Difficult calculus word problem below. Would anyone dare to solve it?

Question

The upper end of a straight ladder is leaning against the vertical side of a building, with its lower end on a horizontal sidewalk. To the consternation of the painter standing on it, the upper end is slipping down the wall at a constant rate of 2 feet/sec. The lower end is, therefore, sliding along the sidewalk. Having had a course in the Calculus, the painter is, of course, interested in how fast the lower end is moving. If the ladder is 29 feet long, how fast is the lower end moving when the upper end is 20 feet above the ground?

possible solutions...

a. 8/3 feet/sec.
b. 3/2 feet/sec
c. 48/7 feet/sec.
d. 7/12 feet/sec.
e. 15/4 feet/sec.
f. 16/15 feet/sec.
g. none of these

Answer 1

Pay no attention to Jess's analysis. Let's use 400 years of mathematical thinking instead.

The hypotenuse of the triangle is of fixed length, 29, of course given by a^2 + b^2 = c^2.

Let a be the length along the ground, and b the length along the building. Thus,

a = sqrt(c^2 - b^2)

We want to find da/dt, which is equal to

da/dt = (da/db) * (db/dt)

We are given db/dt as -2 ft./sec. Let's find da/db, given that c is a constant.

da/db = (1/2)(c^2 - b^2)^(-1/2) * (-2b)
= -b / sqrt(c^2 - b^2)

Now we can put in the numbers. We are interested in the situation when b = 20.

da/db = -20 / sqrt(29^2 - 20^2)

The negative value means that as b gets smaller, a gets larger.
Since we know that db/dt = -2, we have

da/dt = 40 / sqrt(29^2 - 20^2) = 40 / 21

The answer is g, none.

Three of us arrived at the same correct answer using the same analysis. Choose as best answer whichever explanation is clearest to you.

Answer 2

Difficult Calculus Problems

Answer 3

Draw the picture!

The ladder forms a right triangle with the wall and the ground. The hypotenuse is a constant=29 feet. The apex of the right angle is at the origin.

The height =y
So, by the Pythagorean Thm., the base is x = (29^2 - y^2)^0.5

We are looking for the rate of displacement of x , or dx/dt

So, since we know x in terms of y, we can use that expression to find the first derivative by the chain rule::

dx/dt= d[(29^2 - y^2)^0.5] /dt

dx/dt= [0.5 * (29^2 - y^2) ^(-0.5)] * (-2y) * dy/dt

We are given that dy/dt = -2 feet per second and y= 20 feet, so substitute those values, and plug and chug:

Thus, dx/dt= 40/21 feet per second when y=20.

Answer 4

Call the edge length x. V = LWH. L and W = 45 - 2x; H = x. Therefore V = x(45-2x)^2 = x(2025 -180x + 4x^2) = 2025x -180x^2 + 4x^3. We want to maximize V, so first we have to take the derivative of it. V' = 12x^2 - 360x + 2025. Now, we set this equal to zero, making it a quadratic equation. x = (360 +/- sqrt(360^2 - 4*12*2025))/2*12. Solve the square root first: x = (360 +/- sqrt (32400))/2*12 = (360 +/- 180)/24. That means x = 180/12 or 540/12 = 15 or 45. Obviously 45 can't be right, so it must be 15. That makes sense to me, as it makes the box a cube, which is generally ideal for maximizing volume. The length is 15 inches.

Answer 5

Assuming base of the wall as (0, 0), we have the tip of the ladder at (0, y) and base of the ladder at (x, 0). Applying Pythagorus theorem, x^2 + y^2 = 29^2
Differentiating w.r.t. time t, 2x.dx/dt +2y.dy/dt = 0
Equating, dx/dt = - y.dy/dt/x ------(1)
Given: dy/dt = -2 ft/s (in negative y-direction), y = 20 ft
So x = sqrt(29^2 - 20^2) = 21 ft
Plug into (1), dx/dt = -(20).(-2)/(21) = 40/21 ft/s away from the base
Answer: g. none of the above

Answer 6

Think logically. This is a trick question.

The ladder has a fixed length, and is rigid, so the whole ladder must be moving at the same rate. (If the top moved faster than the bottom the ladder would have to be getting shorter, or bending.)
Therefore it's g none of these.

Answer 7

a.8/3 feet/sec.