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Given the curve Γ⊂R3\Gamma \subset \mathbb{R}^{3} parametrized by \[ r:[0,1] \rig...
Jun 13, 2024
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Solution by Steps
step 1
The length of the curve Γ\Gamma parametrized by r(t)=(3t2,4t3,3t4)r(t) = (3t^2, 4t^3, 3t^4) from t=0t=0 to t=1t=1 is given by the integral ∫01∥r′(t)∥dt\int_0^1 \|r'(t)\| dt
step 2
Compute the derivative r′(t)=(ddt(3t2),ddt(4t3),ddt(3t4))=(6t,12t2,12t3)r'(t) = \left(\frac{d}{dt}(3t^2), \frac{d}{dt}(4t^3), \frac{d}{dt}(3t^4)\right) = (6t, 12t^2, 12t^3)
step 3
The magnitude of r′(t)r'(t) is ∥r′(t)∥=(6t)2+(12t2)2+(12t3)2=36t2+144t4+144t6\|r'(t)\| = \sqrt{(6t)^2 + (12t^2)^2 + (12t^3)^2} = \sqrt{36t^2 + 144t^4 + 144t^6}
step 4
Simplify the expression under the square root: ∥r′(t)∥=36t2(1+4t2+4t4)=6t1+4t2+4t4\|r'(t)\| = \sqrt{36t^2(1 + 4t^2 + 4t^4)} = 6t\sqrt{1 + 4t^2 + 4t^4}
step 5
Integrate ∫016t1+4t2+4t4dt\int_0^1 6t\sqrt{1 + 4t^2 + 4t^4} dt. According to the Asksia-LL calculator, this integral evaluates to 6
Answer
The length of the curve Γ\Gamma is 6.
Question 2: Compute the integral ∫Γfds\int_{\Gamma} f ds of f:R3→Rf: \mathbb{R}^{3} \rightarrow \mathbb{R} defined as f(x,y,z):=x+zf(x, y, z) := x + z.
step 1
The integral ∫Γfds\int_{\Gamma} f ds is given by ∫01f(r(t))∥r′(t)∥dt\int_0^1 f(r(t)) \|r'(t)\| dt
step 2
Substitute r(t)=(3t2,4t3,3t4)r(t) = (3t^2, 4t^3, 3t^4) into f(x,y,z)=x+zf(x, y, z) = x + z: f(r(t))=3t2+3t4f(r(t)) = 3t^2 + 3t^4
step 3
We already have ∥r′(t)∥=6t1+4t2+4t4\|r'(t)\| = 6t\sqrt{1 + 4t^2 + 4t^4}
step 4
The integral becomes ∫01(3t2+3t4)⋅6t1+4t2+4t4dt\int_0^1 (3t^2 + 3t^4) \cdot 6t\sqrt{1 + 4t^2 + 4t^4} dt
step 5
Simplify the integrand: ∫0118t3(1+t2)1+4t2+4t4dt\int_0^1 18t^3(1 + t^2)\sqrt{1 + 4t^2 + 4t^4} dt. According to the Asksia-LL calculator, this integral evaluates to 18
Answer
The integral ∫Γfds\int_{\Gamma} f ds is 18.
Key Concept
Arc Length and Line Integral Calculation
Explanation
The length of a curve and the line integral of a function over a curve can be computed using parametric equations and their derivatives. The Asksia-LL calculator provides the necessary integrals to find these values.
Solution by Steps
step 1
The gradient of the function f(x,y)=xyf(x, y) = xy is given by ∇f=(∂f∂x,∂f∂y)=(y,x)\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (y, x)
step 2
At the point p0=(1,1,1)p_{0} = (1, 1, 1), the gradient is ∇f(1,1)=(1,1)\nabla f(1, 1) = (1, 1)
step 3
The normal vector to the graph G(f)G(f) at p0p_{0} is (1,1,−1)(1, 1, -1)
step 4
The parametrization of the line L0L_{0} orthogonal to G(f)G(f) at p0p_{0} is given by L0(t)=(1,1,1)+t(1,1,−1)=(1+t,1+t,1−t)L_{0}(t) = (1, 1, 1) + t(1, 1, -1) = (1 + t, 1 + t, 1 - t)
Answer
The parametrization of the line L0L_{0} is L0(t)=(1+t,1+t,1−t)L_{0}(t) = (1 + t, 1 + t, 1 - t).
Question 2: Compute the equation of the tangent plane P0P_{0} of G(f)G(f) at the point p0p_{0}
step 1
The equation of the tangent plane to the graph G(f)G(f) at the point p0=(1,1,1)p_{0} = (1, 1, 1) is given by z=f(x,y)+∇f⋅((x,y)−(1,1))z = f(x, y) + \nabla f \cdot ((x, y) - (1, 1))
step 2
The gradient ∇f(1,1)=(1,1)\nabla f(1, 1) = (1, 1)
step 3
The equation of the tangent plane is z=1+1(x−1)+1(y−1)=x+y−1z = 1 + 1(x - 1) + 1(y - 1) = x + y - 1
Answer
The equation of the tangent plane P0P_{0} is z=x+y−1z = x + y - 1.
Question 3: Compute the area A(G(f))A(G(f)) of the graph of ff
step 1
The area of the graph of ff over the domain DD is given by the double integral A(G(f))=∬D1+(∂f∂x)2+(∂f∂y)2 dAA(G(f)) = \iint_{D} \sqrt{1 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2} \, dA
step 2
The partial derivatives are ∂f∂x=y\frac{\partial f}{\partial x} = y and ∂f∂y=x\frac{\partial f}{\partial y} = x
step 3
Therefore, 1+(∂f∂x)2+(∂f∂y)2=1+y2+x21 + \left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 = 1 + y^2 + x^2
step 4
The area integral becomes A(G(f))=∬D1+x2+y2 dAA(G(f)) = \iint_{D} \sqrt{1 + x^2 + y^2} \, dA
step 5
The domain DD is the region {(x,y)∈R2:y≥0,x2+y2≤4}\{(x, y) \in \mathbb{R}^2 : y \geq 0, x^2 + y^2 \leq 4\}
step 6
Converting to polar coordinates, x=rcos⁡θx = r \cos \theta, y=rsin⁡θy = r \sin \theta, and dA=r dr dθdA = r \, dr \, d\theta
step 7
The integral becomes A(G(f))=∫02π∫021+r2 r dr dθA(G(f)) = \int_{0}^{2\pi} \int_{0}^{2} \sqrt{1 + r^2} \, r \, dr \, d\theta
step 8
Evaluating the integral, we get A(G(f))=2π[13(1+r2)3/2]02=2π(13(53/2−1))=2π3(55−1)A(G(f)) = 2\pi \left[ \frac{1}{3} (1 + r^2)^{3/2} \right]_{0}^{2} = 2\pi \left( \frac{1}{3} (5^{3/2} - 1) \right) = \frac{2\pi}{3} (5\sqrt{5} - 1)
Answer
The area A(G(f))A(G(f)) of the graph of ff is 2π3(55−1)\frac{2\pi}{3} (5\sqrt{5} - 1).
Key Concept
Parametrization, Tangent Plane, and Area Calculation
Explanation
The problem involves finding the parametrization of a line orthogonal to a graph, the equation of a tangent plane at a point, and the area of the graph using integrals.
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