Question

Need help on this calculus problem!?

A construction worker pulls a 5m plank up the side of a building under construction by means of a rope tied to the end of the plank. The opposite end of the plank is being dragged along the ground. If the worker is pulling at a rate of 15cm/s, how fast is the end of the plank sliding along the ground when it is 2m from the wall of the building? Express answer to 3 decimal places.

Answer 1

`(dx)/(dt) = -0.344"m/s"`

Explanation:

The 5m plank forms a right triangle with the building and the ground, therefore, the Pythagorean Theorem applies to the values of y and x:

`x^2+y^2=(5"m")^2`

Solve for y:

`y = sqrt(25"m"^2-x^2)`

Compute the derivative with respect to x:

`dy/dx = -x/sqrt(25"m"^2-x^2)" [1]"`

Use the chain rule.

`dy/dx = (dy)/(dt)(dt)/(dx)`

Because we want the speed at which the x value is changing, we must solve for `(dx)/(dt)`

`(dx)/(dt) = ((dy)/(dt))/(dy/dx)`

We are given that `dy/dt = 15"cm/s"` but we must convert that to `dy/dt = 0.15"m/s"` and substitute the right side of equation [1] for `dy/dx`:

`(dx)/(dt) = (0.15"m/s")/(-x/sqrt(25"m"^2-x^2))`

`(dx)/(dt) = (0.15"m/s")(sqrt(25"m"^2-x^2))/-x`

We are given that the x coordinate is 2"m":

`(dx)/(dt) = (0.15"m/s")(sqrt(25"m"^2-(2"m")^2))/(-(2"m"))`

`(dx)/(dt) = -0.344"m/s"`

The negative value of the speed that indicates that the orientation of the right triangle is such that the lower edge of the plank is moving to the left as the upper edge is being pulled up the wall.

Answer 2

The end of the plank is sliding along the ground at about `34.369` `cm`/`s`. Please see below.

Explanation:

Draw a sketch of the situation. Label it using the variables, `h` and `g`, shown below.

Variables

Let `h` = the height of the upper end of the plank above the ground

let `g` = the distance between the bottom of the plank and the base of the wall.

(We also have `t` = time in seconds)

Rates of change

`(dh)/dt = 15` `cm`/`s` (or, if you prefer, `= 0.15` `m`/`s`)

We need to find `(dg)/dt` when `g = 2` `m` `= 200` `cm`

(Based on the story this should be a negative number, so our answer to the question will be the absolute value of this number.)

Equation relating the variables

I'll use lengths in `cm`, so the length of the plank is `500` `cm`.

We have a right triangle, so we can use the Theorem of Pythagoras to get

`h^2+g^2 = 500^2`

Finish the problem

Differentiate with respect to `t` (Use implicit differentiation / the chain rule.)
Substitute what we know and solve for what we don't know.

`d/dt(h^2+g^2) = d/dt(500^2)`

`2h (dh)/dt + 2g*(dg)/dt = 0`

Let's make the numbers smaller by dividing by `2`.

`h (dh)/dt + g * (dg)/dt = 0`

We know that `(dh)/dt = 15` and `g = 200`, and we are looking for `(dg)/dt`.

We need to know `h` when `g = 200`

Use the equation: `h^2+g^2 = 500^2` to get

`h = 100sqrt21`

Now we can finish:

`h (dh)/dt + g * (dg)/dt = 0`

`(100sqrt21) (15) + 200 (dg)/dt = 0`

`(dg)/dt = -(15sqrt21)/2` `cm`/s#

To 3 decimal places,

`(dg)/dt ~~ -34.369` `cm`/`s`

The end of the plank is sliding along the ground at about `34.369` `cm`/`s`.