Evaluate the line integral, where C is the given curve. Integral C x sin⁡ y ds, C is the line segment from (0, 3) to (4, 6)

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Nick Camelot · Accepted Answer

Step 1. Parameterize the curve C:Let x=x(t) and y=y(t) be the parametric equations for the line segment. We can choose t to vary from 0 to 1.The x-component of the line segment can be parameterized as x(t)=4t.The y-component of the line segment can be parameterized as y(t)=3+3t.Step 2. Calculate the derivatives:Compute the derivatives dxdt and dydt:dxdt=4 and dydt=3.Step 3. Calculate the line element ds:The line element ds is given by ds=(dxdt)2+(dydt)2dt.Substituting the values, we get:ds=42+32dt=5dt.Step 4. Substitute the parametric equations and line element into the integral:The line integral becomes:∫Cxsinyds=∫01(4t)sin(3+3t)·5dt.Step 5. Simplify and evaluate the integral:We can simplify the integral as follows:∫Cxsinyds=20∫01tsin(3+3t)dt.At this point, you can evaluate the integral numerically using numerical integration techniques or software.

Eliza Beth13 · Accepted Answer

Result:∫Cxsinyds=20∫01tsin(3+3t)dtSolution:Let&#039;s denote the parameter t that ranges from 0 to 1 along the line segment C. We can express the x and y coordinates of C in terms of t as follows:x=x(t)=(1−t)·x0+t·x1=(1−t)·0+t·4=4ty=y(t)=(1−t)·y0+t·y1=(1−t)·3+t·6=3+3tNext, we need to find the derivative of x(t) and y(t) with respect to t. The derivative represents the tangent vector to the curve C:dxdt=4dydt=3The differential arc length ds along the curve C can be calculated using the Pythagorean theorem:ds=(dxdt)2+(dydt)2dt=42+32dt=5dtNow we can rewrite the line integral in terms of the parameter t:∫Cxsinyds=∫01(4t)sin(3+3t)·5dtSimplifying this expression, we have:∫Cxsinyds=20∫01tsin(3+3t)dt

madeleinejames20 · Accepted Answer

To evaluate the line integral ∫Cxsinyds, where C is the line segment from (0,3) to (4,6), we can parametrize the curve C as 𝐫(t)=(4t,3+3t) for 0≤t≤1. Then the line integral becomes:∫Cxsinyds=∫01(4t)sin(3+3t)|𝐫′(t)|dtwhere |𝐫′(t)| denotes the magnitude of the derivative of 𝐫(t). Now, we differentiate 𝐫(t) with respect to t to obtain 𝐫′(t)=(4,3).Substituting these values, the line integral simplifies to:∫Cxsinyds=∫01(4t)sin(3+3t)|(4,3)|dt=∫017tsin(3+3t)dtWe can now evaluate this integral to obtain the final result.

Evaluate the line integral, where C is the given curve. Integral C x \sin

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