Expert: Steve Holleran - 8/27/2007

Question
QUESTION: Hi, im in an undergraduate first year calculus course, just stuck on a question in a text book, wondering if you could point me in the direction of how to go about it, its quite lengthy so i apologize.

The effect of flight maneruvers on heat. The amount of work done by the heart's main pumping chamber is given by the equation:

W = PV + V.delta.v^2 / 2g,   where W is the work per unit time, P isd the average blood pressure, V is the volume of blood pumped out during the unit of time, delta is is weight density of the blood, v is the average velocity of the exiting blood, and g is the acceleration of gravity. When P,V,delta, and v remain constant, W becomes a function of g, and the equation takes the simplified form

W = a + b/g  (a,b constant).

Want to know how sensitive W is to apparent changes in g caused by flight maneuvers, and this depends on the inital change of g. As part of your investigation, you decide to compare the ffect on W of a given change (dg) on the moon, where g= 5.2 ft/sec^2, with the effect the same change (dg) would have on Earth, where g = 32 ft/sec^2. Use the simplified equation above to find the ratio of dW (moon) to dW (earth).

Wow, any help appreciated.
Cheers.
Scott.

ANSWER: Hi Scott,

Hey, they sure gave you a winner early on in the course!  Actually, its not as difficult as they make it sound.

Let's try this, to simplify matters:

Let's use the a + b/g form by letting PV = a, and
V * delta * v^2 / 2 = b.

Then the equation is:  W = a + b/g = a + b * g^-1

So, differentiating, the "a" goes out to zero, and

dW/dg = -bg^-2 = -b / g^2

So then dW/dg(moon) / dW/dg(earth) =

         [ -b/(5.2^2)] / [-b / (32^2)]

this becomes     1/(5.2^2) / 1/32^2 = 32^2 / 5.2^2

         = 1024 / 27.04 = 37.87

See, if all those other values are constants, its easier to deal with the simplified form, because they all either drop out by differentiating, or cancel out when we lose the "b's".

I hope I interpreted this correctly, and that it helps.

Steve Holleran

---------- FOLLOW-UP ----------

QUESTION: just wondering if you could clear up a little bit of confusion on another question, i've done most of it, just  having problems understanding the end of it.

A balloon is rising vertically above a level,straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bike moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bike and the balloon increasing 3 sec later?

so i've established dx/dt = 17 ft/s, dy/dt = 1 ft/s

by pythagoras, s^2=x^2 + y^2
ds/dt = 2x.dx/dt + 2y.dy/dt

when t=3... stuck from here

Answer
Hi Scott,

Good, you've got it set up right so far.

Now, at t = 3, y = 65 + 3(1 ft/sec) = 68 ft
         x = 3(17 ft/sec) = 51 ft

and since s^2 = x^2 + y^2, at t=3, s^2 = 51^2 + 68^2 and s comes out to 85 ft

Also, your differential equation should be:

     2 * s * ds/dt = 2 * x * dx/dt + 2 * y * dy/dt

Then all the 2's cancel , and you have
         
         s * ds/dt = x * dx/dt + y * dy/dt

Substituting all the known values,

         85 * ds/dt = 51 * 17 + 68 * 1

so          ds/dt = 11 ft/sec.

Steve Holleran