Implicit & Parametric Differentiation - IB Mathematics Analysis and Approaches

1. The curve known as the lemniscate of Bernoulli is given by $(x^2 + y^2)^2 = 2(x^2 - y^2)$. Which of the following is an expression for $\frac{dy}{dx}$ in terms of $x$ and $y$?
   Answer: $\frac{x(1 - x^2 - y^2)}{y(1 + x^2 + y^2)}$ Differentiate the equation implicitly with respect to $x$. $2(x^2+y^2)(2x+2y\frac{dy}{dx}) = 2(2x - 2y\frac{dy}{dx})$. Divide by 4: $(x^2+y^2)(x+y\frac{dy}{dx}) = x - y\frac{dy}{dx}$. Expand: $x(x^2+y^2) + y(x^2+y^2)\frac{dy}{dx} = x - y\frac{dy}{dx}$. Group terms with $\frac{dy}{dx}$: $\frac{dy}{dx}(y(x^2+y^2) + y) = x - x(x^2+y^2)$. Factor out common terms: $\frac{dy}{dx} \cdot y(x^2+y^2+1) = x(1 - (x^2+y^2))$. Finally, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{x(1 - x^2 - y^2)}{y(1 + x^2 + y^2)}$. The other options are distractors based on sign errors or incorrect factoring during the algebraic manipulation.
2. A particle's position is given by $x(t) = 3t^2 - 2$ and $y(t) = t^3 - 3t$. For which interval(s) of $t$ is the particle moving to the right and downwards?
   Answer: $0 < t < 1$ The particle moves to the right when its horizontal velocity, $\frac{dx}{dt}$, is positive. $\frac{dx}{dt} = 6t$. So, $6t > 0 \implies t > 0$. The particle moves downwards when its vertical velocity, $\frac{dy}{dt}$, is negative. $\frac{dy}{dt} = 3t^2 - 3$. So, $3t^2 - 3 < 0 \implies t^2 < 1 \implies -1 < t < 1$. 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 $t > 0$ and $-1 < t < 1$. The intersection is $0 < t < 1$.
3. A curve is defined by the parametric equations $x = \sec(t)$ and $y = \tan(t)$. Find $\frac{dy}{dx}$ in terms of $t$.
   Answer: $\csc(t)$ First, find the derivatives with respect to the parameter $t$. $\frac{dx}{dt} = \frac{d}{dt}(\sec(t)) = \sec(t)\tan(t)$. $\frac{dy}{dt} = \frac{d}{dt}(\tan(t)) = \sec^2(t)$. The derivative $\frac{dy}{dx}$ is given by the ratio $\frac{dy/dt}{dx/dt}$. So, $\frac{dy}{dx} = \frac{\sec^2(t)}{\sec(t)\tan(t)} = \frac{\sec(t)}{\tan(t)}$. To simplify, we can write this in terms of sine and cosine: $\frac{1/\cos(t)}{\sin(t)/\cos(t)} = \frac{1}{\sin(t)} = \csc(t)$. Distractor B, $\cos(t)$, is $dx/dy$. Distractors A and D arise from incorrect simplification of the trigonometric expression.
4. What is the slope of the tangent line to the curve defined by the parametric equations $x=t^2+1, y=\arctan(t)$ at the point $(2, \pi/4)$?
   Answer: 1/4 First, find the value of the parameter $t$ corresponding to the point $(2, \pi/4)$. From $x=t^2+1$, we have $2=t^2+1 \implies t^2=1 \implies t=\pm 1$. From $y=\arctan(t)$, we have $\pi/4=\arctan(t) \implies t=\tan(\pi/4)=1$. So we use $t=1$. Next, find $\frac{dy}{dx}$. $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = \frac{1}{1+t^2}$. Then $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/(1+t^2)}{2t} = \frac{1}{2t(1+t^2)}$. Evaluate the slope at $t=1$: $\frac{dy}{dx} = \frac{1}{2(1)(1+1^2)} = \frac{1}{2(2)} = \frac{1}{4}$.
5. A curve is parametrically defined by $x = \sin^3(t)$ and $y = \cos^3(t)$. Find the slope of the tangent line at $t = \pi/4$.
   Answer: -1 First, we find the derivatives with respect to $t$ using the chain rule. $\frac{dx}{dt} = 3\sin^2(t) \cdot \cos(t)$. $\frac{dy}{dt} = 3\cos^2(t) \cdot (-\sin(t)) = -3\cos^2(t)\sin(t)$. The slope of the tangent line is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3\cos^2(t)\sin(t)}{3\sin^2(t)\cos(t)}$. For $t$ where $\sin(t)$ and $\cos(t)$ are non-zero, this simplifies to $-\frac{\cos(t)}{\sin(t)} = -\cot(t)$. Now, we evaluate the slope at $t = \pi/4$: $m = -\cot(\pi/4) = -1$.
6. Find the equation of the normal to the curve $x^2 + 3y^2 = 13$ at the point $(2, \sqrt{3})$.
   Answer: $y = \frac{3\sqrt{3}}{2}x - 2\sqrt{3}$ First, find the slope of the tangent by implicit differentiation: $2x + 6y\frac{dy}{dx} = 0$, which gives $\frac{dy}{dx} = -\frac{2x}{6y} = -\frac{x}{3y}$. At the point $(2, \sqrt{3})$, the slope of the tangent is $m_{tan} = -\frac{2}{3\sqrt{3}}$. The slope of the normal is the negative reciprocal of the tangent's slope: $m_{norm} = -\frac{1}{m_{tan}} = \frac{3\sqrt{3}}{2}$. Now, use the point-slope form for the equation of the normal line: $y - y_1 = m_{norm}(x - x_1)$, which is $y - \sqrt{3} = \frac{3\sqrt{3}}{2}(x - 2)$. Simplifying gives $y - \sqrt{3} = \frac{3\sqrt{3}}{2}x - 3\sqrt{3}$, so $y = \frac{3\sqrt{3}}{2}x - 2\sqrt{3}$. Distractor A uses the tangent slope. Distractors B and D result from calculation errors with the slopes or the point-slope formula.
7. A curve is given by $x = a(t - \sin t)$ and $y = a(1 - \cos t)$, where $a > 0$. For what value of $t$ in $(0, 2\pi)$ is the slope of the tangent line equal to 1?
   Answer: $\pi/2$ First, we find $\frac{dy}{dx}$. $\frac{dx}{dt} = a(1 - \cos t)$ and $\frac{dy}{dt} = a\sin t$. The slope is $\frac{dy}{dx} = \frac{a\sin t}{a(1 - \cos t)} = \frac{\sin t}{1 - \cos t}$. We set the slope equal to 1: $\frac{\sin t}{1 - \cos t} = 1$, which means $\sin t = 1 - \cos t$. Squaring both sides: $\sin^2 t = (1 - \cos t)^2 = 1 - 2\cos t + \cos^2 t$. Since $\sin^2 t + \cos^2 t = 1$, we have $1 - \cos^2 t = 1 - 2\cos t + \cos^2 t$, which simplifies to $2\cos^2 t - 2\cos t = 0$, or $2\cos t(\cos t - 1) = 0$. This gives $\cos t = 0$ or $\cos t = 1$. For $t \in (0, 2\pi)$, $\cos t = 0$ gives $t = \pi/2$ or $t = 3\pi/2$, and $\cos t = 1$ gives $t = 2\pi$ (excluded). Checking: at $t = \pi/2$, $\sin t = 1$ and $1 - \cos t = 1$, so the slope is 1. At $t = 3\pi/2$, $\sin t = -1$ and $1 - \cos t = 1$, so the slope is -1. Therefore, $t = \pi/2$.
8. Consider the curve defined by $\sin(x) + \cos(y) = 1$. Which statement is true about the tangent to the curve at the point $(\pi/6, \pi/3)$?
   Answer: The tangent line has a slope of 1. Differentiate implicitly with respect to $x$: $\cos(x) - \sin(y)\frac{dy}{dx} = 0$. Solving for $\frac{dy}{dx}$ gives $\frac{dy}{dx} = \frac{\cos(x)}{\sin(y)}$. Now, evaluate the slope at the point $(\pi/6, \pi/3)$: $\frac{dy}{dx} = \frac{\cos(\pi/6)}{\sin(\pi/3)} = \frac{\sqrt{3}/2}{\sqrt{3}/2} = 1$. So, the tangent line has a slope of 1. A horizontal tangent would mean the slope is 0, which would require $\cos(x)=0$. A vertical tangent would mean the slope is undefined, which would require $\sin(y)=0$. A slope of -1 is a plausible distractor if a sign error is made or trig values are mixed up.
9. A curve is defined by the parametric equations $x = t^3 - 3t$ and $y = t^2 - 2$. Find the coordinates of the points where the tangent to the curve is horizontal.
   Answer: (0, -2) A horizontal tangent occurs when $\frac{dy}{dx} = 0$, which means $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$. First, find the derivatives with respect to $t$: $\frac{dx}{dt} = 3t^2 - 3$ and $\frac{dy}{dt} = 2t$. Set $\frac{dy}{dt} = 0$, which gives $2t = 0 \implies t = 0$. We must check that $\frac{dx}{dt} \neq 0$ at $t=0$. $\frac{dx}{dt}$ at $t=0$ is $3(0)^2 - 3 = -3 \neq 0$. So there is a horizontal tangent at $t=0$. The coordinates of this point are $x = 0^3 - 3(0) = 0$ and $y = 0^2 - 2 = -2$. The point is $(0, -2)$. Distractor A corresponds to the points of vertical tangency, where $\frac{dx}{dt} = 0 \implies 3t^2 - 3 = 0 \implies t = \pm 1$. At $t=1$, $(x,y) = (-2,-1)$. At $t=-1$, $(x,y) = (2,-1)$.
10. For the curve defined by $e^{xy} = x - y$, what is the value of $\frac{dy}{dx}$ at the point $(1, 0)$?
    Answer: 1/2 We use implicit differentiation. Differentiating both sides with respect to $x$, we apply the product rule to the term $xy$ inside the exponential's chain rule: $e^{xy}(y + x\frac{dy}{dx}) = 1 - \frac{dy}{dx}$. Now, we rearrange the equation to solve for $\frac{dy}{dx}$: $ye^{xy} + xe^{xy}\frac{dy}{dx} = 1 - \frac{dy}{dx}$. $xe^{xy}\frac{dy}{dx} + \frac{dy}{dx} = 1 - ye^{xy}$. $\frac{dy}{dx}(xe^{xy} + 1) = 1 - ye^{xy}$. $\frac{dy}{dx} = \frac{1 - ye^{xy}}{xe^{xy} + 1}$. Substitute the point $(1, 0)$: $\frac{dy}{dx} = \frac{1 - 0 \cdot e^{1 \cdot 0}}{1 \cdot e^{1 \cdot 0} + 1} = \frac{1 - 0}{1 \cdot 1 + 1} = \frac{1}{2}$. 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 $\frac{dy}{dx}$.
11. The tangent to the curve $y = \ln(x^2 + y^2)$ at the point $(1, 0)$ is horizontal. What can be concluded about $\frac{dy}{dx}$ at this point?
    Answer: The statement is false; the tangent is not horizontal. Let's find the slope $\frac{dy}{dx}$ by implicit differentiation. $\frac{dy}{dx} = \frac{1}{x^2+y^2}(2x+2y\frac{dy}{dx})$. Rearranging to solve for $\frac{dy}{dx}$: $\frac{dy}{dx}(1 - \frac{2y}{x^2+y^2}) = \frac{2x}{x^2+y^2}$. $\frac{dy}{dx}(\frac{x^2+y^2-2y}{x^2+y^2}) = \frac{2x}{x^2+y^2}$. $\frac{dy}{dx} = \frac{2x}{x^2+y^2-2y}$. Now evaluate at the point $(1, 0)$: $\frac{dy}{dx} = \frac{2(1)}{1^2+0^2-2(0)} = \frac{2}{1} = 2$. 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.
12. A curve is described by the parametric equations $x = 4 + 2\ln(t)$ and $y = t^2 + 1$. Find the equation of the tangent line to the curve at the point where $t=1$.
    Answer: $y = x - 2$ First, find the point on the curve when $t=1$. $x(1) = 4 + 2\ln(1) = 4 + 0 = 4$. $y(1) = 1^2 + 1 = 2$. The point is $(4, 2)$. Next, find the slope of the tangent, $\frac{dy}{dx}$. $\frac{dx}{dt} = \frac{2}{t}$ and $\frac{dy}{dt} = 2t$. So, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t}{2/t} = t^2$. At $t=1$, the slope is $m = 1^2 = 1$. Now, use the point-slope form of a line: $y - y_1 = m(x - x_1)$. $y - 2 = 1(x - 4)$, which simplifies to $y = x - 2$.

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