Question

can i get help on #60 and 110

Answer 1

To find our velocities, merely evaluate the function as \(t=1,5,10\)... for the accelerations, remember that acceleration is merely the rate of change of velocity, i.e. its derivative \(a=v'\).

Answer 2

For 110, recall that the surface area of a cube of edge length \(x\) is \(A=6x^2\), which should make sense given it has 6 square faces each with area \(x^2\). Now, we're told that our edges are expanding at a rate of 5 cm/s, which we write as \(\frac{dx}{dt}=5\); we're asked for the rate of change of the surface area, i.e. \(\frac{dA}{dt}\).
We need to take the derivative of our area function w.r.t. t:$$\frac{dA}{dt}=6(2x)\frac{dx}{dt}=12x\frac{dx}{dt}$$Using the information given, i.e. \(\frac{dx}{dt}=5\) and that our edges are 4.5 cm, \(x=4.5\), we reduce as follows:$$\frac{dA}{dt}=12(4.5)(5)=270$$... which means our surface area changes at a rate of 270 cm^2/s.

Answer 3

yay ! thank you! your awesome !! (: