A tumor is approximately modeled by the shape of a sphere, given by the following equation: 33 V= πr (1) 4 where V , is the volume of the spherical tumor and r, the radius. Data shows that, when the diameter of a spherical tumor is 16 mm then it is growing at a rate of 0.4 mm a day. How fast is the volume of the tumor changing at that time, mathematically this is symbolised by the derivative dV ? dt Solve the above problem through the following steps: (a) Apply the chain rule to equation 1 to find dV after writing r as r (t), because r is a function of dt time, t. Your answer will include r (t) = dt .[4] ′ dr (b) Information of the size of this tumor is given in diameter measurement, convert this to radius ′ dr measurement for r and r (t) = dt .[2] (c) Lastly, substitute the answers of (b) into the “answer equation” that you found in (a), to calculate dV dt .[2
A tumor is approximately modeled by the shape of a sphere, given by the following equation:
33
V= πr (1)
4
where V , is the volume of the spherical tumor and r, the radius. Data shows that, when the diameter
of a spherical tumor is 16 mm then it is growing at a rate of 0.4 mm a day. How fast is the volume of
the tumor changing at that time, mathematically this is symbolised by the derivative dV ? dt
Solve the above problem through the following steps:
(a) Apply the chain rule to equation 1 to find dV after writing r as r (t), because r is a function of
dt time, t. Your answer will include r (t) = dt .[4]
′ dr
(b) Information of the size of this tumor is given in diameter measurement, convert this to radius
′ dr measurement for r and r (t) = dt .[2]
(c) Lastly, substitute the answers of (b) into the “answer equation” that you found in (a), to calculate
dV
dt .[2
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