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 2 years ago
Dear, Mathtalent!
The slope of the tangent line to the graph of the function y=3x^3 at the point (381) is limx=>3 (3x^3−81)/(x3.) By trying values of x near 3, find the slope of the tangent line
 2 years ago
Dear, Mathtalent! The slope of the tangent line to the graph of the function y=3x^3 at the point (381) is limx=>3 (3x^3−81)/(x3.) By trying values of x near 3, find the slope of the tangent line

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ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Take you calculator and try x=2.9, 2.99, 2.999, to get a good idea where this is going to...

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0ok,Thank you. I have a calculator with me.. but why x=2.9??

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0oh i get it beacuse its near 3....

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Well, you want to know the limit for x > 3. You are not supposed to do all kinds of math trickery to do this, but just try a few values near 3. So 2.9, or 3.1 and then even closer: 2.99 or 3.01, or any other value, as long as it is "near" 3...

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Once I gt all answers from equation. what shall I do?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Can you guess what the number is that you are getting closer and closer to?

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0hmmmm I will try put 2.9 , 2.999 in the equation first Please hold!

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, So I got 78.33 for x=2.9 and 80.7303 for x=2.99... I guess it gets closer to 81?

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0cuze at the point (3, 81)

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1It does, so it's a safe bet 81 is the slope of the tangent line!

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1(3,81) begin the point on the graph is in itself not of importance. By coincidence, the ycoordinate is also 81.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Omg, thats prety cool! thank you very much :) are you faimilar with Tagent Velocity:

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1You mean the velocity of a moving point along a curve is the tangent vector? Or is it a book? (havent't got a clue here..)

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1I could be familiar with the stuff in the book ;)

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0cool. The displacement (in meters) of a particle moving in a straight line is given by s=3t^3 where t is measured in seconds. 1)Find the average velocity of the particle over the time interval [7,10]. 2)Find the (instantaneous) velocity of the particle when t=7

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Do you know the difference between average velocity and instantaneous velocity?

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0mmm No. I have no idea. av velo =displacement/ time elapsed?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1It is, so that one is easy. Just calculate (s(10s(7))/(107).

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1For instanteneous velocity, you'll have to do the limitstuff as in the first problem. Can you write down the limit?

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0so s(10)s(7)/3? for the fist one?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, because s(10s(7) is the distance you've travelled.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0and jusy plug s=3t^3 in s?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Don't get what you're saying there...

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1s(10)=3*10³=3000, s(7)=3*7³=1029 (30001029)/3=657 m/s (you may not be using meters, but feet...)

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0For instanteneous velocity, I have to do limit thing? x=>7?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Write down the limit:\[v(7)=\lim_{t \rightarrow 7}\frac{ s(t)s(7) }{ t7 }\] and try numbers near 7.

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0OK, I tired to 6.99. s(6,99)=(6.99)^3*3=1024.59 S(7)= 1029 1024.591029/10297?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1You have to divide by t7, which is 6.997=0.01

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1441 is good. Just use brackets...

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Oh, now I get it ;) If you set t=6.999 you'll be even closer to 441, so that will be the answer

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Oh! So limit is how you get closer to x or function?

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, it is the value of a formula as you get closer and closer to a certain number. Only, because of the special kind of formula, it is not possible to try that number directly. As in the formula above: if you want the speed at t=7, you can't just put t=7 and get the right answer. If you try it, you'll get 0/0, which is undefined.

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1It is possible however, to do more advanced calculations with the formula to get the limit. It is called differentiation... Once you know how to do that, you don't need to try all these cumbersome numbers "near" to a certain value!

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Wow, thank you so much for your detailed explaination. It is really fun doing math with you!

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.1Thanks for the compliment! I really appreciate it.
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