average rate, velocity type questions

[Guest]

It is estimated that t years from now, the amount of waste accumulated, Q, in tonnes, will be Q(t)= 10^4(t^2+15t+70) tonnes, t is greater than or equal to 1, t is also less than or equal to 10.

b) What will the average rate of change of this quauntity over the next three years?

Is this how you're suppose to figure out the answer to this? avg rate=lim h --> 0 [f(x+h)-f(x)]/x

then sub in 3 for 'x' or is this not the correct thing you do? cuz i tried doing it this way but got the wrong answer, however I may have had just made a mistake.

c) What is the present rate of change of this quantity?<--I don't know what I'm suppose to do for this question.

and d) When will the rate of change reach 3.0 * 10^5 tonnes per year?


4.The height (in metres) that an object has fallen from a height of 180 m is given by the position function s(t)= -5t^2+180, where t is greater than or equal to 0 and t is in seconds.

c) At what velovity will the object hit the ground <--( similar to the previous question I think)

THANKS SO MUCH FOR THE HELP AND TAKING YOUR TIME TO RESPOND =)

 

[soroban]

Hello, bittersweet!

]It is estimated that t years from now, the amount of waste accumulated, Q, in tonnes, will be:
\(\displaystyle \;\;\;\;Q(t)\:=\: 10^4(t^2\,+\,15t\,+\,70)\) tonnes, \(\displaystyle \,1\,\leq\, t\,\leq 10\).

b) What will the average rate of change of this quantity over the next three years?

For average rate of change, we don't need Calculus . . . just arithmetic.

Right now \(\displaystyle (t\,=\,0):\;Q(0)\:=\:10^4(0^2\,+\,15\cdot0\,+\,70)\:=\:70\cdot10^4\) tonnes.

In 3 years \(\displaystyle (t\,=\,3):\;Q(3)\:=\:10^4(3^2\,+\,15\cdot3\,+\,70)\:=\:124\cdot10^4\) tonnes.

The change is: \(\displaystyle \,124\cdot10^4\,-\,70\cdot10^4\:=\:54\cdot10^4\:=\:540,000\) tonnes.

Over 3 years, the average change is: \(\displaystyle \,\frac{540,000}{3}\:=\:180,000\) tonnes per year.

c) What is the present rate of change of this quantity?

It asks for the rate of change right now (instantaneous rate of change).

We're expected to know that the derivative gives us this rate of change.

The derivative is: \(\displaystyle \:Q'(t)\:=\:10^4(2t\,+\,15)\)

Therefore: \(\displaystyle \:Q'(0)\;=\;10^4(2\cdot0\,+\,15)\;=\;150,000\) tonnes per year.

d) When will the rate of change reach \(\displaystyle 3\cdot10^5\) tonnes per year?

When will \(\displaystyle \,Q'(x)\;=\;3\cdot10^5\) ?

\(\displaystyle \;\;10^4(2t\,+\,15)\:=\:3\cdot10^5\;\;\Rightarrow\;\;2t\,+\,15\:=\:30\;\;\Rightarrow\;\;t\,=\,7.5\) years.

 

[Guest]

thanks, but is there another thing you could use instead of using the derivative? We havn't learned about that yet.

 

[stapel]

thanks, but is there another thing you could use instead of using the derivative? We havn't learned about that yet.

Then what calculus technique are you supposed to be using?

Thank you.

Eliz.

 

[Guest]

I'm not sure, but I guess if you can only use der. for that question then that type of question won't be on the test tomorrow.

oh this is what my teacher has said on chatt..

For pg 115 #6c - Find the instantaneous velocity at t=0 since the function is based on t yrs from now, now would be time 0. so evaluate ..(she has put down the equation so I guess Ill try to work that out