Asked by Natasha
a cone has an altitude of 12 cm and a base radius which increases at the rate of 3 cm per minute. When the radius is 6 cm,how fast is the vertex angle increasing?
Answers
Answered by
Reiny
Make a sketch of the cross-section
let the radius be r, let the vertex angle be 2Ø
we have a right-angled triangles with angle Ø, opposite is r, and adjacent is 12
tanØ = r/12
12tanØ = r
12 sec^2 Ø (dØ/dt) = dr/dt
when r = 6 , hypotenuse = √180 = 6√5
then secØ = 12/(6√5) = 2/√5
sec^2 = 4/5
12(4/5) dØ/dt = 3
dØ/dt = 15/48 or 5/16 radians/min
check my arithmetic
let the radius be r, let the vertex angle be 2Ø
we have a right-angled triangles with angle Ø, opposite is r, and adjacent is 12
tanØ = r/12
12tanØ = r
12 sec^2 Ø (dØ/dt) = dr/dt
when r = 6 , hypotenuse = √180 = 6√5
then secØ = 12/(6√5) = 2/√5
sec^2 = 4/5
12(4/5) dØ/dt = 3
dØ/dt = 15/48 or 5/16 radians/min
check my arithmetic
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