+564 Okay so this is a two part question. I solved the first part but need help solving the second.
The first part goes as:
The height of a certain object is given by $$h(t)=t^2 - 1$$. Find the average velocity over the time period [2,5].
Somebody helped me with a similar problem like this before (That awesome person knows who they are ;) ). The answer I got for the one was and average velocity 7
The second part of the question is:
In the previous problem, find the exact value of the instantaneous velocity at t=2. One way is to evaluate the limit with a=2
+118687 Hi chilledz3non,
Well the exact instantaneous velocity is given by h'(t) = 2t so when t=2 velocity=2*2=4 units per time.
Now perhaps you are meant to do it from first principles so lets think about that.
Average velocity between t=some value, I'll call it t1 and t=2 would be $$\frac{f(t_1)-f(2)}{t_1-2}$$
The instantaneous velocity would be given by this as t1 tends to 2
That is
$$\\\lim_{t_1\rightarrow2}\;\frac{f(t_1)-f(2)}{t_1-2}\\\\
f(t_1)=(t_1)^2-1\\
f(2)=2^2-1=3\\\\
$Instantaneous velocity at t=2 is $\lim_{t_1\rightarrow2}\;\frac{f(t_1)-f(2)}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{(t_1)^2-1-3}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{(t_1)^2-4}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{((t_1)-2)((t_1)+2)}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{(t_1-2)(t_1+2)}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;(t_1+2)\\\\
=4$$
Maybe Alan or Heureka could tell me why my limits are not displaying properly. 
+118687 Hi chilledz3non,
Well the exact instantaneous velocity is given by h'(t) = 2t so when t=2 velocity=2*2=4 units per time.
Now perhaps you are meant to do it from first principles so lets think about that.
Average velocity between t=some value, I'll call it t1 and t=2 would be $$\frac{f(t_1)-f(2)}{t_1-2}$$
The instantaneous velocity would be given by this as t1 tends to 2
That is
$$\\\lim_{t_1\rightarrow2}\;\frac{f(t_1)-f(2)}{t_1-2}\\\\
f(t_1)=(t_1)^2-1\\
f(2)=2^2-1=3\\\\
$Instantaneous velocity at t=2 is $\lim_{t_1\rightarrow2}\;\frac{f(t_1)-f(2)}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{(t_1)^2-1-3}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{(t_1)^2-4}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{((t_1)-2)((t_1)+2)}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;\frac{(t_1-2)(t_1+2)}{t_1-2}\\\\
=\lim_{t_1\rightarrow2}\;(t_1+2)\\\\
=4$$
Maybe Alan or Heureka could tell me why my limits are not displaying properly.
