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?

1 Answer
sente
Oct 8, 2016

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.

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