16. The radius of the base of a certain cone is increasing at the rate of 3 cm per minute and the
altitude is decreasing at the rate of 4 cm per minute. Find the rate of change of total surface of
the cone when the radius is 7 cm and the altitude is 24 cm.
Answers
Answer:
The rate of change of total surface area of the cone is 301.44 cm²/min.
Step-by-step explanation:
Given:
The increasing rate of change of radius of a cone, dr/dt = 3 cm/min
The decreasing rate of change of height of a cone, dh/dt = -4 cm/min
Radius of the cone, r = 7 cm
Altitude/height of the cone, h = 24 cm
Slant height of the cone, l = √(r²+h² ) = √(7²+24² ) = √(625 ) = 25 cm
Now,
Total surface area of a cone, S = πr² + πrl = πr² + πr√(r²+h² ) ………. (i)
To find the rate of change of total surface of the cone we need to differentiate (i) with respect to t.
Therefore,
dS/dt = 2πrdr/dt + π [{dr/dt *(√(r²+h² )}+r*{(2r dr/dt + 2hdh/dt)/(2(√(r²+h² ))}]
Substituting the given values
dS/dt = 2π*7*3 + π [ {3 *25 + 7 * {(2*7*3 + 2*24*(-4)) / (2*25)}]
dS/dt = 42 π + π [75 + 7 * (-75/25)]
dS/dt = 42 π + π [75 + (-21)]
dS/dt= 42 π + 54 π
dS/dt= 96 π cm²/min
putting the value of π = 3.14
dS/dt = 96 * 3.14 = 301.44 cm²/min
