Zadanie 2e, Paper 3, listopad 2022, Matematyka AA IB HL

(i) A vertical stretch by factor k about the x-axis. (ii) x=±r. (iii)$y\prime=-kx/\sqrt{r^2-x^2}$. (iv)$y\sqrt{1+y\prime^2}=k\sqrt{r^2-x^2}\sqrt{1+k^2x^2/(r^2-x^2)}=\sqrt{k^2r^2+k^2(k^2-1)x^2}$. Therefore $A=2\pi\int_{-r}^{r}\sqrt{k^2r^2+k^2(k^2-1)x^2}\,dx$. The expression is interpreted by continuous limits at the arc endpoints. (v) The axial semilength is r=15 cm, while kr=9 cm gives k=3/5. Evaluate $2\pi\int_{-15}^{15}\sqrt{81-0.2304x^2}\,dx=1492.137557...$ cm². To four significant figures this is 1.492 × 10³ cm².

(i) A1:vertical stretch/compression. A1:factor k. (ii) A1:both ±r. (iii) M1:differentiate the profile using the chain rule or implicit method. A1:correct derivative. (iv) A1:correct substitution into the supplied surface integral. M1:simplify the radical product. A1:limits −r,r, independently of the method mark. A1:p(x)=k²r²+k²(k²−1)x². (v) A1:r=15. A1:k=3/5. M1:attempt to form the corresponding definite surface-area integral. A1:1.492×10³ cm², with four significant figures and scientific notation. For (v), reversing axial and radial measurements gives r=9, k=15/9: award A0 A1FT M1 A0 when the integral follows this choice. Correct equivalent transformation descriptions earn the marks in (i).