Second Derivatives: Parametric - Calculus 2

If $x=t^3+t$ and $y=t^2-1$ , what is $\frac{d^2y}{dx^2}$ at $t=1$ ?
Answer: -1/16 When you encounter parametric equations and need to find the second derivative $\frac{d^2y}{dx^2}$ , 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 $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . With $x = t^3 + t$ and $y = t^2 - 1$ , you get $\frac{dx}{dt} = 3t^2 + 1$ and $\frac{dy}{dt} = 2t$ . Therefore, $\frac{dy}{dx} = \frac{2t}{3t^2 + 1}$ . For the second derivative, use $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d/dt\left(\frac{dy}{dx}\right)}{dx/dt}$ . Differentiate $\frac{dy}{dx} = \frac{2t}{3t^2 + 1}$ with respect to t using the quotient rule: $\frac{d}{dt}\left(\frac{2t}{3t^2 + 1}\right) = \frac{2(3t^2 + 1) - 2t(6t)}{(3t^2 + 1)^2} = \frac{6t^2 + 2 - 12t^2}{(3t^2 + 1)^2} = \frac{2 - 6t^2}{(3t^2 + 1)^2}$ Now divide by $\frac{dx}{dt} = 3t^2 + 1$ : $\frac{d^2y}{dx^2} = \frac{2 - 6t^2}{(3t^2 + 1)^3}$ At $t = 1$ : $\frac{d^2y}{dx^2} = \frac{2 - 6(1)^2}{(3(1)^2 + 1)^3} = \frac{-4}{4^3} = \frac{-4}{64} = -\frac{1}{16}$ 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 $(3t^2 + 1)^3$ as $(3t^2 + 1)^2$ . Study tip: Always double-check your quotient rule application and remember that the denominator becomes cubed, not squared, in this parametric second derivative formula.
For which interval of $t$ is the curve given by $x = t^2 - 4t$ and $y = t^3$ concave down?
Answer: $(-\infty, 0) \cup (2, 4)$ First, we find $\frac{d^2y}{dx^2}$ . We have $dx/dt = 2t - 4$ and $dy/dt = 3t^2$ . The first derivative is $\frac{dy}{dx} = \frac{3t^2}{2t - 4}$ . Next, we find the derivative of $\frac{dy}{dx}$ with respect to $t$ using the quotient rule: $\frac{d}{dt}(\frac{dy}{dx}) = \frac{6t(2t - 4) - 3t^2(2)}{(2t - 4)^2} = \frac{12t^2 - 24t - 6t^2}{(2t - 4)^2} = \frac{6t^2 - 24t}{(2t - 4)^2} = \frac{6t(t-4)}{4(t-2)^2}$ . The second derivative is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt} = \frac{6t(t-4)}{4(t-2)^2} \cdot \frac{1}{2t-4} = \frac{6t(t-4)}{8(t-2)^3}$ . The curve is concave down when $\frac{d^2y}{dx^2} < 0$ . We analyze the sign of the expression $\frac{t(t-4)}{(t-2)^3}$ . The critical points are $t=0, 2, 4$ . For $t<0$ , the expression is negative. For $0<t<2$ , it's positive. For $2<t<4$ , it's negative. For $t>4$ , it's positive. Thus, the curve is concave down on $(-\infty, 0) \cup (2, 4)$ .
The curve defined by $x = t^2$ and $y = t^3 - 12t$ has a point of inflection where the concavity changes. What is the $y$ -coordinate of this inflection point?
Answer: 0 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 $x = f(t)$ and $y = g(t)$ , the second derivative is found using: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy/dt}{dx/dt}\right)$ First, find the first derivatives: $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2 - 12$ So $\frac{dy}{dx} = \frac{3t^2 - 12}{2t}$ For the second derivative: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{3t^2 - 12}{2t}\right) \cdot \frac{dt}{dx}$ Simplifying $\frac{3t^2 - 12}{2t} = \frac{3t}{2} - \frac{6}{t}$ , we get $\frac{d}{dt}\left(\frac{3t}{2} - \frac{6}{t}\right) = \frac{3}{2} + \frac{6}{t^2}$ Since $\frac{dt}{dx} = \frac{1}{2t}$ : $\frac{d^2y}{dx^2} = \left(\frac{3}{2} + \frac{6}{t^2}\right) \cdot \frac{1}{2t} = \frac{3}{4t} + \frac{3}{t^3}$ This second derivative is always positive for $t \neq 0$ (both terms are positive), meaning the curve is always concave up and never changes concavity. However, there's a cusp at $t = 0$ where $\frac{dx}{dt} = 0$ . At this point, $y = 0^3 - 12(0) = 0$ , giving us answer (C). (A) and (B) correspond to $y$ -values at $t = \pm 2$ , 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 $dx/dt = 0$ in addition to where $d^2y/dx^2 = 0$ , as cusps can behave like inflection points.
A parametric curve is given by $x(t) = \int_0^t \cos(u^2) du$ and $y(t) = \int_0^t \sin(u^2) du$ . Find $\frac{d^2y}{dx^2}$ in terms of $t$ .
Answer: $2t\sec^3(t^2)$ By the Fundamental Theorem of Calculus Part 1, $dx/dt = \cos(t^2)$ and $dy/dt = \sin(t^2)$ . The first derivative is $\frac{dy}{dx} = \frac{\sin(t^2)}{\cos(t^2)} = \tan(t^2)$ . Next, we differentiate $\frac{dy}{dx}$ with respect to $t$ using the chain rule: $\frac{d}{dt}(\tan(t^2)) = \sec^2(t^2) \cdot 2t = 2t\sec^2(t^2)$ . The second derivative is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt} = \frac{2t\sec^2(t^2)}{\cos(t^2)} = 2t\frac{1/\cos^2(t^2)}{\cos(t^2)} = \frac{2t}{\cos^3(t^2)} = 2t\sec^3(t^2)$ .
Let $x=t^p$ and $y=t^q$ for constants $p, q$ . Find an expression for $\frac{d^2y}{dx^2}$ that is independent of $t$ in terms of $y$ .
Answer: $\frac{q(q-p)}{p^2} y^{(q-2p)/q}$ When you encounter parametric equations like $x = t^p$ and $y = t^q$ , finding the second derivative $\frac{d^2y}{dx^2}$ requires careful application of the chain rule and parametric differentiation formulas. Start with the parametric formula: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy/dt}{dx/dt}\right)$ . First, find the first derivatives: $\frac{dx}{dt} = pt^{p-1}$ and $\frac{dy}{dt} = qt^{q-1}$ . So $\frac{dy}{dx} = \frac{qt^{q-1}}{pt^{p-1}} = \frac{q}{p}t^{q-p}$ . For the second derivative, differentiate $\frac{dy}{dx}$ with respect to $x$ : $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{q}{p}t^{q-p}\right) \cdot \frac{dt}{dx} = \frac{q(q-p)}{p}t^{q-p-1} \cdot \frac{1}{pt^{p-1}} = \frac{q(q-p)}{p^2}t^{q-2p}$ . To express this in terms of $y$ , use $y = t^q$ , so $t = y^{1/q}$ . Therefore: $t^{q-2p} = (y^{1/q})^{q-2p} = y^{(q-2p)/q}$ . This gives us $\frac{d^2y}{dx^2} = \frac{q(q-p)}{p^2}y^{(q-2p)/q}$ , which is answer B. Answer A uses incorrect exponents and coefficients. Answer C is missing the factor of $q$ 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.
A curve is defined by $x=f(t)$ and $y=g(t)$ . At $t=t_0$ , it is known that $f'(t_0) = -3$ , $g'(t_0) = 6$ , and $\frac{d^2y}{dx^2} = 2$ . What is the value of $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ at $t=t_0$ ?
Answer: $-6$ The formula for the second derivative of a parametric curve is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt}$ . We are given $\frac{d^2y}{dx^2} = 2$ and $dx/dt = f'(t_0) = -3$ . We need to find the value of the numerator, $\frac{d}{dt}(\frac{dy}{dx})$ . Rearranging the formula, we get $\frac{d}{dt}(\frac{dy}{dx}) = \frac{d^2y}{dx^2} \cdot \frac{dx}{dt}$ . Plugging in the given values at $t=t_0$ , we have $\frac{d}{dt}(\frac{dy}{dx}) = (2) \cdot (-3) = -6$ .
A particle moves along a path described by $x = t^2 + 1$ and $y = \ln(t)$ for $t > 0$ . At what value of $t$ is $\frac{d^2y}{dx^2} = -\frac{1}{2}$ ?
Answer: $t=1$ When you encounter parametric equations and need to find the second derivative $\frac{d^2y}{dx^2}$ , you're working with the chain rule in a more complex form. The key formula is: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy/dt}{dx/dt}\right)$ Let's find the derivatives step by step. From $x = t^2 + 1$ and $y = \ln(t)$ , we get: - $\frac{dx}{dt} = 2t$ - $\frac{dy}{dt} = \frac{1}{t}$ So $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/t}{2t} = \frac{1}{2t^2}$ For the second derivative, we need $\frac{d}{dx}\left(\frac{1}{2t^2}\right)$ . Using the chain rule: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{1}{2t^2}\right) \cdot \frac{dt}{dx}$ Since $\frac{d}{dt}\left(\frac{1}{2t^2}\right) = -\frac{1}{t^3}$ and $\frac{dt}{dx} = \frac{1}{2t}$ : $\frac{d^2y}{dx^2} = -\frac{1}{t^3} \cdot \frac{1}{2t} = -\frac{1}{2t^4}$ Setting this equal to $-\frac{1}{2}$ : $-\frac{1}{2t^4} = -\frac{1}{2}$ $\frac{1}{t^4} = 1$ $t^4 = 1$ $t = 1$ (since $t > 0$ ) This confirms answer C is correct. Option A ( $t = 1/2$ ) gives $-8$ , option B ( $t = 2$ ) gives $-\frac{1}{32}$ , and option D ( $t = e$ ) gives $-\frac{1}{2e^4}$ . Study tip: Always remember that $\frac{dt}{dx} = \frac{1}{dx/dt}$ when converting derivatives between parametric variables. Practice this chain rule application—it's essential for parametric calculus problems.
For the curve defined by $x=t^2+ct$ and $y=t^2$ , where $c$ is a constant, find the set of values for $c$ for which the curve is concave up for all $t > 0$ .
Answer: $c > 0$ When analyzing the concavity of parametric curves, you need to find the second derivative $\frac{d^2y}{dx^2}$ using the chain rule for parametric equations: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ . Starting with the given parametric equations $x = t^2 + ct$ and $y = t^2$ , we find: - $\frac{dx}{dt} = 2t + c$ - $\frac{dy}{dt} = 2t$ - $\frac{dy}{dx} = \frac{2t}{2t + c}$ To find the second derivative, differentiate $\frac{dy}{dx}$ with respect to $t$ : $\frac{d}{dt}\left(\frac{2t}{2t + c}\right) = \frac{2(2t + c) - 2t(2)}{(2t + c)^2} = \frac{2c}{(2t + c)^2}$ Therefore: $\frac{d^2y}{dx^2} = \frac{\frac{2c}{(2t + c)^2}}{2t + c} = \frac{2c}{(2t + c)^3}$ For concavity up when $t > 0$ , we need $\frac{d^2y}{dx^2} > 0$ . Since $t > 0$ and assuming $c > 0$ , we have $2t + c > 0$ , making $(2t + c)^3 > 0$ . Thus $\frac{2c}{(2t + c)^3} > 0$ requires $c > 0$ . Answer choice A is wrong because negative values of $c$ make the second derivative negative. Choice B is wrong because $c < 0$ gives concave down behavior. Choice C is wrong because it includes negative values of $c$ , 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.
For the parametric curve $x = \sec(t)$ , $y = \tan(t)$ on $-\pi/2 < t < \pi/2$ , what is $\frac{d^2y}{dx^2}$ at the point $(\sqrt{2}, 1)$ ?
Answer: -1 When you encounter parametric curves and need to find the second derivative $\frac{d^2y}{dx^2}$ , remember that you can't differentiate $y$ with respect to $x$ directly. Instead, you must use the chain rule and the relationship $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ . First, find the first derivative: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sec^2(t)}{\sec(t)\tan(t)} = \frac{\sec(t)}{\tan(t)} = \csc(t)$ . Next, find the second derivative: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\csc(t))}{\frac{dx}{dt}} = \frac{-\csc(t)\cot(t)}{\sec(t)\tan(t)}$ . Simplifying: $\frac{d^2y}{dx^2} = \frac{-\csc(t)\cot(t)}{\sec(t)\tan(t)} = -\frac{\cos^2(t)}{\sin^2(t)} = -\cot^2(t)$ . To find $t$ at point $(\sqrt{2}, 1)$ : since $x = \sec(t) = \sqrt{2}$ , we have $\cos(t) = \frac{1}{\sqrt{2}}$ , so $t = \pi/4$ . We can verify: $y = \tan(\pi/4) = 1$ ✓ At $t = \pi/4$ : $\cot(\pi/4) = 1$ , so $\frac{d^2y}{dx^2} = -\cot^2(\pi/4) = -1$ . Answer (A) -2 likely comes from incorrectly computing $-2\cot^2(t)$ . Answer (B) 1 results from forgetting the negative sign. Answer (C) $\sqrt{2}$ might come from confusing the $x$ -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.
For a curve defined by $x=f(t)$ and $y=g(t)$ , it is known that for $t$ in an interval $(a,b)$ , we have $f'(t) < 0$ and $\frac{d^2y}{dx^2} < 0$ . Which of the following statements must be true for $t \in (a,b)$ ?
Answer: The slope $\frac{dy}{dx}$ is an increasing function of $t$ . 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 $f'(t) < 0$ and $\frac{d^2y}{dx^2} < 0$ , let's work through what these conditions mean. The second derivative $\frac{d^2y}{dx^2}$ measures the concavity of the curve when plotted in the xy-plane. For parametric curves, this is calculated using: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{g'(t)}{f'(t)}\right)$ Using the chain rule: $\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{g'(t)}{f'(t)}\right) \cdot \frac{dt}{dx} = \frac{d}{dt}\left(\frac{g'(t)}{f'(t)}\right) \cdot \frac{1}{f'(t)}$ Since $\frac{d^2y}{dx^2} < 0$ and $f'(t) < 0$ , we have $\frac{d}{dt}\left(\frac{g'(t)}{f'(t)}\right) \cdot \frac{1}{f'(t)} < 0$ . This means $\frac{d}{dt}\left(\frac{dy}{dx}\right) > 0$ , so the slope $\frac{dy}{dx}$ increases as t increases. Therefore, C is correct. Option A is wrong because $y$ decreasing depends on $g'(t)$ , 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 $\frac{dy}{dx} = \frac{g'(t)}{f'(t)}$ depends on whether $g'(t)$ is positive or negative. Key strategy: Always remember that $\frac{d^2y}{dx^2} < 0$ tells you about concavity in the xy-plane, not about individual derivatives with respect to t.
A particle's motion is described by $x(t) = 2t^3$ and $y(t) = 3t^2$ . Find an expression for $\frac{d^2y}{dx^2}$ in terms of $t$ .
Answer: $-\frac{1}{6t^4}$ When you encounter parametric equations and need to find the second derivative $\frac{d^2y}{dx^2}$ , you're dealing with a two-step process that builds on the chain rule and parametric differentiation. First, find $\frac{dy}{dx}$ using the parametric formula: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ . From $x(t) = 2t^3$ and $y(t) = 3t^2$ , we get $\frac{dx}{dt} = 6t^2$ and $\frac{dy}{dt} = 6t$ . Therefore: $\frac{dy}{dx} = \frac{6t}{6t^2} = \frac{1}{t}$ Next, find the second derivative using: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ . Since $\frac{dy}{dx} = \frac{1}{t} = t^{-1}$ , we have $\frac{d}{dt}\left(\frac{dy}{dx}\right) = -t^{-2} = -\frac{1}{t^2}$ . Substituting: $\frac{d^2y}{dx^2} = \frac{-\frac{1}{t^2}}{6t^2} = -\frac{1}{6t^4}$ Answer D is correct. Answer A, $-\frac{1}{t^2}$ , represents just the derivative of $\frac{dy}{dx}$ with respect to $t$ , but forgets to divide by $\frac{dx}{dt}$ . Answer B, $\frac{1}{t}$ , is actually $\frac{dy}{dx}$ (the first derivative), not the second. Answer C, $\frac{1}{6t^2}$ , has the wrong sign and wrong power of $t$ . Remember: For parametric second derivatives, you must apply the parametric differentiation formula twice. The most common error is stopping after finding $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ and forgetting the final division by $\frac{dx}{dt}$ .
The second derivative $\frac{d^2y}{dx^2}$ of the curve $x=a(t-\sin t)$ , $y=a(1-\cos t)$ (a cycloid) is given by which expression?
Answer: $-\frac{1}{a(1-\cos t)^2}$ 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 $x = f(t)$ and $y = g(t)$ , the second derivative requires careful application of the formula: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ Let's work through this cycloid step by step. First, find the first derivatives: $\frac{dx}{dt} = a(1 - \cos t)$ and $\frac{dy}{dt} = a\sin t$ . So $\frac{dy}{dx} = \frac{a\sin t}{a(1-\cos t)} = \frac{\sin t}{1-\cos t}$ . Now for the second derivative, you need $\frac{d}{dt}\left(\frac{\sin t}{1-\cos t}\right)$ . Using the quotient rule: the numerator becomes $(1-\cos t)(\cos t) - (\sin t)(\sin t) = \cos t - \cos^2 t - \sin^2 t = \cos t - 1$ . The denominator becomes $(1-\cos t)^2$ . Therefore: $\frac{d^2y}{dx^2} = \frac{\cos t - 1}{(1-\cos t)^2} \cdot \frac{1}{a(1-\cos t)} = \frac{-1}{a(1-\cos t)^2}$ This confirms answer B is correct. Answer A has the wrong denominator structure and includes $\sin t$ incorrectly. Answer C, $\cot(t/2)$ , uses a half-angle function that doesn't arise from this calculation. Answer D has $\sin^3 t$ in the denominator, which would come from incorrectly handling the derivatives. Key strategy: For parametric second derivatives, always compute $\frac{dy}{dx}$ first, then carefully apply the quotient rule when differentiating it with respect to $t$ .
Let $x=t^2$ and $y=e^{t^2}$ . The second derivative $\frac{d^2y}{dx^2}$ is:
Answer: $e^{t^2}$ 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 $\frac{d^2y}{dx^2}$ for parametric equations, you first need $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . Given $x = t^2$ and $y = e^{t^2}$ , you have: - $\frac{dx}{dt} = 2t$ - $\frac{dy}{dt} = 2te^{t^2}$ (using chain rule) So $\frac{dy}{dx} = \frac{2te^{t^2}}{2t} = e^{t^2}$ For the second derivative, use $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d/dt\left(\frac{dy}{dx}\right)}{dx/dt}$ Since $\frac{dy}{dx} = e^{t^2}$ , taking its derivative with respect to $t$ : $\frac{d}{dt}(e^{t^2}) = 2te^{t^2}$ Therefore: $\frac{d^2y}{dx^2} = \frac{2te^{t^2}}{2t} = e^{t^2}$ Since $x = t^2$ , we have $e^{t^2} = e^x$ , making answer (A) correct. Looking at the wrong answers: (B) $e^{t^2}$ is technically equivalent to the correct answer but doesn't substitute back to express the result in terms of $x$ . (C) $2e^{t^2}$ likely comes from forgetting to simplify the fraction. (D) $(1+2t^2)e^{t^2}$ suggests incorrectly applying the product rule when finding the second derivative. Study tip: Always express your final answer in terms of the original variable ( $x$ here) unless specifically asked otherwise. Practice the parametric second derivative formula until it becomes automatic.
A curve is defined by the parametric equations $x = \ln(t)$ and $y = t^3$ for $t > 0$ . What is the value of $\frac{d^2y}{dx^2}$ at the point where $x = 1$ ?
Answer: $9e^3$ First, find the value of $t$ corresponding to $x=1$ . We have $x = \ln(t) = 1$ , which gives $t = e$ . Next, we compute the first and second derivatives. The first derivative is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ . Here, $dx/dt = 1/t$ and $dy/dt = 3t^2$ . So, $\frac{dy}{dx} = \frac{3t^2}{1/t} = 3t^3$ . The second derivative is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt}$ . We find $\frac{d}{dt}(\frac{dy}{dx}) = \frac{d}{dt}(3t^3) = 9t^2$ . Therefore, $\frac{d^2y}{dx^2} = \frac{9t^2}{1/t} = 9t^3$ . Evaluating at $t=e$ , we get $9e^3$ .
For what value of the constant $k$ does the curve given by $x = e^{-t}$ and $y = ke^{2t}$ have $\frac{d^2y}{dx^2} = 12$ at the point $(1, k)$ ?
Answer: 2 First, find the value of $t$ that corresponds to the point $(1, k)$ . From $x = e^{-t} = 1$ , we get $-t = 0$ , so $t=0$ . At $t=0$ , $y = ke^{2(0)} = k$ , which confirms this is the correct $t$ value. Now, compute the second derivative. $dx/dt = -e^{-t}$ and $dy/dt = 2ke^{2t}$ . The first derivative is $\frac{dy}{dx} = \frac{2ke^{2t}}{-e^{-t}} = -2ke^{3t}$ . The derivative of this with respect to $t$ is $\frac{d}{dt}(-2ke^{3t}) = -6ke^{3t}$ . The second derivative is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt} = \frac{-6ke^{3t}}{-e^{-t}} = 6ke^{4t}$ . We are given that this equals 12 at the point, which corresponds to $t=0$ . So, $6ke^{4(0)} = 12 \implies 6k = 12 \implies k=2$ .
Find the value of $\frac{d^2y}{dx^2}$ for the curve defined by the parametric equations $x = 3\cos(t)$ and $y = 4\sin(t)$ at the point corresponding to $t = \pi/4$ .
Answer: $-\frac{8\sqrt{2}}{9}$ We first find the derivatives with respect to $t$ : $dx/dt = -3\sin(t)$ and $dy/dt = 4\cos(t)$ . The first derivative is $\frac{dy}{dx} = \frac{4\cos(t)}{-3\sin(t)} = -\frac{4}{3}\cot(t)$ . Next, we differentiate $\frac{dy}{dx}$ with respect to $t$ : $\frac{d}{dt}(-\frac{4}{3}\cot(t)) = -\frac{4}{3}(-\csc^2(t)) = \frac{4}{3}\csc^2(t)$ . The second derivative is $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{dx/dt} = \frac{\frac{4}{3}\csc^2(t)}{-3\sin(t)} = -\frac{4}{9\sin^3(t)}$ . Now we evaluate this expression at $t = \pi/4$ . Since $\sin(\pi/4) = \frac{\sqrt{2}}{2}$ , we have $\sin^3(\pi/4) = (\frac{\sqrt{2}}{2})^3 = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4}$ . Substituting this in, we get $\frac{d^2y}{dx^2} = -\frac{4}{9(\sqrt{2}/4)} = -\frac{16}{9\sqrt{2}} = -\frac{16\sqrt{2}}{18} = -\frac{8\sqrt{2}}{9}$ .
A parametric curve is defined by $x = t^2 + t$ and $y = t^3 - 3t^2 + 2t$ . At how many points does this curve have horizontal inflection points (where $\frac{dy}{dx} = 0$ and $\frac{d^2y}{dx^2} = 0$ )?
Answer: Zero points First, $\frac{dx}{dt} = 2t + 1$ and $\frac{dy}{dt} = 3t^2 - 6t + 2$ . So $\frac{dy}{dx} = \frac{3t^2 - 6t + 2}{2t + 1}$ . For horizontal tangents, we need $\frac{dy}{dx} = 0$ , which occurs when $3t^2 - 6t + 2 = 0$ . Using the quadratic formula: $t = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}$ . For the second derivative, we use $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ . Computing $\frac{d}{dt}\left(\frac{3t^2 - 6t + 2}{2t + 1}\right)$ using the quotient rule: $\frac{(6t - 6)(2t + 1) - (3t^2 - 6t + 2)(2)}{(2t + 1)^2} = \frac{12t^2 - 6t - 6 - 6t^2 + 12t - 4}{(2t + 1)^2} = \frac{6t^2 + 6t - 10}{(2t + 1)^2}$ . Therefore, $\frac{d^2y}{dx^2} = \frac{6t^2 + 6t - 10}{(2t + 1)^3}$ . At the critical points $t = 1 \pm \frac{\sqrt{3}}{3}$ , we can verify that $\frac{d^2y}{dx^2} \neq 0$ , so there are no horizontal inflection points.
For the curve $x=t^2+1, y=t^3$ , which of the following is an expression for $\frac{d^2y}{dx^2}$ in terms of $x$ ? Assume $t>0$ .
Answer: $\frac{3}{4\sqrt{x-1}}$ When you encounter a parametric curve and need to find the second derivative $\frac{d^2y}{dx^2}$ , you're working with parametric differentiation. The key formula is: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ Starting with the parametric equations $x = t^2 + 1$ and $y = t^3$ , first find the first derivative. We have $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2$ , so $\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$ . For the second derivative, differentiate $\frac{dy}{dx}$ with respect to $t$ : $\frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}$ . Then apply the chain rule: $\frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}$ . Now convert to terms of $x$ . From $x = t^2 + 1$ , we get $t^2 = x - 1$ , so $t = \sqrt{x-1}$ (since $t > 0$ ). Substituting: $\frac{d^2y}{dx^2} = \frac{3}{4\sqrt{x-1}}$ . Choice A gives $\frac{3\sqrt{x-1}}{2}$ , which would result from incorrectly multiplying by $t$ instead of dividing. Choice C, $3\sqrt{x-1}$ , makes a similar error with wrong coefficients. Choice D, $\frac{3}{2}$ , 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.

Start this quiz