Worksheet Parametric Equations Practice.tif - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

1. Asymptotes of $x = \frac{1}{t}$, $y = \frac{t}{t+1}$
* Vertical Asymptote: Look at $x$. As $t \to 0$, $x$ goes to infinity. Let's see what $y$ does when $t=0$. $y = \frac{0}{0+1} = 0$. So, as $x$ gets huge, $y$ stays near 0. This means $y=0$ is a horizontal asymptote.
* Horizontal/Other Asymptote: Look at the denominator of $y$. If $t = -1$, $y$ is undefined (goes to infinity). What is $x$ when $t=-1$? $x = \frac{1}{-1} = -1$. So, as $y$ gets huge, $x$ stays near -1. This means $x=-1$ is a vertical asymptote.
* Conclusion: The asymptotes are $x = -1$ and $y = 0$.
* Correct Option: c

2. Acceleration vector for $x = t^3 - 1$ and $y = t^4 - 2t^3$ at $t=1$
* Velocity ($v$): Take the derivative of position with respect to time ($t$).
* $v_x = \frac{dx}{dt} = 3t^2$
* $v_y = \frac{dy}{dt} = 4t^3 - 6t^2$
* Acceleration ($a$): Take the derivative of velocity.
* $a_x = \frac{d}{dt}(3t^2) = 6t$
* $a_y = \frac{d}{dt}(4t^3 - 6t^2) = 12t^2 - 12t$
* Plug in $t=1$:
* $a_x = 6(1) = 6$
* $a_y = 12(1)^2 - 12(1) = 12 - 12 = 0$
* Vector: $(6, 0)$
* Correct Option: d

3. Find $\frac{dy}{dx}$ at $t=1$ for $x=t^3-t$ and $y=\sqrt{3t+1}$
* Formula: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
* Find $\frac{dx}{dt}$:
* $x = t^3 - t \rightarrow \frac{dx}{dt} = 3t^2 - 1$
* At $t=1$: $3(1)^2 - 1 = 2$
* Find $\frac{dy}{dt}$:
* $y = (3t+1)^{1/2} \rightarrow \frac{dy}{dt} = \frac{1}{2}(3t+1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3t+1}}$
* At $t=1$: $\frac{3}{2\sqrt{3(1)+1}} = \frac{3}{2\sqrt{4}} = \frac{3}{2(2)} = \frac{3}{4}$
* Calculate Ratio:
* $\frac{dy}{dx} = \frac{3/4}{2} = \frac{3}{8}$
* Correct Option: b

4. Acceleration vector for $x=e^t+1$ and $y=\ln(2t+3)$
* Velocity:
* $v_x = e^t$
* $v_y = \frac{1}{2t+3} \cdot 2 = \frac{2}{2t+3}$
* Acceleration:
* $a_x = e^t$
* $a_y = \frac{d}{dt}[2(2t+3)^{-1}] = -2(2t+3)^{-2} \cdot 2 = \frac{-4}{(2t+3)^2}$
* Vector: $\left( e^t, \frac{-4}{(2t+3)^2} \right)$
* Correct Option: b

5. Find $\frac{d^2y}{dx^2}$ for $x=t^2+1$ and $y=t^3$
* First derivative $\frac{dy}{dx}$:
* $\frac{dx}{dt} = 2t$
* $\frac{dy}{dt} = 3t^2$
* $\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3}{2}t$
* Second derivative $\frac{d^2y}{dx^2}$:
* Formula: $\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$
* Numerator: $\frac{d}{dt}(\frac{3}{2}t) = \frac{3}{2}$
* Denominator: $\frac{dx}{dt} = 2t$
* Result: $\frac{3/2}{2t} = \frac{3}{4t}$
* Correct Option: a

6. Tangent line equation for $x=t^3+t$ and $y=t^2+2t^2$ at $t=1$
* Note: The problem says $y=t^2+2t^2$, which simplifies to $y=3t^2$.
* Point $(x,y)$ at $t=1$:
* $x = 1^3 + 1 = 2$
* $y = 3(1)^2 = 3$
* Point is $(2, 3)$.
* Slope ($m$) at $t=1$:
* $\frac{dx}{dt} = 3t^2 + 1 \rightarrow 3(1)+1 = 4$
* $\frac{dy}{dt} = 6t \rightarrow 6(1) = 6$
* $m = \frac{6}{4} = \frac{3}{2}$
* Equation:
* $y - y_1 = m(x - x_1)$
* $y - 3 = \frac{3}{2}(x - 2)$
* $y - 3 = \frac{3}{2}x - 3$
* $y = \frac{3}{2}x$
* Correct Option: a

7. Arc length for $x=t^2$ and $y=t$ from $t=0$ to $t=4$
* Formula: $L = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$
* Derivatives:
* $\frac{dx}{dt} = 2t$
* $\frac{dy}{dt} = 1$
* Plug into formula:
* $\sqrt{(2t)^2 + (1)^2} = \sqrt{4t^2 + 1}$
* Integral limits are 0 to 4.
* Expression: $\int_{0}^{4} \sqrt{4t^2 + 1} dt$
* Correct Option: c

8. Slope of tangent for $x=e^t$ and $y=te^{-t}$ where $x=3$
* We need the slope $\frac{dy}{dx}$ at the specific moment when $x=3$.
* First, find $t$ when $x=3$:
* $e^t = 3 \rightarrow t = \ln(3)$
* Find derivatives:
* $\frac{dx}{dt} = e^t$
* $\frac{dy}{dt}$ uses product rule on $t \cdot e^{-t}$:
* $(1)(e^{-t}) + (t)(-e^{-t}) = e^{-t}(1-t)$
* Calculate Slope $\frac{dy}{dx} = \frac{e^{-t}(1-t)}{e^t} = e^{-2t}(1-t)$
* Substitute $t = \ln(3)$:
* $e^{-2\ln(3)} = e^{\ln(3^{-2})} = 3^{-2} = \frac{1}{9}$
* $(1-t) = (1 - \ln(3))$
* Slope $= \frac{1}{9}(1 - \ln(3))$
* Calculate value:
* $\ln(3) \approx 1.0986$
* $1 - 1.0986 = -0.0986$
* $\frac{-0.0986}{9} \approx -0.0109$
* Rounding to 3 decimal places gives $-0.011$.
* Correct Option: d

9. Velocity vector at $t=2$ for position $\vec{r} = \langle \ln(t^2+2t), 2t^3 \rangle$
* Velocity is the derivative of position.
* x-component:
* $x = \ln(t^2+2t)$
* $v_x = \frac{1}{t^2+2t} \cdot (2t+2) = \frac{2(t+1)}{t(t+2)}$
* At $t=2$: $v_x = \frac{2(3)}{2(4)} = \frac{6}{8} = \frac{3}{4}$
* y-component:
* $y = 2t^3$
* $v_y = 6t^2$
* At $t=2$: $v_y = 6(2^2) = 6(4) = 24$
* Vector: $\langle \frac{3}{4}, 24 \rangle$
* Correct Option: a

Final Answer:
1. c
2. d
3. b
4. b
5. a
6. a
7. c
8. d
9. a