A man 2m tall walks away from a lamppot at 1m/s. If te lampatop the post is 5m above the ground, how fast is te tip of is sadow moving? How fast is the length of his shadow changing?

To find out how fast the tip of the man's shadow is moving, we need to recognize that the height of the lamppost, the height of the man, and the length of the man's shadow form similar triangles.

Let's assume that the length of the man's shadow is represented by "x" meters. Since the man's height is 2m, and the height of the lamppost is 5m, we can set up the following equation:

2 / x = 5 / (x + h)

where "h" represents the distance between the man and the lamppost (which is changing since the man is moving away from the lamppost).

We're given that the man is moving away from the lamppost at 1m/s, which means that dx/dt (the rate at which the distance between the man and the lamppost is changing) is 1m/s.

Now, let's differentiate both sides of the equation with respect to time (t):

(d/dt) (2 / x) = (d/dt) (5 / (x + h))

To find the derivative of each term, we can use the chain rule. On the left side, we get:

-2 / x^2 * dx/dt

And on the right side, we get:

-5 / (x + h)^2 * d/dt (x + h)

Since dx/dt is given as 1m/s, we can substitute it in:

-2 / x^2 = -5 / (x + h)^2

Cross-multiplying the equation, we get:

-2(x + h)^2 = -5x^2

Expanding and rearranging, we have:

-2x^2 - 4hx - 2h^2 = -5x^2

Simplifying further, we get:

3x^2 + 4hx + 2h^2 = 0

Now, let's find the derivative of this equation with respect to time:

(d/dt) (3x^2 + 4hx + 2h^2) = 0

Differentiating each term, we get:

6x * dx/dt + 4h * dx/dt + 0 = 0

Substituting the given value of dx/dt as 1m/s, we have:

6x + 4h = 0

Solving for x, we get:

x = -4h / 6 = -2h / 3

Since length cannot be negative, we discard the negative solution.

Therefore, the length of the man's shadow is changing at a rate of -2h / 3 meters per second, which means it is decreasing at a rate of 2h / 3 meters per second.

To find the rate at which the tip of the man's shadow is moving, we need to calculate the derivative of x with respect to t:

(dx/dt) = (dx/dh) * (dh/dt)

We know that dx/dh is -2 / 3 (as derived before) and dh/dt is given as 1 m/s. Substituting these values, we have:

(dx/dt) = (-2 / 3) * 1 = -2 / 3

Therefore, the tip of the man's shadow is moving at a rate of -2 / 3 meters per second, which means it is moving towards the lamppost.