Maths Applications of Derivatives
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New question posted
2 months agoNew answer posted
2 months agoContributor-Level 10
Equation x²/5 + y²/4 = 1 then P (√5cosθ, 2sinθ)
(PQ)² = 5cos²θ + 4 (sinθ+2)² = cos²θ + 16sinθ + 20
= -sin²θ + 16sinθ + 21
= 85 - (sinθ - 8)²
∴ (PQ)²max = 85 - 49 = 36
? (sinθ - 8)² ∈
New answer posted
2 months agoContributor-Level 10
C? → C? - C?
f (θ) = | -sin²θ -1 |
| -cos²θ -1 |
| 12 -2 -2|
= 4 (cos²θ - sin²θ) = 4 (cos2θ), θ ∈ [π/4, π/2]
f (θ)max = M = 0
f (θ)min = m = -4
New answer posted
2 months agoContributor-Level 10
A (α, 0), B (-α, 0)
⇒ D (α, α² − 1)
Area (ABCD) = (AB) (AD)
⇒ S = (2α) (1 − α²) = 2α – 2α³
dS/dα = 2 - 6α²
= 0 ⇒ α² = 1/3
⇒ α = 1/√3
Area = 2α – 2α³ = 2/√3 - 2/ (3√3)
= 4/ (3√3)
New answer posted
2 months agoContributor-Level 10
f (2)=8, f' (2)=5, f' (x) ≥ 1, f' (x) ≥ 4, ∀x ∈ (1,6)
Using LMVT
f' (x) = (f' (5) - f' (2)/ (5-2) ≥ 4 ⇒ f' (5) ≥ 17
f' (x) = (f (5) - f (2)/ (5-2) ≥ 1 ⇒ f (5) ≥ 11
Therefore f' (5) + f (5) ≥ 28
New question posted
2 months agoNew answer posted
2 months agoContributor-Level 10
For (1,2) of y² = 4x => t=1, a=1
normal => tx+y = 2at + at³
=> x+y=3 intersect x-axis at (3,0)
y = e? => dy/dx = e?
tangent => y-e? = e? (x-c)
at (3,0) => 0-e? = e? (3-c) => c=4
New answer posted
2 months agoContributor-Level 10
S = 6a² => dS/dt = 12a * da/dt = 3.6
=> 12 (10) da/dt = 3.6
=> da/dt = 0.03
V = a³ => dV/dt = 3a² * da/dt
= 3 (10)² * (3/100) = 9
New answer posted
2 months agoContributor-Level 10
f (x) = (3x - 7)x²/³
⇒ f (x) = 3x? /³ - 7x²/³
⇒ f' (x) = 5x²/³ - 14/ (3x¹/³)
= (15x - 14) / (3x¹/³) > 0
∴ f' (x) > 0 ∀x ∈ (-∞, 0) U (14/15, ∞)
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