@Michele_Laino
it is none of the above right?
@Michele_Laino
please you have to apply the subsequent rule: \[\frac{ d (x ^{n}) }{ dx }=n*x ^{(n-1)}, n \in \mathbb{N} \] nevertheless n can be any real number
for example: \[\frac{ dx ^{3} }{ dx }=3x ^{2}\] so \[\frac{ d4x ^{3} }{ dx }=12x^{2}\] that's the first term of your answer, please continue that calculus
for example the derivative of the second term is: \[\frac{ d(-3x ^{2}) }{ dx }=-6x\] now, try to calculate the derivative of the third term of your polynomial, please
9 @Michele_Laino
12x^2-6x+9 right?
which would be none of the above(:
perfect! your answer is right, namely noone of the answers listed
I'm ready!
iis it -24 in^2/hour?
the rate of 1.5 inches/hour is referring to an edge of your cube?
noo how fast its melting
I think the formulae are these: \[\frac{ dl }{ dt }=-1.5\] so we can write: \[l(t)=-1.5*t+l _{0}\] where l_0 is the initial length furthermore, we have: \[A(t)=[l(t)]^{2}\] so: \[\frac{ dA }{ dt }=2*l(t)*\frac{ dl }{ dt }=2*(-1.5t+l _{0})*(-1.5)\]
do you know what is the initial length of your cube?
side length of 2 inches?
I think that it must be greater than 2 inches
don't worry I got the solution: in order to find our rate you havet calculate this \[\frac{ dS }{ dt }=2*2*(-1.5)=-6\] we don't needto know what is the initial lenftg of a sidof ur cube, because at the time when the side of o cube has a lenfgth equalss to 2, we can write: l_0-1.5t=2, without need to know what is l_0
@mondona
and then what do we do?
rate of changing (decreasing) area is 6 inches^2/hour
so it would be negative right?
that's right, because during melting the dimensions of our ice cube are decreasing, so also area does that
6 wouldnt be right..
I think that your problem calls surface the total surface of our cube, whereas my calculus is refferring to the surface of only one face of the cube. So in order to get the right answer, you have to multiply -6 by 6, and you will get: \[\frac{ dS }{ dt }=6*(-6)=-36\] inches/hour since, of course a cube has six faces
so the third option is the right answer!
ohhh thats what i thought i was thinking about that
thank you !
thank you!
Join our real-time social learning platform and learn together with your friends!