Maths Applications of Derivatives
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If the point P on the curve, 4x² + 5y² = 20 is farthest from the point Q(0, −4) then PQ² is equal to:
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If the point P on the curve, 4x² + 5y² = 20 is farthest from the point Q(0, −4) then PQ² is equal to:
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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 ∈
Upvote 
If the minimum and the maximum values of the function f: [-π/2, π/2] → R, defined by f(θ) = |-sin²θ -1 -sin²θ; -cos²θ -1 -cos²θ; 12 10 -2| are m and M respectively, then the ordered pair (m, M) is equal to:
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If the minimum and the maximum values of the function f: [-π/2, π/2] → R, defined by f(θ) = |-sin²θ -1 -sin²θ; -cos²θ -1 -cos²θ; 12 10 -2| are m and M respectively, then the ordered pair (m, M) is equal to:
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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
Upvote The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y = x² – 1 below the x-axis, is:
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The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y = x² – 1 below the x-axis, is:
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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)
Upvote Let f be a twice differentiable function on (1,6). If f(2) = 8, f'(2) = 5, f'(x) ≥ 1 and f''(x) ≥ 4, for all x ∈ (1,6), then:
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Let f be a twice differentiable function on (1,6). If f(2) = 8, f'(2) = 5, f'(x) ≥ 1 and f''(x) ≥ 4, for all x ∈ (1,6), then:
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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
Upvote If the tangent to the curve, y = eˣ at a point (c, eᶜ) and the normal to the parabola, y² = 4x at the point (1,2) intersect at the same point on the x-axis, then the value of c is
If the tangent to the curve, y = eˣ at a point (c, eᶜ) and the normal to the parabola, y² = 4x at the point (1,2) intersect at the same point on the x-axis, then the value of c is
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
Upvote If the surface area of a cube is increasing at a rate of 3.6 cm²/sec, retaining its shape; then the rate of change of its volume (in cm³/sec ), when the length of a side of the cube is 10 cm, is:
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If the surface area of a cube is increasing at a rate of 3.6 cm²/sec, retaining its shape; then the rate of change of its volume (in cm³/sec ), when the length of a side of the cube is 10 cm, is:
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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
Upvote The function, f(x) = (3x – 7)x²/³, x ∈ R, is increasing for all x lying in:
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The function, f(x) = (3x – 7)x²/³, x ∈ R, is increasing for all x lying in:
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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, ∞)
Upvote 