Chapter 1, Problem 14REAE

Solution

To determine how far did the flea jump and the maximum height of the jump.

The flea jumped for a distance of $33\text{cm}$ and for a maximum height of about $19.87\text{cm}$ .

Given:

Given the equation modelling the jump of the flea:

  $y=-0.073x\left(x-33\right)$.

Concept Used:

Second derivative test:

Given a real valued function $f\left(x\right)$ in one variable $x$ .

The function will have a maximum at $x=a$ if $f^{\text{'}}\left(a\right)=0$ and $f^{\text{'}\text{'}}\left(a\right)<0$ .

The function will have a minimum at $x=a$ if $f^{\text{'}}\left(a\right)=0$ and $f^{\text{'}\text{'}}\left(a\right)>0$ .

The test fails when $f^{\text{'}\text{'}}\left(a\right)=0$ .

Calculation:

Determine the distance up to which the flea jumped:

Since $y$ represents the height, $y$ is always non-negative.

Also, the distance of the jump is the value of $x$ so that the height is zero.

That is, the distance of the jump is value of $x$ so that $y=0$ .

Substitute $y=0$ in the equation $y=-0.073x\left(x-33\right)$:

  $0=-0.073x\left(x-33\right)\Rightarrow x\left(x-33\right)=0$ .

Now, $x\left(x-33\right)=0\Rightarrow x=0\text{or}x\text{=33}$ .

But, the flea must have jumped for some distance.

That is, $x$ is strictly positive.

Then, it follows that $x=33$ .

That is, the flea jumped for a distance of $33\text{cm}$ .

Find the maximum height of the jump:

Find the first derivative $y^{\text{'}}$ :

  $\begin{array}{l}{y}^{\text{'}}=\left(-0.073x\right)\frac{d}{dx}\left(x-33\right)+\left(x-33\right)\frac{d}{dx}\left(-0.073x\right)\\ y^{\text{'}}=\left(-0.073x\right)\cdot 1+\left(x-33\right)\left(-0.073\right)\\ y^{\text{'}}=-0.073x-0.073x+2.409\\ y^{\text{'}}=-0.146x+2.409\end{array}$.

Equate $y^{\text{'}}$ with zero and solve the equation for $x$ :

  $\begin{array}{c}-0.146x+2.409=0\\ 2.409=0.146x\\ \frac{2.409}{0.146}=x\end{array}$ .

Thus, it is found that $x=\frac{2.409}{0.146}$ .

Now, find the second derivative $y^{\text{'}\text{'}}$ :

  $y^{\text{'}\text{'}}=-0.146$ .

That is, the second derivative is negative everywhere.

In particular the second derivative $y^{\text{'}\text{'}}$ is negative at $x=\frac{2.409}{0.146}$ .

Then, by the second derivative test, $x=\frac{2.409}{0.146}$ is a point corresponding to the maximum value of $y$ .

Find the value of $y$ corresponding to $x=\frac{2.409}{0.146}$ :

  $\begin{array}{l}y=-0.073\cdot \frac{2.409}{0.146}\left(\frac{2.409}{0.146}-33\right)\\ y\approx 19.87\end{array}$.

Thus, the maximum value of $y$ is about $19.87$ .

That is, the maximum height of the jump is about:

  $19.87\text{cm}$ .

Conclusion:

The flea jumped for a distance of $33\text{cm}$ and for a maximum height of about $19.87\text{cm}$ .