▶️ Answer/Explanation
Solution
(a) Tangent at \( x = 3 \)
Curve: \( y = (2x^3 + 10)^{\frac{1}{2}} \)
Derivative: \( \frac{dy}{dx} = \frac{1}{2} (2x^3 + 10)^{-\frac{1}{2}} \cdot 6x^2 = \frac{3x^2}{\sqrt{2x^3 + 10}} \)
At \( x = 3 \):
\( 2 \cdot 3^3 + 10 = 64 \), \( y = \sqrt{64} = 8 \)
Slope: \( \frac{dy}{dx} = \frac{3 \cdot 9}{\sqrt{64}} = \frac{27}{8} \)
Tangent at \( (3, 8) \): \( y – 8 = \frac{27}{8} (x – 3) \)
Multiply by 8: \( 8y – 64 = 27x – 81 \)
\( 27x – 8y – 17 = 0 \)
Answer: \( 27x – 8y – 17 = 0 \)
(b) Volume of solid
Volume: \( V = \pi \int_{1}^{3} (2x^3 + 10) \, dx \)
Integrate: \( \int (2x^3 + 10) \, dx = \frac{x^4}{2} + 10x \)
Evaluate: \( \left[ \frac{x^4}{2} + 10x \right]_{1}^{3} = \left( \frac{81}{2} + 30 \right) – \left( \frac{1}{2} + 10 \right) = 60 \)
\( V = 60\pi \)
Answer: \( 60\pi \)