Re: HD cost Sorry to bump this thread but just a heads up to check your next bill when you do this. Even though I did it well in advance of my billing date (more than 2 weeks) I was still billed for the add on on my bill. Credit though to the BT Live Chat who added a ...
解r(t)=(acost,asint,bt) , r'(t)= {-asin t, acos t, b} , 当 t=π/(3) ,有 r(π/3)=[a/2,(√3)/2a,(πb)/3] r^r(π/3)=\(-(√3)/2a,a/2,b\) 所以切线的方程为 ρ-r(π/(3))=λr(π/(3)) 即ρ=(1-λ√3)/2at_1+(√3+λ)/2ae_2+(π/3+λ)...
解析 证明: ∴=-a sin ∵ .'=b. 螺旋线的切向量为 T=(-asint,acost,b) 与z轴同向的单位向量为 A ={0.0.1} 两向量的夹角余弦为 cosθ=b/(√((-asint)^2+(acost)^2+b^2)=b/(√(a^2-b^2)) 为一定值: 故螺旋线的切线与:轴形成定角。
Home move TV aerial cost Go to solution I’m moving home soon, I’ve started the home move process with BT and paid £30 for aerial installation, however looking online it quotes the cost as £60 have I been charged incorrectly or have I paid for something entirely different? I did...
Prevention of sales calls is the objective of the service. BT did not clarify that it would continue to charge its customers an annual cost of GBP21 for caller display of the service.EBSCO_bspTarifica Alert
曲线r(t)=(acost,asint,bt)在t=0点的切向量为（ 0,a,b ）下面一题，我写个过程给你 切线方程是 x-1=y=z-1 看这个题是填空题出的 如果求法平面方程，就是x-1+y+z-1=0,也就是x+y+z-2=0
3.证明:螺旋线x=acost,y=asint,z=bt的切线与z轴形成定角。 证明:x=-asint,y=acost,z'=b. 螺旋线的切向量为 T=(-a sint,a cost,b). 与z轴同向的单位向量为 k={0,0,1} 两向量的夹角余弦为 Cin..豆 cose √(-asint)2+(acost)2+b2√a2+b2 为一定值。 故螺旋线的切线与z轴形成定角。
证明曲线x=acost,y=asint,z=bt的切线与Oz轴成定角。 相关知识点: 试题来源: 解析 ∵-sinx+(√3)/4=-cos2x∴设切线与Oz轴所围成夹角为αcosα=(4/5)/(√((π/4)+θ/2)^(((π)/3)^3)(π/2)^2=-b/(√(a^2+b^2))a,b都是常数,所以cosα是定值,即切线与Oz轴的夹角也是定值要证明...
【答案】：ydx+zdy+xdz=(-a2sin2t+abtcost+abcost)dt∫cydx+zdy+xdz=∫02π(-a2sin2t+abtcost+abcost)dt=-a2π
【计算题】证明:螈旋线r=(acost,asint,bt)的曲率中心轨迹仍然是螺旋线. 答案: 手机看题 问答题 【计算题】设曲线Γ的方程为r=r(t),其中r∈C(2),,P0(即r(t0))及P(即r(t0+Δt))是Γ上两点,且r′(t0)×r″(t0)≠0.记Γ在P处的切线为l,过P0及l的平面为π′.证明:当P沿Γ趋于P0时,平面...