Combine implicit, inverse-circular and integral methods

Check the prerequisites before combining methods

Here the task is to recognise which quantities depend on x, retain their chain factors, then connect a complete derivative or integral to its requested conclusion

Differentiate before substituting

For e^(2(x−1))+3(x−1)y+4(y−2)=1, the point(1,2) satisfies the equation. Differentiating gives2e^(2(x−1))+3y+[3(x−1)+4]y′=0. At the point, 4y′=−8, so m=−2 and the tangent is y=−2x+4.

For e^(2(x−1))+3(x−1)y+4(y−2)=1 at (1,2), give the tangent gradient

For e^(2(x−1))+3(x−1)y+4(y−2)=1 at (1,2), give the tangent y-intercept

Which derivative of (x−1)y is valid when y depends on x?

Keep the dependent-variable chain factor

Let u=x−h and arcsin(y/b)+cuy/b=ku+eu², with b>0 and |y|<b. Differentiating gives y′/[b√(1−(y/b)²)]+c(y+uy′)/b=k+2eu. Multiply by b and collect y′ before dividing

Work a second derivative at a regular point

For arcsin(y/b)+c(x−h)y/b=k(x−h)+e(x−h)², take h=−1,b=3,c=2,k=−2,e=1. At x=h,y=0, y′=bk=−6. Differentiate again: y″/b+2cy′/b=2e, since the extra inverse-chain term contains y=0. Thus y″=2b(e−ck)=30. The first derivative denominator at this point is 1.

For arcsin(y/3)+2(x+1)y/3=−2(x+1)+(x+1)² at (−1,0), give y′ at x=−1

For arcsin(y/3)+2(x+1)y/3=−2(x+1)+(x+1)² at (−1,0), give y″ at x=−1

What is needed when dividing to isolate y′?

Translate the volume geometry

For y=2arcsin(x/3),0≤x≤3, rotate the region below the graph about the x-axis. The disk radius is y. Therefore V=π∫₀³[2arcsin(x/3)]²dx. The geometry fixes the squared quantity before choosing a substitution

Which quantity supplies the disk radius?

Transform the differential and both limits

For the region below y=2arcsin(x/3) on 0≤x≤3, rotated about the x-axis, put u=arcsin(x/3). Then x=3sin u, dx=3cos u du and the new interval is0≤u≤π/2. The volume becomes 12π∫₀^(π/2)u²cos u du. Integration by parts twice gives u²sin u+2ucos u−2sin u.

Rotate the region below y=2arcsin(x/3), 0≤x≤3, about the x-axis. Write its volume as V=Pπ³+Qπ and give P

Rotate the region below y=2arcsin(x/3), 0≤x≤3, about the x-axis. Write its volume as V=Pπ³+Qπ and give Q

Derive and justify both implicit results, ungraded

Use e^(2(x−1))+3(x−1)y+4(y−2)=1 at (1,2), and arcsin(y/3)+2(x+1)y/3=−2(x+1)+(x+1)² at (−1,0). For each example, verify the supplied point, show the complete first derivative before substituting, obtain the tangent or second derivative, and state why the local division is valid. Explain how missing a dependent-variable chain factor changes the conclusion

Model: regularity and the complete derivative

The exponential example has coefficient 3(x−1)+4=4 at(1,2), giving y′=−[2e^(2(x−1))+3y]/[3(x−1)+4], gradient−2 and tangent y=−2x+4. For the inverse example, y′=[−6+6u−2y]/[1/√(1−(y/3)²)+2u]. At u=0,y=0 this gives−6. The second derivative of arcsin(y/b) is y″/[b√(1−(y/b)²)]+y(y′)²/[b³(1−(y/b)²)^(3/2)]. Together with c(2y′+uy″)/b it yields y″=30. Both points are interior with nonzero collected derivative coefficients. These arguments establish local differentiable branches, not global single-valued solutions

Derive the substituted volume, ungraded

For a,b>0, derive the exact volume generated by rotating the region0≤y≤b arcsin(x/a),0≤x≤a about the x-axis. State the principal-branch restriction, show the transformed limits and both integrations by parts, and verify the primitive by differentiation

Model: connect disks, substitution and parts

For a,b>0, rotate 0≤y≤b arcsin(x/a), 0≤x≤a, about the x-axis. Disks give V=πb²∫₀ᵃ[arcsin(x/a)]²dx. On u∈[0,π/2], x=a sin u increases from 0 to a and dx=a cos u du. Thus V=ab²π∫₀^(π/2)u²cos u du. First parts gives u²sin u−2∫u sin u du; second parts gives ∫u sin u du=−u cos u+sin u. The primitive is u²sin u+2u cos u−2sin u, whose derivative is u²cos u after cancellation. Endpoint evaluation gives V=ab²π(π²/4−2)>0. These steps establish the full symbolic derivation; coefficient fields alone do not assess it