Help! Calculus!?

Question

The length and width of a box are increasing in such a way that the length is always thrice the width, while the height of the box is fixed at 6 inches.
a. How fast is the volume of the box increasing with respect to its width at the instant when the width is 3/2 inches?
b. Use differentials to approximate the increase in the volume of the box when the width increases from 2 inches to 2.01 inches.

2. The position function of a particle moving along the line is given by s=5t^2-(t^3/3)-21t+1, where s feet is the directed distance of the particle from the origin at time t seconds. At what time interval/s is the particle moving to the left and slowing down.

how about letter b?

Answer 1

1.
Find the equation for the volume of the box; namely V = l*w*h.
Since l = 3w, and h = 6, the volume becomes V = 18 w^2
Now it should be easy to solve. If you need more help, add a note.
In case you did want the help, though, dV/dw = 36 w, so at w = 3/2, dV/dw = 54 in^2.
At w = 2, dV/dw = 72, so multiply that by (2.01 - 2), and you get .72 in^3 for your approximation of the increase.
Compared to the actual change, .7218, this estimate is not too bad.

2.
If it is moving left, it is moving towards the negative end, so the first derivative is negative. If it is slowing down while doing so, it is accelerating towards a speed of 0, which means an increase in speed, or a positive acceleration.
Find where the first derivative is negative and the second derivative is positive.

This would be when -t^2 + 10t - 21 < 0 and t < 5
The first inequality is solved when t^2 - 10 t + 21 > 0, or when t < 3 or t > 7.
Therefore, it is slowing and moving left during the interval (-infty, 3)

Answer 2

let the length be L,width be W,height be H,volume be V.
As given,L=3W and H=6.
Now,for a box,V=LxWXH.here,H=6.
Therefore,V=LXWX6.We need to find dV/dW.
But,L=3W.So,V=(3W)XWX6
or,V=18X(W^2).
Differentiating both sides w.r.t W,
dV/dW=18Xd(W^2)/dW
or,dV/dW=18X(2XW)
or,dV/dW=36X(W)
or,dV/dW=36X(3/2)
or,dV/dW=54 sq. inches(Ans)
Now,as width increases from 2 inches to 2.01 inches,so dW(increment in width)=.01 inch.
As,dV/dW=54 and dW=.01 inch,
so,dV=54X.01
or,dV=.54 cubic inches.(Ans)

2)for this, as far as I can understand,it means at what time velocity towards the right becomes zero.
Accln. can be obtained by differentiating s twice w.r.t time(t).
So,differentiating s once,it gives velocity,V=10t-(t^2)-21.
differentiating again,acceleration,a=10-2t.
When V=0,the value of t comes out to be either 3 or 7 secs.
But also as it is slowing down towards right,so 10-2t<0 or t>5.so ,t=3 secs(Ans)

Answer 3

42

Answer 4

For question 2, v =ds/dt=5d[t^2]/dx-(1/3)*d[t^3]/dx-21d[t]/dx+0
=10t-t^2-21
since turning point is when v=o, we let v to be 0
10t-t^2-21=0
t^2-10t+21=0
(t-7)(t-3)=0
t=3sec or 7sec

first turn of direction is made at t=3sec, 2nd at t=7sec