Calculus 1 Extra Credit Help?

Question

Hello, I desperately need help with this sheet. The professor gave out these 3 questions and they are way hard and a little beyond what he even taught. He said it was a challenge and he is even grading it in a weird way (the more people get it right the less points it is worth). Anyway if you can please take a look at it here: http://www.flickr.com/photo_zoom.gne?id=307392680&size=l

Please help in ANY WAY you can, hints tips anything will do.
Thank you again, this community in yahoo is wonderful.

Answer 1

Just an idea:

the two tangency points are related by the fact that the slope from point Pf to Pg is the slope of the line AND the derivatives at those two points.

You have 2 free variables: the x coords of Pf and Pg. From these the y coords are given by the function values and the slopes are given by the derivatives at those values. There should be enough relationships to allow solution.

Answer 2

Problem A

The coordinates where the tangent line intersects f(x) we will call (Xa, Ya)

The coordinates where the tangent line intersects g(x) we will call (Xb, Yb)

The derivative of f(x) = -2x
The derivative of g(x) = -2x + 8

We know that the slope of the tangent line at (Xa, Ya) must be the same as the slope of the tangent line at (Xb, Yb)

Thus -2Xa = -2Xb + 8

Xa = Xb - 4

The slope is also given by (Yb - Ya) / (Xb - Xa)

Ya = 1 - (Xa)^2 = 1 - (Xb - 4)^2 = 1 - (Xb^2 - 8Xb + 16)
= -(Xb)^2 +8Xb - 15
Yb = -(Xb)^2 + 8Xb - 12

So the slope is (Yb - Ya) / (Xb - Xa) = 3/4

-2Xa = 3/4 -----> Xa = -3/8
-2Xb + 8 = 3/4 ----> Xb = 29/8

Ya and Yb are determined by substituting Xa into 1-x^2
and Xb into -x^2 +8x - 12

Finding the equation of the tangent line should be easy at this point.


Problem B

Make a right triangle with one leg the 10 ft pole and the other leg X. The hypotenuse squared (S)^2 = 100 + X^2

Now take the derivative

2S dS/dt = 2X dX/dt
When X = 12, S = sqrt (244) = 15.62
We know dS/dt = 1.5 ft / sec

15.62 (1.5) = 12 * (dX/dt)
dX/dt = 1.9525 ft/sec


Problem C

The profit is given by

P = pq - dL = p (k* C^a* L^b) - dL

dP/dL = bpk * C^a * L^(b-1) - d

Set this equal to zero and solve for L

0 = bpk * C^a * L^(b-1) - d

d = bpk *C^a * L^(b-1)

.....d
------------ = L^(b-1)
bpk C^a

Now raise each side to the power 1/(b-1) to get your answer

Answer 3

Hint
The two parabolas have identicle shape. The only difference is the location. The common tangent is parallel to the line through the highest points in the parabolas.
That gets you the slope of the line.
The slope of the tangent plus the differential of each formula can be combined to determine where the tangent points are.
From those two points then you should be able to find the formula for the common tangent

Answer 4

This could be an approximation The sum is < Int (1,infinity)x^-3/2 dx = 2 so take In(1,n)x^-3/2 = 1.999 = 2(1-n^-1/2) so n^-1/2 =0.0005 and n= 4*10^6 Don´t take it to seriuos

Answer 5

good luck man, those are hard problems. Do-able if you know your stuff but not trivial, for sure.