Parametric Functions - IB Mathematics Analysis and Approaches

1. A parametric curve is defined by $x = 2\sin^2\theta$ and $y = 2\cos\theta$ for $0 \le \theta \le \pi$. Find its Cartesian equation and domain.
   Answer: $y^2 = 2(2-x)$ for $0 \le x \le 2$ We use the identity $\sin^2\theta + \cos^2\theta = 1$. From the parametric equations, we have: $$\sin^2\theta = \frac{x}{2}$$ $$\cos\theta = \frac{y}{2} \implies \cos^2\theta = \frac{y^2}{4}$$ Substitute these into the identity: $$\frac{x}{2} + \frac{y^2}{4} = 1$$ To clear the denominators, multiply by 4: $$2x + y^2 = 4 \implies y^2 = 4 - 2x \implies y^2 = 2(2-x)$$ Now we need to determine the domain for $x$. The parameter $\theta$ is in $[0, \pi]$. For $x = 2\sin^2\theta$, since $0 \le \sin\theta \le 1$ for $\theta \in [0, \pi]$, we have $0 \le \sin^2\theta \le 1$. Thus, $0 \le 2\sin^2\theta \le 2$. So the domain for $x$ is $0 \le x \le 2$. For $y = 2\cos\theta$, since $-1 \le \cos\theta \le 1$ for $\theta \in [0, \pi]$, the range for $y$ is $-2 \le y \le 2$. The equation $y^2 = 2(2-x)$ is consistent with this range for the given domain of $x$. * **B) $x^2 + y^2 = 4$:** This would be the case if $x=2\cos\theta, y=2\sin\theta$. * **C) $y = 2\sqrt{1-x/2}$:** This is only part of the curve. Since $\cos\theta$ can be negative on $[0, \pi]$, we cannot just take the positive square root. * **D) $x = 2-y$:** This is a linear equation and does not represent the parabolic relationship derived.
2. A curve is given by $x = \cos(2t)$ and $y = \sin(t)$ for $0 \le t \le \pi$. Find the coordinates of the point where the tangent to the curve is vertical.
   Answer: (1, 0) When you encounter parametric curves and need to find where the tangent is vertical, you're looking for points where $\frac{dx}{dt} = 0$ while $\frac{dy}{dt} \neq 0$. This is because the slope of the tangent line is $\frac{dy/dt}{dx/dt}$, which becomes undefined (vertical) when the denominator is zero but the numerator isn't. Given $x = \cos(2t)$ and $y = \sin(t)$, let's find the derivatives: - $\frac{dx}{dt} = -2\sin(2t)$ - $\frac{dy}{dt} = \cos(t)$ For a vertical tangent, set $\frac{dx}{dt} = 0$: $-2\sin(2t) = 0$ $\sin(2t) = 0$ On the interval $[0, \pi]$, this occurs when $2t = 0, \pi, 2\pi$, so $t = 0, \frac{\pi}{2}, \pi$. Check that $\frac{dy}{dt} \neq 0$ at these points: - At $t = 0$: $\cos(0) = 1 \neq 0$ ✓ - At $t = \frac{\pi}{2}$: $\cos(\frac{\pi}{2}) = 0$ ✗ - At $t = \pi$: $\cos(\pi) = -1 \neq 0$ ✓ So vertical tangents occur at $t = 0$ and $t = \pi$. The corresponding coordinates are: - $t = 0$: $(1, 0)$ - $t = \pi$: $(-1, 0)$ Answer D gives $(1, 0)$, which corresponds to $t = 0$. Answer A gives $(-1, 1)$, but at $t = \pi$ we get $(-1, 0)$, not $(-1, 1)$. Answer B gives $(0, 1)$, which occurs at $t = \frac{\pi}{2}$ where the tangent is horizontal, not vertical. Answer C incorrectly includes $(-1, 1)$. Remember: vertical tangents require $\frac{dx}{dt} = 0$ AND $\frac{dy}{dt} \neq 0$. Always verify both conditions and calculate the actual coordinates carefully.
3. Find the equation of the normal to the curve defined by $x = 2e^t$ and $y = t^2 + t$ at the point where $t=0$.
   Answer: $y = -2x + 4$ First, find the point of interest by substituting $t=0$: $$x(0) = 2e^0 = 2$$ $$y(0) = 0^2 + 0 = 0$$ The point is $(2, 0)$. Next, find the derivatives with respect to $t$: $$\frac{dx}{dt} = 2e^t$$ $$\frac{dy}{dt} = 2t + 1$$ Evaluate the derivatives at $t=0$: $$\frac{dx}{dt}\bigg|_{t=0} = 2e^0 = 2$$ $$\frac{dy}{dt}\bigg|_{t=0} = 2(0) + 1 = 1$$ Calculate the gradient of the tangent, $m_{\text{tan}} = \frac{dy}{dx}$: $$m_{\text{tan}} = \frac{dy/dt}{dx/dt} = \frac{1}{2}$$ The gradient of the normal is the negative reciprocal of the tangent's gradient: $$m_{\text{norm}} = -\frac{1}{m_{\text{tan}}} = -\frac{1}{1/2} = -2$$ Use the point-slope form $y - y_1 = m(x - x_1)$ with the point $(2,0)$ and $m_{\text{norm}}=-2$: $$y - 0 = -2(x - 2)$$ $$y = -2x + 4$$ * **B) $y = -\frac{1}{2}x + 1$:** This equation uses the negative of the tangent gradient, not the negative reciprocal. * **C) $y = \frac{1}{2}x - 1$:** This is the equation of the tangent line, not the normal. * **D) $y = 2x - 4$:** This equation uses the reciprocal of the tangent gradient, but not the negative reciprocal.
4. For which interval of $t$ is the curve defined by $x = t^2$ and $y = t^3 - t$ concave up?
   Answer: $t > 0$ Concavity is determined by the sign of the second derivative, $\frac{d^2y}{dx^2}$. The curve is concave up when $\frac{d^2y}{dx^2} > 0$. First, find $\frac{dy}{dx}$: $$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2 - 1$$ $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2 - 1}{2t}$$ Next, find $\frac{d^2y}{dx^2}$. We need to differentiate $\frac{dy}{dx}$ with respect to $t$ using the quotient rule: $$\frac{d}{dt}\left(\frac{3t^2 - 1}{2t}\right) = \frac{(6t)(2t) - (3t^2 - 1)(2)}{(2t)^2} = \frac{12t^2 - 6t^2 + 2}{4t^2} = \frac{6t^2 + 2}{4t^2} = \frac{3t^2 + 1}{2t^2}$$ Now, divide by $\frac{dx}{dt}$: $$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{(3t^2 + 1)/(2t^2)}{2t} = \frac{3t^2 + 1}{4t^3}$$ For the curve to be concave up, we need $\frac{d^2y}{dx^2} > 0$: $$\frac{3t^2 + 1}{4t^3} > 0$$ The numerator, $3t^2 + 1$, is always positive for any real $t$. Therefore, the sign of the fraction is determined by the sign of the denominator, $4t^3$. $$4t^3 > 0 \implies t^3 > 0 \implies t > 0$$ So, the curve is concave up for $t > 0$. * **A) $t < 0$:** This is the interval where the curve is concave down, as $4t^3 < 0$. * **C) $-\frac{1}{\sqrt{3}} < t < \frac{1}{\sqrt{3}}$:** This interval comes from solving $\frac{dy}{dt} = 3t^2 - 1 = 0$, which relates to horizontal tangents, not concavity. * **D) For all real $t$:** This would be the case if the sign of $\frac{d^2y}{dx^2}$ was constant, which would happen if the student only looked at the numerator $3t^2+1$ or the term $\frac{d}{dt}(\frac{dy}{dx}) = \frac{3t^2+1}{2t^2}$, which is always positive for $t \neq 0$.
5. At what values of $t$ does the curve defined by $x=t^2-2t$ and $y=t^2+2t$ intersect the line $y=x$?
   Answer: $t=0$ only To find the points of intersection, we set $y=x$ and substitute the parametric expressions: $$y(t) = x(t)$$ $$t^2 + 2t = t^2 - 2t$$ Subtract $t^2$ from both sides: $$2t = -2t$$ Add $2t$ to both sides: $$4t = 0$$ $$t = 0$$ So, the curve intersects the line $y=x$ only at $t=0$. The point of intersection is $x(0)=0^2-2(0)=0$ and $y(0)=0^2+2(0)=0$, which is the origin $(0,0)$. * **B) $t=-1$ only:** This could be a miscalculation such as $t^2+2t=-(t^2-2t)$, which gives $2t^2=0$, still $t=0$. A different error might lead to this. * **C) $t=0$ and $t=1$:** This might result from an error like $t^2+2t-1 = t^2-2t$, leading to $4t=1$. * D) The curve never intersects the line.: This would be the answer if the equation $4t=0$ had no solution, which is incorrect.
6. A curve is defined by the parametric equations $x = 3 + 2\cos t$ and $y = -1 + 2\sin t$. Find the Cartesian equation of the curve.
   Answer: $(x-3)^2 + (y+1)^2 = 4$ The goal is to eliminate the parameter $t$. We can use the trigonometric identity $\cos^2 t + \sin^2 t = 1$. First, rearrange the parametric equations to isolate $\cos t$ and $\sin t$: $$x = 3 + 2\cos t \implies x - 3 = 2\cos t \implies \cos t = \frac{x-3}{2}$$ $$y = -1 + 2\sin t \implies y + 1 = 2\sin t \implies \sin t = \frac{y+1}{2}$$ Now substitute these expressions into the identity $\cos^2 t + \sin^2 t = 1$: $$\left(\frac{x-3}{2}\right)^2 + \left(\frac{y+1}{2}\right)^2 = 1$$ $$\frac{(x-3)^2}{4} + \frac{(y+1)^2}{4} = 1$$ Multiply by 4 to get the standard form of the equation of a circle: $$(x-3)^2 + (y+1)^2 = 4$$ This is the equation of a circle with center $(3, -1)$ and radius 2. * **B) $(x+3)^2 + (y-1)^2 = 4$:** This results from sign errors when rearranging the parametric equations, leading to a circle with the wrong center $(-3, 1)$. * **C) $(x-3)^2 + (y+1)^2 = 2$:** This results from not squaring the radius. The equation is $r^2$, so if $r=2$, then $r^2=4$. * **D) $\frac{(x-3)^2}{4} - \frac{(y+1)^2}{4} = 1$:** This equation represents a hyperbola and would arise from incorrectly using an identity like $\sec^2 t - \tan^2 t = 1$ instead of $\cos^2 t + \sin^2 t = 1$.
7. Find the coordinates of the point where the curve defined by $x = t^3 - 12t$ and $y = 2t^2 - 4t$ has a horizontal tangent.
   Answer: (-11, -2) A horizontal tangent occurs when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$. First, find the derivative of $y$ with respect to $t$: $$\frac{dy}{dt} = 4t - 4$$ Set $\frac{dy}{dt} = 0$ to find the value of $t$: $$4t - 4 = 0 \implies t = 1$$ Next, check that $\frac{dx}{dt} \neq 0$ at $t=1$: $$\frac{dx}{dt} = 3t^2 - 12$$ At $t=1$, $\frac{dx}{dt} = 3(1)^2 - 12 = -9 \neq 0$. So, there is a horizontal tangent at $t=1$. Finally, find the coordinates $(x, y)$ at $t=1$: $$x(1) = (1)^3 - 12(1) = 1 - 12 = -11$$ $$y(1) = 2(1)^2 - 4(1) = 2 - 4 = -2$$ The coordinates of the point are $(-11, -2)$. * A) (-16, 0): This is a point where the tangent is vertical. This occurs when $\frac{dx}{dt} = 0 \Rightarrow 3t^2 - 12 = 0 \Rightarrow t=2$. At $t=2$, $x=-16, y=0$. * C) (0, 0): This point corresponds to $t=0$. At $t=0$, $\frac{dy}{dx} = \frac{-4}{-12} = \frac{1}{3}$. The tangent is not horizontal. * D) (16, 16): This is another point where the tangent is vertical. This occurs when $t=-2$. At $t=-2$, $x=16, y=16$.
8. A curve is given by $x=t+\frac{1}{t}$ and $y=t-\frac{1}{t}$ for $t>0$. Which of the following is the Cartesian equation of the curve?
   Answer: $x^2-y^2=4$ When you encounter parametric equations like these, your goal is to eliminate the parameter $t$ to find the relationship between $x$ and $y$. This process reveals the curve's Cartesian equation. Given $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$, let's find $x^2$ and $y^2$: $x^2 = \left(t + \frac{1}{t}\right)^2 = t^2 + 2 + \frac{1}{t^2}$ $y^2 = \left(t - \frac{1}{t}\right)^2 = t^2 - 2 + \frac{1}{t^2}$ Now subtract: $x^2 - y^2 = (t^2 + 2 + \frac{1}{t^2}) - (t^2 - 2 + \frac{1}{t^2}) = 4$ This gives us $x^2 - y^2 = 4$, which is answer D. Looking at the wrong answers: A) $x^2 + y^2 = 4$ would result from incorrectly adding the squared expressions instead of subtracting them. B) $xy = 1$ comes from a common mistake of trying to multiply the parametric equations directly, but $(t + \frac{1}{t})(t - \frac{1}{t}) = t^2 - \frac{1}{t^2}$, which isn't constant. C) $y^2 - x^2 = 4$ reverses the correct subtraction order. Study tip: When eliminating parameters from equations involving $t + \frac{1}{t}$ and $t - \frac{1}{t}$, always square both expressions and look for the difference of squares pattern. The cross terms will cancel beautifully, leaving you with a clean constant.
9. Find the equation of the tangent to the curve defined by $x = \sqrt{t}$ and $y = t^2 - \frac{1}{t}$ at the point where $t=1$.
   Answer: $y = 6x - 6$ First, find the coordinates of the point of tangency when $t=1$: $$x(1) = \sqrt{1} = 1$$ $$y(1) = 1^2 - \frac{1}{1} = 0$$ The point is $(1, 0)$. Next, find the derivatives with respect to $t$: $$\frac{dx}{dt} = \frac{1}{2\sqrt{t}}$$ $$\frac{dy}{dt} = 2t + \frac{1}{t^2}$$ Evaluate the derivatives at $t=1$: $$\frac{dx}{dt}\bigg|_{t=1} = \frac{1}{2\sqrt{1}} = \frac{1}{2}$$ $$\frac{dy}{dt}\bigg|_{t=1} = 2(1) + \frac{1}{1^2} = 3$$ Calculate the gradient of the tangent, $m = \frac{dy}{dx}$: $$m = \frac{dy/dt}{dx/dt} = \frac{3}{1/2} = 6$$ Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$: $$y - 0 = 6(x - 1)$$ $$y = 6x - 6$$ * **A) $y = \frac{1}{6}x - \frac{1}{6}$:** This equation uses the reciprocal of the correct gradient, $m = \frac{dx/dt}{dy/dt} = \frac{1}{6}$. * **B) $y = \frac{3}{2}x - \frac{3}{2}$:** This uses an incorrect gradient calculation, possibly from multiplying the derivatives instead of dividing: $m = (dy/dt) \times (dx/dt) = 3 \times \frac{1}{2} = \frac{3}{2}$. * **D) $x + 6y = 1$:** This is the equation of the normal to the curve at the point. The normal has a gradient of $-\frac{1}{m} = -\frac{1}{6}$.
10. A particle moves along a curve with its position at time $t$ given by $x(t) = \ln(t)$ and $y(t) = t^2$. Find the speed of the particle at $t=2$.
    Answer: $\frac{\sqrt{65}}{2}$ The speed of a particle moving along a parametric curve is given by the formula $s = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$. First, find the derivatives of $x(t)$ and $y(t)$ with respect to $t$: $$\frac{dx}{dt} = \frac{1}{t}$$ $$\frac{dy}{dt} = 2t$$ Now, evaluate these derivatives at $t=2$: $$\frac{dx}{dt}\bigg|_{t=2} = \frac{1}{2}$$ $$\frac{dy}{dt}\bigg|_{t=2} = 2(2) = 4$$ Substitute these values into the speed formula: $$s = \sqrt{\left(\frac{1}{2}\right)^2 + (4)^2} = \sqrt{\frac{1}{4} + 16} = \sqrt{\frac{1}{4} + \frac{64}{4}} = \sqrt{\frac{65}{4}} = \frac{\sqrt{65}}{2}$$ * A) 8: This is the value of the gradient of the tangent, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4}{1/2} = 8$, not the speed. * **B) $\frac{\sqrt{65}}{4}$:** This is the magnitude of the acceleration vector. $\frac{d^2x}{dt^2} = -\frac{1}{t^2}$ and $\frac{d^2y}{dt^2} = 2$. At $t=2$, the acceleration vector is $(-\frac{1}{4}, 2)$. Its magnitude is $\sqrt{(-\frac{1}{4})^2 + 2^2} = \sqrt{\frac{1}{16} + 4} = \sqrt{\frac{65}{16}} = \frac{\sqrt{65}}{4}$. * **C) $\frac{9}{2}$:** This is the sum of the velocity components, $\frac{1}{2} + 4 = \frac{9}{2}$, not the magnitude of the velocity vector.
11. A curve is defined by the parametric equations $x = t^2$ and $y = t^3 - 3t$. Find the value of $\frac{d^2y}{dx^2}$ at the point where $t=2$.
    Answer: $\frac{15}{32}$ First, find the first derivatives with respect to $t$: $$\frac{dx}{dt} = 2t$$ $$\frac{dy}{dt} = 3t^2 - 3$$ Next, find $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2 - 3}{2t} = \frac{3}{2}t - \frac{3}{2t}$$ Now, differentiate $\frac{dy}{dx}$ with respect to $t$: $$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2} + \frac{3}{2t^2}$$ Finally, use the formula for the second derivative: $$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac{3}{2} + \frac{3}{2t^2}}{2t} = \frac{3(t^2+1)}{2t^2} \cdot \frac{1}{2t} = \frac{3(t^2+1)}{4t^3}$$ Substitute $t=2$ into the expression for $\frac{d^2y}{dx^2}$: $$\frac{d^2y}{dx^2}\bigg|_{t=2} = \frac{3(2^2+1)}{4(2^3)} = \frac{3(5)}{4(8)} = \frac{15}{32}$$ * **A) $\frac{9}{32}$:** This result comes from a sign error when differentiating $-\frac{3}{2t}$, getting $-\frac{3}{2t^2}$ instead of $+\frac{3}{2t^2}$. This gives $\frac{d}{dt}(\frac{dy}{dx}) = \frac{3}{2} - \frac{3}{2t^2}$, which is $\frac{9}{8}$ at $t=2$. Then, dividing by $\frac{dx}{dt}=4$ gives $\frac{9}{32}$. * **C) $\frac{15}{8}$:** This is the value of $\frac{d}{dt}(\frac{dy}{dx})$ at $t=2$, but without the final step of dividing by $\frac{dx}{dt}$. * D) 6: This result comes from calculating the incorrect formula $\frac{d^2y/dt^2}{d^2x/dt^2}$. Here, $\frac{d^2y}{dt^2} = 6t$ and $\frac{d^2x}{dt^2} = 2$. At $t=2$, this would be $\frac{12}{2} = 6$.
12. Consider the curve given by $x = \sec(t)$ and $y = \tan(t)$. Find the gradient of the tangent at the point where $t = \pi/4$.
    Answer: $\sqrt{2}$ The gradient of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. First, find the derivatives with respect to $t$: $$\frac{dx}{dt} = \sec(t)\tan(t)$$ $$\frac{dy}{dt} = \sec^2(t)$$ Now, form the expression for $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{\sec^2(t)}{\sec(t)\tan(t)} = \frac{\sec(t)}{\tan(t)}$$ This can be simplified: $$\frac{dy}{dx} = \frac{1/\cos(t)}{\sin(t)/\cos(t)} = \frac{1}{\sin(t)} = \csc(t)$$ Evaluate the gradient at $t = \pi/4$: $$m = \csc(\pi/4) = \frac{1}{\sin(\pi/4)} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$$ * **A) $\frac{1}{\sqrt{2}}$:** This is the value of $\sin(\pi/4)$ or $\cos(\pi/4)$, not the final gradient. * B) 1: This is the value of $\tan(\pi/4)$. * D) 2: This is the value of $\sec^2(\pi/4)$.
13. Find the coordinates of the point(s) on the curve $x = t^2 - 1, y = t^3 - 3t$ where the tangent is parallel to the x-axis.
    Answer: (0, -2) and (0, 2) A tangent parallel to the x-axis is a horizontal tangent. This occurs when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$. First, find the derivatives: $$\frac{dy}{dt} = 3t^2 - 3$$ $$\frac{dx}{dt} = 2t$$ Set $\frac{dy}{dt} = 0$: $$3t^2 - 3 = 0 \implies 3(t^2 - 1) = 0 \implies t = \pm 1$$ Now, check $\frac{dx}{dt}$ at these values: * At $t=1$: $\frac{dx}{dt} = 2(1) = 2 \neq 0$. So there is a horizontal tangent. * At $t=-1$: $\frac{dx}{dt} = 2(-1) = -2 \neq 0$. So there is another horizontal tangent. Now find the coordinates for each value of $t$: * For $t=1$: $x = 1^2 - 1 = 0$, $y = 1^3 - 3(1) = -2$. Point: $(0, -2)$. * For $t=-1$: $x = (-1)^2 - 1 = 0$, $y = (-1)^3 - 3(-1) = -1 + 3 = 2$. Point: $(0, 2)$. So the points are $(0, -2)$ and $(0, 2)$. * B) (-1, 0): This is the point where the tangent is vertical ($\frac{dx}{dt}=0$ at $t=0$). * C) (2, 2) and (2, -2): These points correspond to $t=\pm\sqrt{3}$, which are the x-intercepts of the curve ($$y=t(t^2-3)=0$$). * D) (0, 2): This is only one of the two points with a horizontal tangent.
14. The curve defined by $x = t^2 - 1$ and $y = t^3 - t$ intersects itself at a point P. Find the coordinates of P.
    Answer: (0, 0) A self-intersection occurs at a point $(x, y)$ that is generated by two different parameter values, say $t_1$ and $t_2$, where $t_1 \neq t_2$. We need to solve the system of equations: $$x(t_1) = x(t_2) \implies t_1^2 - 1 = t_2^2 - 1$$ $$y(t_1) = y(t_2) \implies t_1^3 - t_1 = t_2^3 - t_2$$ From the first equation: $$t_1^2 = t_2^2 \implies t_1 = \pm t_2$$ Since $t_1 \neq t_2$, we must have $t_1 = -t_2$. Now substitute $t_2 = -t_1$ into the second equation: $$t_1^3 - t_1 = (-t_1)^3 - (-t_1)$$ $$t_1^3 - t_1 = -t_1^3 + t_1$$ $$2t_1^3 - 2t_1 = 0$$ $$2t_1(t_1^2 - 1) = 0$$ $$2t_1(t_1 - 1)(t_1 + 1) = 0$$ This gives possible values $t_1 = 0, 1, -1$. If $t_1 = 0$, then $t_2 = -0 = 0$, which means $t_1 = t_2$. This is not a self-intersection. If $t_1 = 1$, then $t_2 = -1$. These are distinct values. If $t_1 = -1$, then $t_2 = 1$. This is the same pair of values. So, the self-intersection occurs when $t = 1$ and $t = -1$. Let's find the coordinates of the point P by substituting either value into the parametric equations: For $t = 1$: $$x(1) = 1^2 - 1 = 0$$ $$y(1) = 1^3 - 1 = 0$$ For $t = -1$: $$x(-1) = (-1)^2 - 1 = 0$$ $$y(-1) = (-1)^3 - (-1) = -1 + 1 = 0$$ Both parameter values give the point $(0, 0)$. Therefore, the coordinates of P are $(0, 0)$. * A) (-1, 0): This is the point on the curve when $t=0$. * C) (0, 2): This point is not on the curve. It might result from an algebraic error. * D) (3, 0): This is the point on the related curve $y=t^3-4t$ when $t=2$.
15. The motion of a particle is described by $x(t) = 3t^2$ and $y(t) = 2t^3$. For what value of $t>0$ is the tangent to the path parallel to the line $y=2x$?
    Answer: 2 The line $y=2x$ has a gradient of 2. We need to find the value of $t$ for which the gradient of the tangent to the parametric curve, $\frac{dy}{dx}$, is equal to 2. First, find the derivatives with respect to $t$: $$\frac{dx}{dt} = 6t$$ $$\frac{dy}{dt} = 6t^2$$ Next, find the gradient $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t^2}{6t} = t$$ We want the gradient to be 2, so we set $\frac{dy}{dx} = 2$: $$t = 2$$ Since the question specifies $t>0$, this is the correct solution. * A) 1: This would mean the gradient is 1. * **B) $\sqrt{2}$:** This value might arise from a calculation error. * D) 4: This could arise if one equated $y/x = 2$, i.e., $2t^3 / (3t^2) = 2 \Rightarrow 2t/3 = 2 \Rightarrow t=3$. Or perhaps from another calculation error.

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