Applications of Derivatives Question - None

Question

The curve y = e^x sin(π/2 − x) = e^x cos x. The slope of the tangent S = e^x(cos x − sin x). For what value of x in [0, 2π] is the slope of the tangent minimum?

Accepted Answer

Step 1: Find S'(x) where S(x) = e^x(cos x - sin x)

Using product rule: S'(x) = e^x(cos x - sin x) + e^x(-sin x - cos x)

S'(x) = e^x[(cos x - sin x) - (sin x + cos x)]

S'(x) = e^x(-2sin x)

Step 2: Set S'(x) = 0. Since e^x > 0 always:

-2sin x = 0 ⟹ sin x = 0

In [0, 2π]: x = 0, π, 2π

Step 3: Evaluate S(x) at critical points:

• S(0) = e^0(cos 0 - sin 0) = 1(1 - 0) = 1

• S(π) = e^π(cos π - sin π) = e^π(-1 - 0) = -e^π ≈ -23.14

• S(2π) = e^{2π}(cos 2π - sin 2π) = e^{2π}(1 - 0) = e^{2π}

Step 4: Check sign of S'(x) = -2e^x sin x:

• For x ∈ (0, π): sin x > 0 ⟹ S'(x) < 0 (S decreasing)

• For x ∈ (π, 2π): sin x < 0 ⟹ S'(x) > 0 (S increasing)

Therefore S(x) is minimum at x = π.

∴ Answer: x = π

Question:

Step-by-Step Solution

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