Chapter 10.2, Problem 50E

$\text{(a)}$

To determine

To find: The magnitude of the velocity vector at $t=4$ .

Answer to Problem 50E

The answer: $\sqrt{1088}$

Explanation of Solution

Given information:

  $\begin{array}{c}x(t)=t^2-3\\ y(t)=\frac{2}{3}{t}^3\end{array}$

Calculation:

  $\begin{array}{c}\overrightarrow{v}(t)=\frac{dx(t)}{dt},\frac{dy(t)}{dt}\\ \overrightarrow{v}(t)=2t,2t^2\\ \overrightarrow{v}(4)=2\cdot 4,2\cdot 4^2\\ =(8,32)\\ |\overrightarrow{v}(4)|=\sqrt{8^2+{32}^2}\\ =\sqrt{1088}\end{array}$

Therefore, the required magnitude of the velocity vector at $t=4$ is $\sqrt{1088}$ .

$\text{(b)}$

To determine

To find: The total distance travelled by the particle from $t=0$

to $t=4$ .

Answer to Problem 50E

The answer: $\approx 46.062$

Explanation of Solution

Given information:

  $\begin{array}{c}x(t)=t^2-3\\ y(t)=\frac{2}{3}{t}^3\end{array}$

Calculation:

Know that,

Know that,

  

The distance that the particle from $t=0$ to

  $t=4$ is equal to:

  

To use the previous result,

  

According to the previous result distance that the particle from $t=0$ to

  $t=4$ is equal to:

  

Therefore, the required total distance travelled by the particle from $t=0$

to $t=4$ is $\approx 46.062$ .

$(c)$

To determine

To find: The value of $dy/dx$ as the function of $x$ .

Answer to Problem 50E

The answer: $\frac{dy}{dx}=\sqrt{x+3}$

Explanation of Solution

Given information:

  $\begin{array}{c}x(t)=t^2-3\\ y(t)=\frac{2}{3}{t}^3\end{array}$

Calculation:

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\\ =\frac{2t^2}{2t}\\ =t\end{array}$

To get in term $x$ , solve $x(t)$ for $t$ :

  $\begin{array}{c}x=t^2-3\\ \Rightarrow x+3=t^2\\ \Rightarrow t=\sqrt{x+3}\\ \frac{dy}{dx}=t\\ =\sqrt{x+3}\end{array}$

Therefore, the required value of $dy/dx$ as the function of $x$ is $\frac{dy}{dx}=\sqrt{x+3}$ .