# Intergrate/differentiate?

#### Willo142

could someone explain too me the answer to this question?

water is emptied from a water tank using a tap at its base. The depth of the water in the tank, d cm, can be modelled by the function
$$\displaystyle d(t) = (((t-150)^2)/1000)+1$$

a) show it can be written as
d(t)=0.001t^2 - 0.3t + 23.5

b) find value of d'(80)

any help would be appreciated, im really stuck. even just an explination of how to go about this.

#### AllanCuz

could someone explain too me the answer to this question?

water is emptied from a water tank using a tap at its base. The depth of the water in the tank, d cm, can be modelled by the function
$$\displaystyle d(t) = (((t-150)^2)/1000)+1$$

a) show it can be written as
d(t)=0.001t^2 - 0.3t + 23.5

b) find value of d'(80)

any help would be appreciated, im really stuck. even just an explination of how to go about this.
Part A is asking you to expand,

$$\displaystyle d(t) = \frac{ (t-150)^2)}{1000}+1 = \frac{ t^2 - 300t + 150^2 }{1000} + 1 = 0.001t^2 - 0.3t + 23.5$$

Part B is asking for the derivative and to evaluate at 80.

$$\displaystyle d(t) = 0.002t -.3$$

$$\displaystyle d(80) = 0.002(8) -. 3$$

#### AllanCuz

could someone explain too me the answer to this question?

water is emptied from a water tank using a tap at its base. The depth of the water in the tank, d cm, can be modelled by the function
$$\displaystyle d(t) = (((t-150)^2)/1000)+1$$

a) show it can be written as
d(t)=0.001t^2 - 0.3t + 23.5

b) find value of d'(80)

any help would be appreciated, im really stuck. even just an explination of how to go about this.
Part A is asking you to expand,

$$\displaystyle d(t) = \frac{ (t-150)^2)}{1000}+1 = \frac{ t^2 - 300t + 150^2 }{1000} + 1 = 0.001t^2 - 0.3t + 23.5$$

Part B is asking for the derivative and to evaluate at 80.

$$\displaystyle d(t) = 0.002t -.3$$

$$\displaystyle d(80) = 0.002(8) -. 3$$

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