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2條a maths Differentiation問題
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1) It is given that the curve C: x^2+4y^2=72(a) Find dy/dx(b) Find the points on C at which the tangent to C passes through (4,4)2) Given a curve C: y=x^3-3x^2+2x-1. A(1,-1) and B(a,b) are two points on the curve and O is the origin.(a) Prove that the tangent to the curve at A passes through O.(b) If... 顯示更多 1) It is given that the curve C: x^2+4y^2=72 (a) Find dy/dx (b) Find the points on C at which the tangent to C passes through (4,4) 2) Given a curve C: y=x^3-3x^2+2x-1. A(1,-1) and B(a,b) are two points on the curve and O is the origin. (a) Prove that the tangent to the curve at A passes through O. (b) If the tangent to the curve at B passes through O,show that 2a^3-3a^2+1=0 (c) Hence,find the equations of the tangents drawn from O to the curve C
1(a)x2+4y2=72 Diff. both sides w.r.t x 2x+8y(dy/dx)=0 dy/dx=-x/4y 1(b)Let the required point be (a,b) a2+4b2=72--------(1) & (b-4)/(a-4)=-a/4b-------(2) From (2) a(4-a)=4b(b-4) 4a-a2=4b2-16b 4a+16b=a2+4b2 4a+16b=72(From (1)) a+4b=18 a=18-4b--------(3) Sub(3)in (1) (18-4b)2+4b2=72 16b2-144b+324+4b2=72 20b2-144b+252=0 5b2-36b+63=0 (5b-21)(b-3)=0 b=3 or 21/5 when b=3, a=6 when b=21/5, a=6/5 ∴The points are (6,3) or (-6,3) or (6/5, 21/5) or (-6/5, 21/5) 2(a)y=x3-3x2+2x-1 dy/dx=3x2-6x+2 dy/dx|(1,-1)=-1 Slope of AO=(-1-0)/(1-0)=-1=dy/dx|(1,-1) ∴The tangent to the curve at A passes through O. 2(b)Slope of BO=b/a Sub B(a,b) into C y=x3-3x2+2x+1 dy/dx=3x2-6x+2 dy/dx|(a,b)=3a2-6a+2 ∴3a2-6a+2=b/a 3a3-6a2+2a=a3-3a2+2a-1 (∵b=a3-3a2+2a-1) ∴2a3-3a2+1=0 2(c)Let the equations of the tangent drawn from O to C by y=mx Sub y=mx in C x3-3x2+(2-m)x-1=0 dy/dx=3x2
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2條a maths Differentiation問題
發問:
1) It is given that the curve C: x^2+4y^2=72(a) Find dy/dx(b) Find the points on C at which the tangent to C passes through (4,4)2) Given a curve C: y=x^3-3x^2+2x-1. A(1,-1) and B(a,b) are two points on the curve and O is the origin.(a) Prove that the tangent to the curve at A passes through O.(b) If... 顯示更多 1) It is given that the curve C: x^2+4y^2=72 (a) Find dy/dx (b) Find the points on C at which the tangent to C passes through (4,4) 2) Given a curve C: y=x^3-3x^2+2x-1. A(1,-1) and B(a,b) are two points on the curve and O is the origin. (a) Prove that the tangent to the curve at A passes through O. (b) If the tangent to the curve at B passes through O,show that 2a^3-3a^2+1=0 (c) Hence,find the equations of the tangents drawn from O to the curve C
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最佳解答:1(a)x2+4y2=72 Diff. both sides w.r.t x 2x+8y(dy/dx)=0 dy/dx=-x/4y 1(b)Let the required point be (a,b) a2+4b2=72--------(1) & (b-4)/(a-4)=-a/4b-------(2) From (2) a(4-a)=4b(b-4) 4a-a2=4b2-16b 4a+16b=a2+4b2 4a+16b=72(From (1)) a+4b=18 a=18-4b--------(3) Sub(3)in (1) (18-4b)2+4b2=72 16b2-144b+324+4b2=72 20b2-144b+252=0 5b2-36b+63=0 (5b-21)(b-3)=0 b=3 or 21/5 when b=3, a=6 when b=21/5, a=6/5 ∴The points are (6,3) or (-6,3) or (6/5, 21/5) or (-6/5, 21/5) 2(a)y=x3-3x2+2x-1 dy/dx=3x2-6x+2 dy/dx|(1,-1)=-1 Slope of AO=(-1-0)/(1-0)=-1=dy/dx|(1,-1) ∴The tangent to the curve at A passes through O. 2(b)Slope of BO=b/a Sub B(a,b) into C y=x3-3x2+2x+1 dy/dx=3x2-6x+2 dy/dx|(a,b)=3a2-6a+2 ∴3a2-6a+2=b/a 3a3-6a2+2a=a3-3a2+2a-1 (∵b=a3-3a2+2a-1) ∴2a3-3a2+1=0 2(c)Let the equations of the tangent drawn from O to C by y=mx Sub y=mx in C x3-3x2+(2-m)x-1=0 dy/dx=3x2
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