# Let x(t)=t+1 and y(t)=t^2 a. Graph (x(t),y(t) for -1 \leq 1 \leq 4 b. Find \frac {dx}{dt}, \frac...

## Question:

Let {eq}x(t)=t+1 {/eq} and {eq}y(t)=t^2 {/eq}

a. Graph {eq}(x(t),y(t)) {/eq} for {eq}-1 \leq 1 \leq 4 {/eq}

b. Find {eq}\frac {dx}{dt}, \frac {dy}{dt} {/eq} the tangent slope {eq}\frac {dy}{dx} {/eq} and speed when {eq}t=1 {/eq} and {eq}t=4. {/eq}

## Parametric Functions

On Calculus 2D, if we have an x-axis position vs time and y-axis position vs time functions, we have a **paramteric function** where its parameter is the time.

## Answer and Explanation:

a.

To graph a parametric function we need to create a value table:

For {eq}t=-1 \\ \displaystyle x(-1)= (-1)+1 \; \Rightarrow \; x(-1)=0 \; \text{ & } \; y(-1)=(-1)^2 \; \Rightarrow \; y(-1)=1 \\ {/eq}

t | x | y |

-1 | 0 | 1 |

0 | 1 | 0 |

1 | 2 | 1 |

2 | 3 | 4 |

3 | 4 | 9 |

4 | 5 | 16 |

Graph

b.

Find {eq}\frac {dx}{dt}, \frac {dy}{dt} {/eq} the tangent slope {eq}\frac {dy}{dx} {/eq} and speed when {eq}t=1 {/eq} and {eq}t=4. {/eq}

So,

{eq}\displaystyle \frac{dx}{dt} \left[ x(t) \right] = \frac{dx}{dt} \left[ t+1 \right] \\ \displaystyle \frac{dx}{dt} \left[ t+1 \right] = 1 \\ {/eq}

Also

{eq}\displaystyle \frac{dy}{dt} \left[ y(t) \right] = \frac{dy}{dt} \left[ t^2 \right] \\ \displaystyle \frac{dy}{dt} \left[ t^2 \right] = 2t \\ {/eq}

And

{eq}\displaystyle \frac{dy}{dx}= \frac{ dy/dt }{ dx/dt } = \\ \displaystyle \frac{ dy/dt }{ dx/dt } = \frac{ 2t }{1 } \\ \displaystyle \frac{ dy/dt }{ dx/dt } =2t \\ {/eq}

Therefore, tangent slope at {eq}x=1 {/eq} is

{eq}\displaystyle 2(1)=2 {/eq}

And, tangent slope at {eq}x=4 {/eq} is

{eq}\displaystyle 2(4)=8 {/eq}

Speed:

If {eq}x(t)=t+1 {/eq} is a x-axis position vs time, then x-axis speed vs time function is {eq}\displaystyle v_x(t)=1 {/eq}

And, If {eq}y(t)=t^2 {/eq} is a y-axis position vs time, then y-axis speed vs time function is {eq}\displaystyle v_y(t)=2t {/eq}

thereofre: {eq}\displaystyle v_x(1)=1 \; \text{ & } \; v_x(1)=1 \\ {/eq}

And

{eq}\displaystyle v_y(1)=2(1) \; \Rightarrow \; v_y(1)=2 \\ \; \text{ & } \; \\ \displaystyle v_y(4)=2(4) \; \Rightarrow \; v_y(2)=8 \\ {/eq}

#### Learn more about this topic:

from Precalculus: High School

Chapter 24 / Lesson 3