Q 87 - Calculus

Question

In Example $4$ of Section $2.1$, we saw that a watermelon dropped from a height of $100$ feet will be $s\left(t\right)=-16t^2+100$ feet above the ground $t$ seconds after it is dropped. In that example, we approximated the velocity of such a watermelon at time $t=1$ by calculating a sequence of average rates of change. Now we can calculate this velocity exactly, using the definition of the derivative. Do so, and compare the exact answer to the approximation we found earlier. 

Step-by-Step Solution

Answer

By using derivative the exact velocity of the given function at $t=1$ is $-32$.

By comparing this exact velocity with the previously find velocity $-33.6$ of the same function at at the same time, we find that there is a difference of $-1.6$ between the both velocities.

Step 1. Given Information

We have given that height of water melon from ground after $t$ seconds is :-

$s\left(t\right)=-16t^2+100$.

We have to find the exact velocity of the function for $t=1$ by using derivative.

Also in previous section we calculate the velocity for the same function and for same time  by using continuous sequence as $-33.6$.

We have to compare this velocity with the exact velocity.

Step 2. Use derivative to find velocity

The given function is :-

$s\left(t\right)=-16t^2+100$.

We know that velocity of a function is first derivative of the function at a particular point.

We have to find the velocity of given function at $t=1$.

That is we have to find first derivative of the function at $t=1$.

We know that the derivative of a function $f\left(x\right)$ is defined as :-

$\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

So the exact velocity of the given function at $t=1$ is :-

$\underset{h\to 0}{\lim }\frac{s(1+h)-s\left(1\right)}{h}$

Put all the values, then we have :-

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{[-16{(1+h)}^2+100]-[-16{\left(1\right)}^2+100]}{h} \\ =\underset{h\to 0}{\lim }\frac{[-16(1+2h+h^2)+100]-[-16+100]}{h} \\ =\underset{h\to 0}{\lim }\frac{-16-32h-16h^2+16-100}{h} \\ =\underset{h\to 0}{\lim }\frac{-32h-16h^2}{h} \\ =\underset{h\to 0}{\lim }\frac{h(-32-16h)}{h} \\ =\underset{h\to 0}{\lim }-32-16h \\ =-32-16\left(0\right) \\ =-32-0 \\ =-32 \end{gathered}$$

So the exact velocity at $t=1$ is $-32$.

Step 3. Compare the velocities

In previous section by using continuous sequence, we calculate the velocity of the given function at $t=1$ is $-33.6$.

Here by using derivative, we calculate that the exact velocity of the function at $t=1$ is $-32$.

By comparing both velocities, we can see that there is a difference between both velocities is

 $$\begin{gathered} -33.6-(-32) \\ =-33.6+32 \\ =-1.6 \end{gathered}$$