The depth of water in a container is $$h=20-0.4t^2,\qquad 0\le t\le5,$$ where $h$ is in cm. A

Question

The depth of water in a container is $$h=20-0.4t^2,\qquad 0\le t\le5,$$ where $h$ is in cm. At $t=3$, find: $\dfrac{\mathrm{d}h}{\mathrm{d}t}$; the rate of decrease of the depth.

Accepted Answer

(a) $\frac{\mathrm{d}h}{\mathrm{d}t}=-2.4\text{ cm/s}$; (b) Rate of decrease $=2.4\text{ cm/s}$.

Differentiation

Differentiation

Solution:

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