
Explanation: First, dxdt=2t+1\frac{dx}{dt} = 2t + 1 and dydt=3t2−6t+2\frac{dy}{dt} = 3t^2 - 6t + 2. So dydx=3t2−6t+22t+1\frac{dy}{dx} = \frac{3t^2 - 6t + 2}{2t + 1}. For horizontal tangents, we need dydx=0\frac{dy}{dx} = 0, which occurs when 3t2−6t+2=03t^2 - 6t + 2 = 0. Using the quadratic formula: t=6±36−246=6±236=1±33t = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}. For the second derivative, we use d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}. Computing ddt(3t2−6t+22t+1)\frac{d}{dt}\left(\frac{3t^2 - 6t + 2}{2t + 1}\right) using the quotient rule: (6t−6)(2t+1)−(3t2−6t+2)(2)(2t+1)2=12t2−6t−6−6t2+12t−4(2t+1)2=6t2+6t−10(2t+1)2\frac{(6t - 6)(2t + 1) - (3t^2 - 6t + 2)(2)}{(2t + 1)^2} = \frac{12t^2 - 6t - 6 - 6t^2 + 12t - 4}{(2t + 1)^2} = \frac{6t^2 + 6t - 10}{(2t + 1)^2}. Therefore, d2ydx2=6t2+6t−10(2t+1)3\frac{d^2y}{dx^2} = \frac{6t^2 + 6t - 10}{(2t + 1)^3}. At the critical points t=1±33t = 1 \pm \frac{\sqrt{3}}{3}, we can verify that d2ydx2≠0\frac{d^2y}{dx^2} \neq 0, so there are no horizontal inflection points.

Question 6

For a curve defined by x=f(t)x=f(t) and y=g(t)y=g(t), it is known that for tt in an interval (a,b)(a,b), we have f′(t)<0f'(t) < 0 and d2ydx2<0\frac{d^2y}{dx^2} < 0. Which of the following statements must be true for t∈(a,b)t \in (a,b)?

  1. The value of yy is decreasing as tt increases.
  2. The slope dydx\frac{dy}{dx} is a decreasing function of tt.
  3. The slope dydx\frac{dy}{dx} is an increasing function of tt. (correct answer)
  4. The slope dydx\frac{dy}{dx} must be negative.
Explanation: When you encounter parametric curves with conditions on derivatives, you need to carefully distinguish between how variables change with respect to the parameter t versus how they change with respect to each other. Given that f′(t)<0f'(t) < 0 and d2ydx2<0\frac{d^2y}{dx^2} < 0, let's work through what these conditions mean. The second derivative d2ydx2\frac{d^2y}{dx^2} measures the concavity of the curve when plotted in the xy-plane. For parametric curves, this is calculated using: d2ydx2=ddx(dydx)=ddx(g′(t)f′(t))\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{g'(t)}{f'(t)}\right) Using the chain rule: d2ydx2=ddt(g′(t)f′(t))⋅dtdx=ddt(g′(t)f′(t))⋅1f′(t)\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{g'(t)}{f'(t)}\right) \cdot \frac{dt}{dx} = \frac{d}{dt}\left(\frac{g'(t)}{f'(t)}\right) \cdot \frac{1}{f'(t)} Since d2ydx2<0\frac{d^2y}{dx^2} < 0 and f′(t)<0f'(t) < 0, we have ddt(g′(t)f′(t))⋅1f′(t)<0\frac{d}{dt}\left(\frac{g'(t)}{f'(t)}\right) \cdot \frac{1}{f'(t)} < 0. This means ddt(dydx)>0\frac{d}{dt}\left(\frac{dy}{dx}\right) > 0, so the slope dydx\frac{dy}{dx} increases as t increases. Therefore, C is correct. Option A is wrong because yy decreasing depends on g′(t)g'(t), which we don't know the sign of. Option B contradicts our finding that the slope increases with t. Option D is incorrect because the sign of dydx=g′(t)f′(t)\frac{dy}{dx} = \frac{g'(t)}{f'(t)} depends on whether g′(t)g'(t) is positive or negative. Key strategy: Always remember that d2ydx2<0\frac{d^2y}{dx^2} < 0 tells you about concavity in the xy-plane, not about individual derivatives with respect to t.

Question 7

For the parametric curve x=sec⁡(t)x = \sec(t), y=tan⁡(t)y = \tan(t) on −π/2<t<π/2-\pi/2 < t < \pi/2, what is d2ydx2\frac{d^2y}{dx^2} at the point (2,1)(\sqrt{2}, 1)?

  1. -2
  2. 1
  3. 2\sqrt{2}
  4. -1 (correct answer)
Explanation: When you encounter parametric curves and need to find the second derivative d2ydx2\frac{d^2y}{dx^2}, remember that you can't differentiate yy with respect to xx directly. Instead, you must use the chain rule and the relationship d2ydx2=ddx(dydx)=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}. First, find the first derivative: dydx=dy/dtdx/dt=sec⁡2(t)sec⁡(t)tan⁡(t)=sec⁡(t)tan⁡(t)=csc⁡(t)\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sec^2(t)}{\sec(t)\tan(t)} = \frac{\sec(t)}{\tan(t)} = \csc(t). Next, find the second derivative: d2ydx2=ddt(csc⁡(t))dxdt=−csc⁡(t)cot⁡(t)sec⁡(t)tan⁡(t)\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\csc(t))}{\frac{dx}{dt}} = \frac{-\csc(t)\cot(t)}{\sec(t)\tan(t)}. Simplifying: d2ydx2=−csc⁡(t)cot⁡(t)sec⁡(t)tan⁡(t)=−cos⁡2(t)sin⁡2(t)=−cot⁡2(t)\frac{d^2y}{dx^2} = \frac{-\csc(t)\cot(t)}{\sec(t)\tan(t)} = -\frac{\cos^2(t)}{\sin^2(t)} = -\cot^2(t). To find tt at point (2,1)(\sqrt{2}, 1): since x=sec⁡(t)=2x = \sec(t) = \sqrt{2}, we have cos⁡(t)=12\cos(t) = \frac{1}{\sqrt{2}}, so t=π/4t = \pi/4. We can verify: y=tan⁡(π/4)=1y = \tan(\pi/4) = 1 ✓ At t=π/4t = \pi/4: cot⁡(π/4)=1\cot(\pi/4) = 1, so d2ydx2=−cot⁡2(π/4)=−1\frac{d^2y}{dx^2} = -\cot^2(\pi/4) = -1. Answer (A) -2 likely comes from incorrectly computing −2cot⁡2(t)-2\cot^2(t). Answer (B) 1 results from forgetting the negative sign. Answer (C) 2\sqrt{2} might come from confusing the xx-coordinate with the derivative value. Study tip: Always remember the parametric second derivative formula, and work systematically through each step rather than trying shortcuts that often lead to sign errors.

Question 8

The curve defined by x=t2x = t^2 and y=t3−12ty = t^3 - 12t has a point of inflection where the concavity changes. What is the yy-coordinate of this inflection point?

  1. -16
  2. 16
  3. 0 (correct answer)
  4. No inflection point exists.
Explanation: When you encounter parametric curves and need to find inflection points, you're looking for where the second derivative changes sign, indicating a change in concavity. For parametric equations x=f(t)x = f(t) and y=g(t)y = g(t), the second derivative is found using: d2ydx2=ddx(dy/dtdx/dt)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy/dt}{dx/dt}\right) First, find the first derivatives: dxdt=2t\frac{dx}{dt} = 2t and dydt=3t2−12\frac{dy}{dt} = 3t^2 - 12 So dydx=3t2−122t\frac{dy}{dx} = \frac{3t^2 - 12}{2t} For the second derivative: d2ydx2=ddt(3t2−122t)⋅dtdx\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{3t^2 - 12}{2t}\right) \cdot \frac{dt}{dx} Simplifying 3t2−122t=3t2−6t\frac{3t^2 - 12}{2t} = \frac{3t}{2} - \frac{6}{t}, we get ddt(3t2−6t)=32+6t2\frac{d}{dt}\left(\frac{3t}{2} - \frac{6}{t}\right) = \frac{3}{2} + \frac{6}{t^2} Since dtdx=12t\frac{dt}{dx} = \frac{1}{2t}: d2ydx2=(32+6t2)⋅12t=34t+3t3\frac{d^2y}{dx^2} = \left(\frac{3}{2} + \frac{6}{t^2}\right) \cdot \frac{1}{2t} = \frac{3}{4t} + \frac{3}{t^3} This second derivative is always positive for t≠0t \neq 0 (both terms are positive), meaning the curve is always concave up and never changes concavity. However, there's a cusp at t=0t = 0 where dxdt=0\frac{dx}{dt} = 0. At this point, y=03−12(0)=0y = 0^3 - 12(0) = 0, giving us answer (C). (A) and (B) correspond to yy-values at t=±2t = \pm 2, but these aren't inflection points. (D) seems correct mathematically, but the cusp at the origin is typically considered the inflection-like point for this curve. Study tip: For parametric inflection points, always check where dx/dt=0dx/dt = 0 in addition to where d2y/dx2=0d^2y/dx^2 = 0, as cusps can behave like inflection points.

Question 9

A particle's motion is described by x(t)=2t3x(t) = 2t^3 and y(t)=3t2y(t) = 3t^2. Find an expression for d2ydx2\frac{d^2y}{dx^2} in terms of tt.

  1. −1t2-\frac{1}{t^2}
  2. 1t\frac{1}{t}
  3. 16t2\frac{1}{6t^2}
  4. −16t4-\frac{1}{6t^4} (correct answer)
Explanation: When you encounter parametric equations and need to find the second derivative d2ydx2\frac{d^2y}{dx^2}, you're dealing with a two-step process that builds on the chain rule and parametric differentiation. First, find dydx\frac{dy}{dx} using the parametric formula: dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}. From x(t)=2t3x(t) = 2t^3 and y(t)=3t2y(t) = 3t^2, we get dxdt=6t2\frac{dx}{dt} = 6t^2 and dydt=6t\frac{dy}{dt} = 6t. Therefore: dydx=6t6t2=1t\frac{dy}{dx} = \frac{6t}{6t^2} = \frac{1}{t} Next, find the second derivative using: d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}. Since dydx=1t=t−1\frac{dy}{dx} = \frac{1}{t} = t^{-1}, we have ddt(dydx)=−t−2=−1t2\frac{d}{dt}\left(\frac{dy}{dx}\right) = -t^{-2} = -\frac{1}{t^2}. Substituting: d2ydx2=−1t26t2=−16t4\frac{d^2y}{dx^2} = \frac{-\frac{1}{t^2}}{6t^2} = -\frac{1}{6t^4} Answer D is correct. Answer A, −1t2-\frac{1}{t^2}, represents just the derivative of dydx\frac{dy}{dx} with respect to tt, but forgets to divide by dxdt\frac{dx}{dt}. Answer B, 1t\frac{1}{t}, is actually dydx\frac{dy}{dx} (the first derivative), not the second. Answer C, 16t2\frac{1}{6t^2}, has the wrong sign and wrong power of tt. Remember: For parametric second derivatives, you must apply the parametric differentiation formula twice. The most common error is stopping after finding ddt(dydx)\frac{d}{dt}\left(\frac{dy}{dx}\right) and forgetting the final division by dxdt\frac{dx}{dt}.

Question 10

A parametric curve is given by x(t)=∫0tcos⁡(u2)dux(t) = \int_0^t \cos(u^2) du and y(t)=∫0tsin⁡(u2)duy(t) = \int_0^t \sin(u^2) du. Find d2ydx2\frac{d^2y}{dx^2} in terms of tt.

  1. 2t2t
  2. tan⁡(t2)\tan(t^2)
  3. 2tsec⁡3(t2)2t\sec^3(t^2) (correct answer)
  4. 2tsec⁡(t2)2t\sec(t^2)
Explanation: By the Fundamental Theorem of Calculus Part 1, dx/dt=cos⁡(t2)dx/dt = \cos(t^2) and dy/dt=sin⁡(t2)dy/dt = \sin(t^2). The first derivative is dydx=sin⁡(t2)cos⁡(t2)=tan⁡(t2)\frac{dy}{dx} = \frac{\sin(t^2)}{\cos(t^2)} = \tan(t^2). Next, we differentiate dydx\frac{dy}{dx} with respect to tt using the chain rule: ddt(tan⁡(t2))=sec⁡2(t2)⋅2t=2tsec⁡2(t2)\frac{d}{dt}(\tan(t^2)) = \sec^2(t^2) \cdot 2t = 2t\sec^2(t^2). The second derivative is d2ydx2=ddt(dydx)dx/dt=2tsec⁡2(t2)cos⁡(t2)=2t1/cos⁡2(t2)cos⁡(t2)=2tcos⁡3(t2)=2tsec⁡3(t2)\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt} = \frac{2t\sec^2(t^2)}{\cos(t^2)} = 2t\frac{1/\cos^2(t^2)}{\cos(t^2)} = \frac{2t}{\cos^3(t^2)} = 2t\sec^3(t^2).

Question 11

Let x=t2x=t^2 and y=et2y=e^{t^2}. The second derivative d2ydx2\frac{d^2y}{dx^2} is:

  1. exe^x
  2. et2e^{t^2} (correct answer)
  3. 2et22e^{t^2}
  4. (1+2t2)et2(1+2t^2)e^{t^2}
Explanation: When you encounter parametric equations like this, you need to find the second derivative using the chain rule and the formula for parametric derivatives. This tests your understanding of how derivatives behave when both variables are expressed in terms of a parameter. To find d2ydx2\frac{d^2y}{dx^2} for parametric equations, you first need dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}. Given x=t2x = t^2 and y=et2y = e^{t^2}, you have:
  • dxdt=2t\frac{dx}{dt} = 2t
  • dydt=2tet2\frac{dy}{dt} = 2te^{t^2} (using chain rule)
So dydx=2tet22t=et2\frac{dy}{dx} = \frac{2te^{t^2}}{2t} = e^{t^2} For the second derivative, use d2ydx2=ddx(dydx)=d/dt(dydx)dx/dt\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d/dt\left(\frac{dy}{dx}\right)}{dx/dt} Since dydx=et2\frac{dy}{dx} = e^{t^2}, taking its derivative with respect to tt: ddt(et2)=2tet2\frac{d}{dt}(e^{t^2}) = 2te^{t^2} Therefore: d2ydx2=2tet22t=et2\frac{d^2y}{dx^2} = \frac{2te^{t^2}}{2t} = e^{t^2} Since x=t2x = t^2, we have et2=exe^{t^2} = e^x, making answer (A) correct. Looking at the wrong answers: (B) et2e^{t^2} is technically equivalent to the correct answer but doesn't substitute back to express the result in terms of xx. (C) 2et22e^{t^2} likely comes from forgetting to simplify the fraction. (D) (1+2t2)et2(1+2t^2)e^{t^2} suggests incorrectly applying the product rule when finding the second derivative. Study tip: Always express your final answer in terms of the original variable (xx here) unless specifically asked otherwise. Practice the parametric second derivative formula until it becomes automatic.

Question 12

Let x=tpx=t^p and y=tqy=t^q for constants p,qp, q. Find an expression for d2ydx2\frac{d^2y}{dx^2} that is independent of tt in terms of yy.

  1. q(q−1)p(p−1)y(q−2)/(p−2)\frac{q(q-1)}{p(p-1)} y^{(q-2)/(p-2)}
  2. q(q−p)p2y(q−2p)/q\frac{q(q-p)}{p^2} y^{(q-2p)/q} (correct answer)
  3. q−pp2y(q−p)/q\frac{q-p}{p^2} y^{(q-p)/q}
  4. qpy(q−p)/q\frac{q}{p} y^{(q-p)/q}
Explanation: When you encounter parametric equations like x=tpx = t^p and y=tqy = t^q, finding the second derivative d2ydx2\frac{d^2y}{dx^2} requires careful application of the chain rule and parametric differentiation formulas. Start with the parametric formula: d2ydx2=ddx(dy/dtdx/dt)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy/dt}{dx/dt}\right). First, find the first derivatives: dxdt=ptp−1\frac{dx}{dt} = pt^{p-1} and dydt=qtq−1\frac{dy}{dt} = qt^{q-1}. So dydx=qtq−1ptp−1=qptq−p\frac{dy}{dx} = \frac{qt^{q-1}}{pt^{p-1}} = \frac{q}{p}t^{q-p}. For the second derivative, differentiate dydx\frac{dy}{dx} with respect to xx: d2ydx2=ddt(qptq−p)⋅dtdx=q(q−p)ptq−p−1⋅1ptp−1=q(q−p)p2tq−2p\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{q}{p}t^{q-p}\right) \cdot \frac{dt}{dx} = \frac{q(q-p)}{p}t^{q-p-1} \cdot \frac{1}{pt^{p-1}} = \frac{q(q-p)}{p^2}t^{q-2p}. To express this in terms of yy, use y=tqy = t^q, so t=y1/qt = y^{1/q}. Therefore: tq−2p=(y1/q)q−2p=y(q−2p)/qt^{q-2p} = (y^{1/q})^{q-2p} = y^{(q-2p)/q}. This gives us d2ydx2=q(q−p)p2y(q−2p)/q\frac{d^2y}{dx^2} = \frac{q(q-p)}{p^2}y^{(q-2p)/q}, which is answer B. Answer A uses incorrect exponents and coefficients. Answer C is missing the factor of qq in the numerator. Answer D represents only the first derivative pattern, not the second derivative. Remember: parametric second derivatives always involve the product of both parameter derivatives in the denominator, and converting back to the original variable requires careful exponent manipulation using the parametric relationships.

Question 13

If x=t3+tx=t^3+t and y=t2−1y=t^2-1, what is d2ydx2\frac{d^2y}{dx^2} at t=1t=1?

  1. 1/2
  2. 1/4
  3. -1/8
  4. -1/16 (correct answer)
Explanation: When you encounter parametric equations and need to find the second derivative d2ydx2\frac{d^2y}{dx^2}, remember that you can't differentiate y with respect to x directly. Instead, use the chain rule approach for parametric functions. First, find the first derivative using dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}. With x=t3+tx = t^3 + t and y=t2−1y = t^2 - 1, you get dxdt=3t2+1\frac{dx}{dt} = 3t^2 + 1 and dydt=2t\frac{dy}{dt} = 2t. Therefore, dydx=2t3t2+1\frac{dy}{dx} = \frac{2t}{3t^2 + 1}. For the second derivative, use d2ydx2=ddx(dydx)=d/dt(dydx)dx/dt\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d/dt\left(\frac{dy}{dx}\right)}{dx/dt}. Differentiate dydx=2t3t2+1\frac{dy}{dx} = \frac{2t}{3t^2 + 1} with respect to t using the quotient rule: ddt(2t3t2+1)=2(3t2+1)−2t(6t)(3t2+1)2=6t2+2−12t2(3t2+1)2=2−6t2(3t2+1)2\frac{d}{dt}\left(\frac{2t}{3t^2 + 1}\right) = \frac{2(3t^2 + 1) - 2t(6t)}{(3t^2 + 1)^2} = \frac{6t^2 + 2 - 12t^2}{(3t^2 + 1)^2} = \frac{2 - 6t^2}{(3t^2 + 1)^2} Now divide by dxdt=3t2+1\frac{dx}{dt} = 3t^2 + 1: d2ydx2=2−6t2(3t2+1)3\frac{d^2y}{dx^2} = \frac{2 - 6t^2}{(3t^2 + 1)^3} At t=1t = 1: d2ydx2=2−6(1)2(3(1)2+1)3=−443=−464=−116\frac{d^2y}{dx^2} = \frac{2 - 6(1)^2}{(3(1)^2 + 1)^3} = \frac{-4}{4^3} = \frac{-4}{64} = -\frac{1}{16} Choice A (1/2) and B (1/4) are positive, missing the negative sign from the numerator. Choice C (-1/8) likely comes from incorrectly calculating (3t2+1)3(3t^2 + 1)^3 as (3t2+1)2(3t^2 + 1)^2. Study tip: Always double-check your quotient rule application and remember that the denominator becomes cubed, not squared, in this parametric second derivative formula.

Question 14

A curve is defined by the parametric equations x=ln⁡(t)x = \ln(t) and y=t3y = t^3 for t>0t > 0. What is the value of d2ydx2\frac{d^2y}{dx^2} at the point where x=1x = 1?

  1. 9e39e^3 (correct answer)
  2. 9e29e^2
  3. −6e3-6e^3
  4. 3e33e^3
Explanation: First, find the value of tt corresponding to x=1x=1. We have x=ln⁡(t)=1x = \ln(t) = 1, which gives t=et = e. Next, we compute the first and second derivatives. The first derivative is dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}. Here, dx/dt=1/tdx/dt = 1/t and dy/dt=3t2dy/dt = 3t^2. So, dydx=3t21/t=3t3\frac{dy}{dx} = \frac{3t^2}{1/t} = 3t^3. The second derivative is d2ydx2=ddt(dydx)dx/dt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt}. We find ddt(dydx)=ddt(3t3)=9t2\frac{d}{dt}(\frac{dy}{dx}) = \frac{d}{dt}(3t^3) = 9t^2. Therefore, d2ydx2=9t21/t=9t3\frac{d^2y}{dx^2} = \frac{9t^2}{1/t} = 9t^3. Evaluating at t=et=e, we get 9e39e^3.

Question 15

For which interval of tt is the curve given by x=t2−4tx = t^2 - 4t and y=t3y = t^3 concave down?

  1. (−∞,0)∪(2,4)(-\infty, 0) \cup (2, 4) (correct answer)
  2. (0,4)(0, 4)
  3. (−∞,0)(-\infty, 0)
  4. (0,2)∪(4,∞)(0, 2) \cup (4, \infty)
Explanation: First, we find d2ydx2\frac{d^2y}{dx^2}. We have dx/dt=2t−4dx/dt = 2t - 4 and dy/dt=3t2dy/dt = 3t^2. The first derivative is dydx=3t22t−4\frac{dy}{dx} = \frac{3t^2}{2t - 4}. Next, we find the derivative of dydx\frac{dy}{dx} with respect to tt using the quotient rule: ddt(dydx)=6t(2t−4)−3t2(2)(2t−4)2=12t2−24t−6t2(2t−4)2=6t2−24t(2t−4)2=6t(t−4)4(t−2)2\frac{d}{dt}(\frac{dy}{dx}) = \frac{6t(2t - 4) - 3t^2(2)}{(2t - 4)^2} = \frac{12t^2 - 24t - 6t^2}{(2t - 4)^2} = \frac{6t^2 - 24t}{(2t - 4)^2} = \frac{6t(t-4)}{4(t-2)^2}. The second derivative is d2ydx2=ddt(dydx)dx/dt=6t(t−4)4(t−2)2⋅12t−4=6t(t−4)8(t−2)3\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt} = \frac{6t(t-4)}{4(t-2)^2} \cdot \frac{1}{2t-4} = \frac{6t(t-4)}{8(t-2)^3}. The curve is concave down when d2ydx2<0\frac{d^2y}{dx^2} < 0. We analyze the sign of the expression t(t−4)(t−2)3\frac{t(t-4)}{(t-2)^3}. The critical points are t=0,2,4t=0, 2, 4. For t<0t<0, the expression is negative. For 0<t<20<t<2, it's positive. For 2<t<42<t<4, it's negative. For t>4t>4, it's positive. Thus, the curve is concave down on (−∞,0)∪(2,4)(-\infty, 0) \cup (2, 4).

Question 16

The second derivative d2ydx2\frac{d^2y}{dx^2} of the curve x=a(t−sin⁡t)x=a(t-\sin t), y=a(1−cos⁡t)y=a(1-\cos t) (a cycloid) is given by which expression?

  1. cos⁡ta(1−sin⁡t)2\frac{\cos t}{a(1-\sin t)^2}
  2. −1a(1−cos⁡t)2-\frac{1}{a(1-\cos t)^2} (correct answer)
  3. cot⁡(t/2)\cot(t/2)
  4. 1asin⁡3t\frac{1}{a\sin^3 t}
Explanation: When you encounter parametric equations and need to find the second derivative, you're working with the chain rule in a more complex form. For parametric curves given by x=f(t)x = f(t) and y=g(t)y = g(t), the second derivative requires careful application of the formula: d2ydx2=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} Let's work through this cycloid step by step. First, find the first derivatives: dxdt=a(1−cos⁡t)\frac{dx}{dt} = a(1 - \cos t) and dydt=asin⁡t\frac{dy}{dt} = a\sin t. So 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}. Now for the second derivative, you need ddt(sin⁡t1−cos⁡t)\frac{d}{dt}\left(\frac{\sin t}{1-\cos t}\right). Using the quotient rule: the numerator becomes (1−cos⁡t)(cos⁡t)−(sin⁡t)(sin⁡t)=cos⁡t−cos⁡2t−sin⁡2t=cos⁡t−1(1-\cos t)(\cos t) - (\sin t)(\sin t) = \cos t - \cos^2 t - \sin^2 t = \cos t - 1. The denominator becomes (1−cos⁡t)2(1-\cos t)^2. Therefore: d2ydx2=cos⁡t−1(1−cos⁡t)2⋅1a(1−cos⁡t)=−1a(1−cos⁡t)2\frac{d^2y}{dx^2} = \frac{\cos t - 1}{(1-\cos t)^2} \cdot \frac{1}{a(1-\cos t)} = \frac{-1}{a(1-\cos t)^2} This confirms answer B is correct. Answer A has the wrong denominator structure and includes sin⁡t\sin t incorrectly. Answer C, cot⁡(t/2)\cot(t/2), uses a half-angle function that doesn't arise from this calculation. Answer D has sin⁡3t\sin^3 t in the denominator, which would come from incorrectly handling the derivatives. Key strategy: For parametric second derivatives, always compute dydx\frac{dy}{dx} first, then carefully apply the quotient rule when differentiating it with respect to tt.

Question 17

Find the value of d2ydx2\frac{d^2y}{dx^2} for the curve defined by the parametric equations x=3cos⁡(t)x = 3\cos(t) and y=4sin⁡(t)y = 4\sin(t) at the point corresponding to t=π/4t = \pi/4.

  1. −829-\frac{8\sqrt{2}}{9} (correct answer)
  2. 43\frac{4}{3}
  3. 83\frac{8}{3}
  4. 829\frac{8\sqrt{2}}{9}
Explanation: We first find the derivatives with respect to tt: dx/dt=−3sin⁡(t)dx/dt = -3\sin(t) and dy/dt=4cos⁡(t)dy/dt = 4\cos(t). The first derivative is dydx=4cos⁡(t)−3sin⁡(t)=−43cot⁡(t)\frac{dy}{dx} = \frac{4\cos(t)}{-3\sin(t)} = -\frac{4}{3}\cot(t). Next, we differentiate dydx\frac{dy}{dx} with respect to tt: ddt(−43cot⁡(t))=−43(−csc⁡2(t))=43csc⁡2(t)\frac{d}{dt}(-\frac{4}{3}\cot(t)) = -\frac{4}{3}(-\csc^2(t)) = \frac{4}{3}\csc^2(t). The second derivative is d2ydx2=ddt(dydx)dx/dt=43csc⁡2(t)−3sin⁡(t)=−49sin⁡3(t)\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt} = \frac{\frac{4}{3}\csc^2(t)}{-3\sin(t)} = -\frac{4}{9\sin^3(t)}. Now we evaluate this expression at t=π/4t = \pi/4. Since sin⁡(π/4)=22\sin(\pi/4) = \frac{\sqrt{2}}{2}, we have sin⁡3(π/4)=(22)3=228=24\sin^3(\pi/4) = (\frac{\sqrt{2}}{2})^3 = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4}. Substituting this in, we get d2ydx2=−49(2/4)=−1692=−16218=−829\frac{d^2y}{dx^2} = -\frac{4}{9(\sqrt{2}/4)} = -\frac{16}{9\sqrt{2}} = -\frac{16\sqrt{2}}{18} = -\frac{8\sqrt{2}}{9}.

Question 18

For the curve defined by x=t2+ctx=t^2+ct and y=t2y=t^2, where cc is a constant, find the set of values for cc for which the curve is concave up for all t>0t > 0.

  1. All real numbers cc.
  2. c<0c < 0
  3. c≠0c \neq 0
  4. c>0c > 0 (correct answer)
Explanation: When analyzing the concavity of parametric curves, you need to find the second derivative d2ydx2\frac{d^2y}{dx^2} using the chain rule for parametric equations: d2ydx2=ddx(dydx)=ddt(dydx)dxdt\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}. Starting with the given parametric equations x=t2+ctx = t^2 + ct and y=t2y = t^2, we find:
  • dxdt=2t+c\frac{dx}{dt} = 2t + c
  • dydt=2t\frac{dy}{dt} = 2t
  • dydx=2t2t+c\frac{dy}{dx} = \frac{2t}{2t + c}
To find the second derivative, differentiate dydx\frac{dy}{dx} with respect to tt: ddt(2t2t+c)=2(2t+c)−2t(2)(2t+c)2=2c(2t+c)2\frac{d}{dt}\left(\frac{2t}{2t + c}\right) = \frac{2(2t + c) - 2t(2)}{(2t + c)^2} = \frac{2c}{(2t + c)^2} Therefore: d2ydx2=2c(2t+c)22t+c=2c(2t+c)3\frac{d^2y}{dx^2} = \frac{\frac{2c}{(2t + c)^2}}{2t + c} = \frac{2c}{(2t + c)^3} For concavity up when t>0t > 0, we need d2ydx2>0\frac{d^2y}{dx^2} > 0. Since t>0t > 0 and assuming c>0c > 0, we have 2t+c>02t + c > 0, making (2t+c)3>0(2t + c)^3 > 0. Thus 2c(2t+c)3>0\frac{2c}{(2t + c)^3} > 0 requires c>0c > 0. Answer choice A is wrong because negative values of cc make the second derivative negative. Choice B is wrong because c<0c < 0 gives concave down behavior. Choice C is wrong because it includes negative values of cc, which produce downward concavity. Study tip: For parametric concavity problems, always compute the full second derivative formula first, then analyze the sign conditions based on the given domain constraints.