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# Need help with calculus problems?

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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

chilledz3non  Sep 18, 2014

#1
+92206
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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.

Melody  Sep 19, 2014
Sort:

#1
+92206
+10

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.

Melody  Sep 19, 2014
#2
+564
0

Thanks Melody!

chilledz3non  Sep 19, 2014

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