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

answer: b.48 for

step-by-step explanation:

$12. that is my answer.