Area Normal 46Fa53

1. **State the problem:**\nWe have the curve $$y = 3\sqrt{2x + 1}$$ and a point $$R(4, 9)$$ on it. We want to find:\n(a) The area of the shaded region bounded by the curve, the axes, and the line $$y = a$$ up to $$x=4$$.\n(b) The exact value of $$a$$ such that the triangular area enclosed by the y-axis, the normal to the curve at $$R$$, and the line $$y = a$$ equals the area found in (a).\n\n2. **Find the area of the shaded region (part a):**\nThe shaded region is bounded by the curve, the x-axis, the y-axis, and the line $$y = a$$ (which is horizontal). Since the curve passes through $$R(4,9)$$, the upper limit for integration is $$x=4$$.\n\nThe area under the curve from $$x=0$$ to $$x=4$$ is:\n$$\text{Area} = \int_0^4 3\sqrt{2x + 1} \, dx$$\n\n3. **Calculate the integral:**\nLet $$u = 2x + 1 \Rightarrow du = 2 dx \Rightarrow dx = \frac{du}{2}$$\nWhen $$x=0, u=1$$; when $$x=4, u=9$$.\n\nRewrite the integral:\n$$\int_0^4 3\sqrt{2x + 1} \, dx = 3 \int_1^9 \sqrt{u} \cdot \frac{du}{2} = \frac{3}{2} \int_1^9 u^{1/2} du$$\n\nIntegrate:\n$$\frac{3}{2} \cdot \left[ \frac{u^{3/2}}{\frac{3}{2}} \right]_1^9 = \left[u^{3/2}\right]_1^9 = 9^{3/2} - 1^{3/2}$$\n\nCalculate powers:\n$$9^{3/2} = (9^{1/2})^3 = 3^3 = 27$$\n$$1^{3/2} = 1$$\n\nSo area is:\n$$27 - 1 = 26$$\n\n4. **Answer for (a):**\nThe area of the shaded region is $$\boxed{26}$$.\n\n5. **Find the exact value of $$a$$ (part b):**\nWe need to find $$a$$ such that the triangular area enclosed by the y-axis, the normal to the curve at $$R$$, and the line $$y = a$$ equals 26.\n\n6. **Find the slope of the curve at $$R$$:**\nGiven $$y = 3\sqrt{2x + 1} = 3(2x+1)^{1/2}$$\n\nDifferentiate:\n$$\frac{dy}{dx} = 3 \cdot \frac{1}{2} (2x+1)^{-1/2} \cdot 2 = 3 (2x+1)^{-1/2}$$\n\nAt $$x=4$$:\n$$\frac{dy}{dx} = 3 (2(4)+1)^{-1/2} = 3 (9)^{-1/2} = 3 \cdot \frac{1}{3} = 1$$\n\n7. **Find the slope of the normal at $$R$$:**\nThe normal slope is the negative reciprocal of the tangent slope:\n$$m_{normal} = -\frac{1}{1} = -1$$\n\n8. **Equation of the normal line at $$R(4,9)$$:**\nUsing point-slope form:\n$$y - 9 = -1 (x - 4) \Rightarrow y = -x + 4 + 9 = -x + 13$$\n\n9. **Find the intercepts of the normal line:**\n- y-intercept: set $$x=0$$, $$y = -0 + 13 = 13$$\n- x-intercept: set $$y=0$$, $$0 = -x + 13 \Rightarrow x = 13$$\n\n10. **Find the triangular area enclosed by the y-axis, the normal, and the line $$y = a$$:**\nThe triangle is bounded by:\n- y-axis (x=0)\n- normal line $$y = -x + 13$$\n- horizontal line $$y = a$$\n\nThe triangle's vertices are:\n- $$A = (0, a)$$ (intersection of y-axis and $$y=a$$)\n- $$B = (0, 13)$$ (y-intercept of normal)\n- $$C = (13 - a, a)$$ (intersection of normal and $$y=a$$) because setting $$y=a$$ in normal line gives $$a = -x + 13 \Rightarrow x = 13 - a$$\n\n11. **Calculate the area of the triangle:**\nThe base is vertical segment $$AB$$ with length $$13 - a$$. The height is horizontal segment $$AC$$ with length $$13 - a$$. But since $$A$$ and $$C$$ share the same y-coordinate $$a$$, the base is vertical segment $$AB = 13 - a$$ and the height is horizontal segment $$AC = 13 - a$$.\n\nArea of triangle:\n$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (13 - a)(13 - a) = \frac{(13 - a)^2}{2}$$\n\n12. **Set the triangular area equal to the shaded area 26:**\n$$\frac{(13 - a)^2}{2} = 26$$\nMultiply both sides by 2:\n$$ (13 - a)^2 = 52$$\n\nTake square root:\n$$13 - a = \pm \sqrt{52} = \pm 2\sqrt{13}$$\n\n13. **Solve for $$a$$:**\n$$a = 13 \mp 2\sqrt{13}$$\nSince $$a$$ is positive and less than 13 (because the line $$y=a$$ intersects the normal between y=0 and y=13), we take the smaller value:\n$$a = 13 - 2\sqrt{13}$$\n\n14. **Final answers:**\n(a) Area of shaded region = $$\boxed{26}$$\n(b) Exact value of $$a$$ = $$\boxed{13 - 2\sqrt{13}}$$