Does the vector function F––(t)=ti–+2tcostj–+2tsintk––𝐹_(𝑡)=𝑡𝑖_+2𝑡𝑐𝑜𝑠𝑡𝑗_+2𝑡𝑠𝑖𝑛𝑡𝑘_ lie on a paraboloid?
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Solution 1
The vector function F(t) = ti + 2tcostj + 2tsintk does not lie on a paraboloid.
A paraboloid is a type of surface in three-dimensional Euclidean space. It is described by a quadratic equation in three variables. The general form of a paraboloid is given by the equation z = x^2 + y^2.
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Does the vector function F––(t)=ti–+2tcostj–+2tsintk––𝐹_(𝑡)=𝑡𝑖_+2𝑡𝑐𝑜𝑠𝑡𝑗_+2𝑡𝑠𝑖𝑛𝑡𝑘_ lie on a paraboloid?
The domain of the vector function F––(t)=t−3−−−−√i–+ln(10−2t−−−−−−√)–j+1t−3√k––𝐹_(𝑡)=𝑡−3𝑖_+𝑙𝑛(10−2𝑡)_𝑗+1𝑡−3𝑘_ is given by the interval
Let F––(t)=t2i–+tj–+k––𝐹_(𝑡)=𝑡2𝑖_+𝑡𝑗_+𝑘_ and G––(t)=i–+tj–+t2k––𝐺_(𝑡)=𝑖_+𝑡𝑗_+𝑡2𝑘_ be two vector functions defined for each t∈R𝑡∈𝑅.
Let F––(t)=2ti–−5j–+t2k––𝐹_(𝑡)=2𝑡𝑖_−5𝑗_+𝑡2𝑘_ and G––(t)=(1−t)i–+1tk––𝐺_(𝑡)=(1−𝑡)𝑖_+1𝑡𝑘_ where t∈R+𝑡∈𝑅+.
The work done on a particle enacted to move from (0,0)(0,0) to (2,4)(2,4) along the parabola y=x2𝑦=𝑥2 by the force F––(t)=x2i–+y2j–𝐹_(𝑡)=𝑥2𝑖_+𝑦2𝑗_ is given by
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