A community for students.
Here's the question you clicked on:
 0 viewing
101Ryan101
 4 years ago
If the length of one a side of a cube is increasing at a rate of 4m/s how fast is the surface area of the cube increasing when the length is 16m.
101Ryan101
 4 years ago
If the length of one a side of a cube is increasing at a rate of 4m/s how fast is the surface area of the cube increasing when the length is 16m.

This Question is Closed

pk51
 4 years ago
Best ResponseYou've already chosen the best response.0This is related rates, right?

pk51
 4 years ago
Best ResponseYou've already chosen the best response.0I get 768 also. Hint. Relate the information, take a derivative, plug in what you know.

Alchemista
 4 years ago
Best ResponseYou've already chosen the best response.1Its surface area not volume: \[A = 6\cdot l^2\]\[\frac{dl}{dt} = 4\]\[\frac{dA}{dl} = 12 \cdot l \cdot \frac{dl}{dt}\]\[12 \cdot 16 \cdot 4 = 768\]

101Ryan101
 4 years ago
Best ResponseYou've already chosen the best response.0so if I want to know the rate of increase of volume it's increasing much faster than the surface area so ... (I'm a bit confused as to taking the derivative of V here...) I know dL/dt = 4m/s so for volume...when the length is 16m V = L^3 dV/dt =(3L^2)dL/dt = (3)(16)^2)(4) = 3072m/s I assume that's correct. But I'm not used to seeing an extra dL/dt. I guess the explanation for that is that it's a dependent variable.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.