The gradient of a curve which passes through P(2, 7) is given by dy/dx = 15/sqrt(7x + 2). a) Find the equation of the curve. b) Find the equation of the normal to the curve at P.

Question 10: The gradient of a curve which passes through P(2, 7) is given by (dy)/(dx) = (15)/(sqrt(7x + 2)). a) Find the equation of the curve. Step 1: Integrate the given gradient function to find the equation of the curve. y = (15)/(sqrt(7x + 2))\,dx Rewrite the integrand using exponent notation: y = 15(7x + 2)^-1/2\,dx Step 2: Apply the integration rule (ax+b)^n\,dx = (ax+b)^n+1a(n+1) + C. Here, a=7, n=-1/2. So, n+1 = 1/2. y = 15 · (7x + 2)^1/27 · (1/2) + C y = 15 · (2)/(7) (7x + 2)^1/2 + C y = (30)/(7) sqrt(7x + 2) + C Step 3: Use the given point P(2, 7) to find the value of the constant C. Substitute x=2 and y=7 into the equation. 7 = (30)/(7) sqrt(7(2) + 2) + C 7 = (30)/(7) sqrt(14 + 2) + C 7 = (30)/(7) sqrt(16) + C 7 = (30)/(7) · 4 + C 7 = (120)/(7) + C C = 7 - (120)/(7) C = (49)/(7) - (120)/(7) C = -(71)/(7) Step 4: Substitute the value of C back into the curve's equation. y = (30)/(7) sqrt(7x + 2) - (71)/(7) b) Find the equation of the normal to the curve at P. Step 1: Find the gradient of the tangent to the curve at point P(2, 7) by substituting x=2 into the given (dy)/(dx). m_t = (dy)/(dx) |_x=2 = (15)/(sqrt(7(2) + 2)) m_t = (15)/(sqrt(14 + 2)) m_t = (15)/(sqrt(16)) m_t = (15)/(4) Step 2: Find the gradient of the normal, m_n. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of the tangent's gradient. m_n = -(1)/(m_t) m_n = -(1)/(15/4) m_n = -(4)/(15) Step 3: Use the point-slope form of a linear equation, y - y_1 = m(x - x_1), with point P(2, 7) and m_n = -(4)/(15). y - 7 = -(4)/(15)(x - 2) Step 4: Multiply both sides by 15 to eliminate the fraction and rearrange the equation into the general form Ax + By + C = 0. 15(y - 7) = -4(x - 2) 15y - 105 = -4x + 8 4x + 15y - 105 - 8 = 0 4x + 15y - 113 = 0 3 done, 2 left today. You're making progress.