Given, the equation that represents the height of an object:
[tex]y(t)=100t-16t^2[/tex]
First, we will find the velocity of the object which is the first derivative of the height using the method of the limits
[tex]\frac{dy}{dt}=\lim_{h\to a}\frac{y(3+h)-y(3)}{(3+h)-(3)}[/tex]
We will find the value of the function y(t) when t = 3, and when t = 3+h
[tex]\begin{gathered} y(3+h)=100(3+h)-16(3+h)^2 \\ y(3+h)=300+300h-16(9+6h+h^2) \\ y(3+h)=300+300h-144-96h-16h^2 \\ y(3+h)=156+4h-16h^2 \\ y(3)=100(3)-16(3)^2=156 \end{gathered}[/tex]
Substitute y(3+h) and y(3) into the expression of the limit
[tex]\begin{gathered} \frac{dy}{dt}|_{t=3}=\lim_{h\to a}\frac{156+4h-16h^2-156}{3+h-3}=\lim_{h\to a}\frac{4h-16h^2}{h} \\ \\ \frac{dy}{dt}|_{t=3}=\lim_{h\to a}(4-16h) \end{gathered}[/tex]
Where a = 0
d) compute the instantaneous velocity at t = 3
[tex]\frac{dy}{dt}|_{t=3}=100-16*2*3=4[/tex]
So, the answer will be:
[tex]\begin{gathered} \frac{dy}{dt}|_{t=3}=\lim_{h\to a}(4-16h) \\ a=0 \\ \\ \frac{dy}{dt}|_{t=3}=4\text{ ft/sec} \end{gathered}[/tex]
