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Q: Some calculus questions just for fun! ( Answered 5 out of 5 stars,   0 Comments )
Subject: Some calculus questions just for fun!
Category: Reference, Education and News > Homework Help
Asked by: alexinor-ga
List Price: $25.00
Posted: 02 May 2005 12:56 PDT
Expires: 01 Jun 2005 12:56 PDT
Question ID: 516890
Hi!  I?m trying to teach myself some calculus for fun and was
wondering if you could help walk me through a few problems.  Anyone is
welcome to offer their input.  Thanks!

How would I find dy/dx by implicit differentiation for:

1. (6x^2) + (8xy) + (y^2) = 6

2.  x + ln[y] = (x^2)(y^3)

3. (e^((x^2)(y))) = 5 x + 4 y + 2

Suppose x and y are functions of t.  How would I evaluate dy/dt for:

4. xy ? 5x + 2y^3 = -70

5. xe^y = 3 + Ln[x]

Also, some problems that I?m trying to figure out:

6.  ?A 13 ft ladder is leaning against the wall.  When the top is 5 ft
above the ground it slips down the wall at a rate of 2 ft/sec. 
Determine how fast the foot will be moving away from the wall when the
top is 5 ft above the ground.?

7.  ?The cost to produce x units per month is given by C = .1x^2 +
10000.  Determine rate of change of cost per month (that is (dC/dt)
when the production is changing at the rate of 10 units per month
(that is dx/dt = 10) and that the production level at that instant is
100 units (that is x = 100).?

8. ?An orchard is increasing its production of apples at the rate of
50 boxes a day (that is dq/dt = 50).  All boxes produced can be sold. 
The daily demand function is given by p=50-(q/200) where q is the
number of boxes produced (and sold) and p is price in dollars.  When
the daily product is 200 boxes determine the rate of change of revenue
per day (that is find dR/dt).?

9. ?A spherical balloon is to be deflated so that its radius decreases
at a constant rate of 15 cm per minute.  Determine the rate per minute
the air is removed when the radius of the balloon is 9 cm.?
Subject: Re: Some calculus questions just for fun!
Answered By: elmarto-ga on 02 May 2005 16:42 PDT
Rated:5 out of 5 stars
Hi alexinor!
Here are the answers to your questions.

First of all, you may want to visit the following link, which has a
tutorial for implicit differentiation.

Implicit Differentiation

Question 1

The procedure to solve questions 1-3 is exactly the same. Basically,
we must find the derivative with respect to x of both sides of the
equality. Thus, in this question, we start by finding:

d [(6x^2) + (8xy) + (y^2)]
--------------------------   (*)


d [6]
-----  (**)

Let's start with (*). The derivative with respect to x of this function is:

6*2x + 8(x(dy/dx) + y) + 2y(dy/dx)

I assume here that you're already familiar with the multiplication
rule and the chain rule for derivatives. If you don't understand how I
arrived to this result, please request clarification; although it
should be clear if you are familiar with those concepts.

Now, clearly (**) gives zero, as the derivative of a constant is
always zero. So we now have the equation:

   (6x^2) + (8xy) + (y^2) = 6

d [(6x^2) + (8xy) + (y^2)]    d[6]
-------------------------- = ------
          dx                   dx

6*2x + 8(x(dy/dx) + y) + 2y(dy/dx) = 0

So now we simply isolte (dy/dx) in this equation and that will be the answer:

12x + 8y + x(dy/dx) + 2y(dy/dx) = 0

(dy/dx)(x + 2y) = -12x -8y

dy/dx = (-12x -8y)/(x + 2y)

We've now found dy/dx. As you can see, the procedure is quite simple.
Let's repeat it for questions 2 and 3.

Question 2

We have the equation

x + ln[y] = (x^2)(y^3)

Again, we find the derivative of both sides with respect to x to get:

1 + (1/y)(dy/dx) = (2x)(y^3) + (x^2)(3y^2)(dy/dx)

And again, we isolte dy/dx from this equation.

(dy/dx)(1/y - (x^2)(3y^2)) = (2x)(y^3) - 1

dy/dx = [(2x)(y^3) - 1]/[1/y - (x^2)(3y^2)]

Question 3

Same procedure here:

e^(y*x^2) = 5 x + 4 y + 2

[e^(y*x^2)][(dy/dx)(x^2) + y*2x] = 5 + 4(dy/dx)

(dy/dx)*(x^2)*e^(y*x^2) + y*2x*e^(y*x^2) = 5 + 4(dy/dx)

(dy/dx)*[(x^2)*e^(y*x^2) - 4] = 5 - y*2x*e^(y*x^2)

dy/dx = [5 - y*2x*e^(y*x^2)] / [(x^2)*e^(y*x^2) - 4]

Question 4

This question and the following are a bit simpler than the previous
ones. For simplicity, I'll use the notation x' instead of dx/dt, and
y' instead of dy/dt. The idea here is, once again, to find the
derivative of both sides of the equality, this time with respect t. So
in this question we have:

xy ? 5x + 2y^3 = -70

(recall that x and y are functions x(t) and y(t) )

We take the derivative of both sides with respect t:

x'*y + x*y' - 5x' + (6y^2)y' = 0

And now we simply isolate y':

x*y' + (6y^2)y' = 5x' - x'*y

y'(x + 6y^2) = 5x' - x'*y

y' = [5x' - x'*y]/[x + 6y^2]

Thus, we've found dy/dt. The procedure is of course the same for question 5

Question 5

The equation is:

xe^y = 3 + Ln[x]

We find the derivative with respect to t:

x'*(e^y) + x*(e^y)*y' = 0 + (1/x)*x'

And we isolte y':

x*(e^y)*y' = (1/x)*x' - x'*(e^y)

y' = [(1/x)*x' - x'*(e^y)]/[x*(e^y)]

Question 6

The following questions are an application of the procedure we used in
questions 4 and 5.

In this case, think of the ladder, the wall, and the floor as forming
a right triangle. The right angle is, of course, the angle between the
floor and the wall. The ladder would be the hypotenuse (13 ft long).
The top being 5 ft above the floor means that the height of this
triangle is 5 ft:

  | \
  |  \
  |   \  13
h |    \
  |     \
  |      \
  |       \

We know that h, initially, is 5 ft, although it changes as the ladder
slips away. We don't know y, the distance between the foot of the
ladder and the wall. Notice that this question asks us to find the
rate at which y changes over time. So basically, this problem is
asking us for dy/dt.

Now we need and equation for y. Since this is a right triangle, we use
Pythagoras Theorem:

h^2 + y^2 = 13^2 = 169

(so if h=5, then y=12)

We know that both h and y are a function of time t, as the ladder
slips away, but this equation always holds. Therefore, as in questions
4 and 5, we find the derivative with respect to t:

2h*(dh/dt) + 2y*(dy/dt) = 0

dy/dt = (dh/dt)*(-2h/2y)

dy/dt = -(h/y)*(dh/dt)

Now we need to find what values to plug in for h, y and dh/dt. We want
to know the speed at which the foot moves away when the top is 5ft
above the ground, so h=5. We also know that, at this point, the top of
the ladder is slipping away at 2 ft/sec. Therefore, dh/dt=-2 (this
side is getting smaller, hence the negative sign). Finally, we need to
know y at this point. We've already seen that when h=5, then y=12. So,
we plug these values:

dy/dt = -(h/y)*(dh/dt) = -(5/12)*(-2) = 5/6

Therefore, the foot is moving at 5/6 ft/sec away from the wall.

Question 7

In order to solve this problem, we first find dC/dt:

C = .1x^2 + 10000

dC/dt = 0.2x*(dx/dt)

Since we know that dx/dt = 10, and that x = 100, we pluf these values to get:

dC/dt = 0.2*100*10 = 200

Therefore, the cost is increasing at $200 per month.

Question 8

Here we must first find the revenue (R) function. Revenue is simply
price times quantity sold. We already know that p=50-(q/200).
Therefore, the revenue function is:

R = p*q = (50 - q/200)*q

R = 50q - (q^2)/200

Now we find dR/dt in the usual way:

dR/dt = 50dq/dt - (1/200)(2q)(dq/dt)

Now, since we know that dq/dt = 50, and q = 200, we can find the answer:

dR/dt = 50*50 - (1/200)(2*200)(50) = 2400

Therefore, revenue is increasing at $2,400 per day.

Question 9

In order to answer this, we need a measure of the quantity of air
inside the balloon, because the question asks us to find the rate at
which the air is removed from the balloon. I'll measure it in terms of
volume. For the sake of simplicity, I'll assume that the volume of air
is equal to the volume of the balloon. Since the balloon is a sphere,
it's volume is determined by the follinwg formula:

V = (4/3)*pi*r^3
(r is for radius)

Since we want to know the rate at which V changes over time, we have
to find the derivative of V with respect to t. We get:

dV/dt = (4/3)(pi)(3r^2)(dr/dt)

We already know that dr/dt is -15, because the radius is decreasing at
15 cm per second. Since we want to find the rate when the radius is 9
cm, we replace r=9 in the previous equation to get:

dV/dt = (4/3)(pi)(3*9^2)(-15) = -15268.14

Therefore, air is being removed from this balloon at a rate of
15268.14 cm^3 when its radius is 9 cm.

Google search terms
implicit differentiation

I hope this helps! If you have any questions regarding my answer,
please don't hesitate to request a clarification. Otherwise I await
your rating and final comments.

Best wishes!
alexinor-ga rated this answer:5 out of 5 stars
You do a great job outlining everything.  The online links were very
helpful.  Thank you so much!

There are no comments at this time.

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