Question

The movement of a certain glacier can be modelled by d(t) = 0.01t^2 + 0.5t, where d is the distance in metres, that a stake on the glacier has moved, relative to a fixed position, t days after the first measurement was made. Question?

Estimate the rate at which the glacier is moving after 20 days?

Answer 1

After `20` days, the glacier is moving at `0.9` meters per day.

Explanation:

The rate of motion of the glacier is a measure of its change in distance per change in time: `(triangled(t))/(trianglet)`. Viewed at a single moment, that is, as the change in time approaches `0`, this reduces to the derivative of the distance function with respect to time: `d/dt d(t) = d'(t)`. With that, we can actually read the question as asking for `d'(20)`.

To find `d'(20)`, we will apply three commonly used rules of differentiation:

• `(f(x)+g(x))' = f'(x)+g'(x)`

• `(cf(x))' = cf'(x)`

• `(x^n)' = nx^(n-1)`

Applying those, we have

`d'(t) = (0.01t^2+0.5t)'`

`=(0.01t^2)'+(0.5t)'`

`=0.01(t^2)'+0.5(t^1)'`

`=0.01(2t^1) + 0.5(1t^0)`

`=0.02t + 0.5`

Evaluating this at `t=20`, we get our answer:

`d'(20) = 0.02(20)+0.5 = 0.9`

After `20` days, the glacier is moving at `0.9` meters per day.