Find the length of the curve.r(t)=<2t,t^2, 13t^3>, 0≤ t≤ 1 相关知识点: 试题来源: 解析 73. r(t)=<2t,t^2, 13t^3> ⇒ \ r'(t)=<2,2t,t^2> ⇒|r'(t)|=√ (2^2+(2t)^2+(t^2)^2)=√ (4+4t^2+t^4)=√ ((2+t^2)^2)=2+t^2 for 0≤ t≤ 1....
Find the length of the curve \int_{0}^{y}\sqrt[]{9\sec^4 t-1}dt, on -\frac{\pi}{4}\leq y \leq \frac{\pi}{4} Find the length of the curve y = x e^{x^7}, \quad 0 \leq x \leq 1 Find the length of the curve. r(t) = (9t, 3 \cos t,3 \...
Find the length of the curver(t)=⟨2t32,cos2t,sin2t⟩. Question: Find the length of the curver(t)=⟨2t32,cos2t,sin2t⟩. Arc Length Formula: The formula for calculating the arc length for the space curve whose vector equation is given in t...
Find the length of the curve r(t) = \langle 8t, t^2, \frac{1}{12} t^3 \rangle in the interval 0 \leq t \leq 1 Find the length of the curve r(t)= \langle 2t^{2/3}, \cos 4t, \sin 4t \rangle in the interval 0 \...
Find the length of the curve x = 4y from y = -1 to y = 1. Find the length of the curve r(t) = 6t i + 3\sqrt{2}t^2 j + 2t^3 k as 0 \le t \le 1. Find the length of the curve r=a \sin^3\left(\frac{\theta}{3}\right) . ...
Find the length of the curve.(t)=+t^2+t^3, 0≤ t≤ 1 相关知识点: 试题来源: 解析 1(27)(13^(32)-8) (t)=+t^2+t^3 ⇒ '(t)=2t+3t^2 ⇒ |'(t)|=√ (4t^2+9t^4)=t√ (4+9t^2) [since t≥q 0]. Then L=\int ^{1}_{0}|\mathrm{r}'(t)|\d t=\...
回答：变速刚和我妈,几个
Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. r(t)=(7cost)i+(7sint)j+(52t)k,0≤t≤π5 Choose the correct answer for the unit tangent vector ofr(t). A.T(t)=(-76218sint)i-...
Function to find the extreme point of a curveDemetris T. Christopoulos
Question: Help Entering Answers(1 point) Find the length of the curve:r(t)=-2i+4t2j+t3kwhere 0≤t≤1Length = Help Entering Answers (1point)Find the length of the curve: r(t)=-2i+4tj+tk where0≤t≤1 Length=...