3 The equation of a curve is such that \( \frac{\mathrm{dy}}{\mathrm{dt}}=3(4 x-7)^{\frac{1}{2}}-4 x^{-\frac{1}{2}} \). It is given that the curve passes throug the point \( \left(4, \frac{5}{2}\right) \). Find the equation of the curve.

We are given that   dy/dt = 3√(4x – 7) – 4/√x and that the curve passes through the point (4, 5/2). (Note: Although the derivative is written as dy/dt, the right‐hand side is expressed in terms of x. In many such problems the independent variable is taken to be x rather than t. Hence, we interpret the given equation as   dy/dx = 3√(4x – 7) – 4/√x.) Our goal is to find y as a function of x. Step 1. Write the derivative explicitly:   dy/dx = 3(4x – 7)^(1/2) – 4x^(–1/2) Step 2. Integrate both sides with respect to x:   ∫ dy/dx dx = ∫ [3(4x – 7)^(1/2) – 4x^(–1/2)] dx Thus,   y = ∫ 3√(4x – 7) dx – ∫ 4x^(–1/2) dx  + C We now treat the two integrals separately. Part A. Evaluate I₁ = ∫ 3√(4x – 7) dx Let u = 4x – 7, so that   du/dx = 4  ⟹  dx = du/4. Then   I₁ = 3 ∫ u^(1/2) (du/4) = (3/4) ∫ u^(1/2) du Integrate u^(1/2):   ∫ u^(1/2) du = (2/3) u^(3/2) Thus,   I₁ = (3/4) × (2/3) u^(3/2) = (1/2) u^(3/2) Substitute back u = 4x – 7:   I₁ = (1/2) (4x – 7)^(3/2) Part B. Evaluate I₂ = ∫ 4x^(–1/2) dx Recall that   ∫ x^(–1/2) dx = 2√x Thus,   I₂ = 4 × (2√x) = 8√x Step 3. Write the antiderivative for y:   y = (1/2)(4x – 7)^(3/2) – 8√x + C Step 4. Use the initial condition to determine C. The curve passes through (x, y) = (4, 5/2): Plug x = 4 into the antiderivative:   Calculate (4x – 7) for x = 4:     4×4 – 7 = 16 – 7 = 9 Then,   (4x – 7)^(3/2) = 9^(3/2) Recall that 9^(3/2) = (√9)³ = 3³ = 27. Also, √4 = 2. So, substitute into the equation:   y = (1/2)(27) – 8(2) + C = (13.5) – 16 + C = –2.5 + C We are given that y = 5/2 = 2.5 when x = 4. Hence,   2.5 = –2.5 + C  ⟹  C = 2.5 + 2.5 = 5. Step 5. Write the final answer: The equation of the curve is   y = (1/2)(4x – 7)^(3/2) – 8√x + 5. This is the required equation of the curve.