Approximate the change in the lateral surface area (excluding the area of the base) of a right circular cone of fixed height of h = 6 m when its radius decreases from r = 14 m to r = 13.9 m (S = (pi) (r) sqrt (r^2+h^2) ? S = __?

( - 42.8π/√232) m²

Step-by-step explanation:

For any function, y = f (x), we approximately from a to a + ∆x

We can make use of the formula

∆y = f' (a) ∆x

Where S = πr√ (r² + h²)

We are asked to find S

h = 6m

r = decreases from 14m to 13.9

S = d/dr[ πr√ (r² + h²) ] * (13.9 - 14)

Where h = 36m

S = π d/dr [ r √ (r² + 6²) ] * (13.9 - 14)

S = π d/dr [ r √ (r² + 36) ] * (13.9 - 14)

Where r = 14m

S = π [√ (r² + 36) + r²/√ (r² + 36) ] * (13.9 - 14)

S = π [√ (14² + 36) + 14²/√ (14² + 36) ] * (13.9 - 14)

S = π [√ (196 + 36) + 196/√ (196 + 36) ] * (13.9 - 14)

S = π[√232 + 196/√232] * (13.9 - 14)

S = π * - 0. 1 [√232 + 196/√232]

Collecting like terms

S = π * - 0. 1 * 1/√232 [232 + 196]

S = π * - 0. 1 * 1/√232[428]

S = ( - 42.8π/√232) m²