Determine whether each red path in the vector field is positive, negative, or zero

Mathematics

2 answers:

ololo11 [35]3 years ago

7 0

Please answer please please thank you thank

anygoal [31]3 years ago

6 0

If the path is perpendicular to the field it is zero.

If the path is along the field it is positive or negative depending on it's direction.

See the attached picture.

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Find sin4a and cos4α, if tanα=3

Mashcka [7]

Answer:

<h2>See the explanation.</h2>

Step-by-step explanation:

Sin2α = 2SinαCosα.

Cos2α = (Cosα)^2 - (Sinα)^2.

Sin4α can be written as Sin2(2α).

Similarly Cos4α can be written as Cos2(2α).

$Sin 4\alpha = 2Sin 2\alpha \times Cos 2\alpha = 2Sin 2\alpha \times(Cos^{2} \alpha - Sin^{2} \alpha ) = 4Sin \alpha \times Cos \alpha \times (Cos^{2} \alpha - Sin^{2} \alpha )$

It is given that tanα=3.

Sinα(Sinα) = $\frac{9}{10}$ [ $Sec^{2} \alpha = 1 + 3^{2} = 10.\\Cos^{2} \alpha = \frac{1}{10} \\Sin^{2} \alpha = 1 - \frac{1}{10} = \frac{9}{10}$ ], the values either be both positive or both negative, as the value of tanα is positive only on first and third quadrant.

Sin4α = $4\times \sqrt{(\frac{9}{10} )} \times \sqrt{(\frac{1}{10} )} \times (\frac{1 - 9}{10} ) = -\frac{4\times8\times3}{100} = -\frac{96}{100}$ .

Cos4α = $Cos^{2} 2\alpha - Sin^{2} 2\alpha = (Cos^{2} \alpha - Sin^{2} \alpha )^{2} - Sin^2 {2\alpha } = (Cos^{2} \alpha - Sin^{2} \alpha )^{2} - (2Sin \alpha Cos \alpha )^{2} = (-\frac{8}{10} )^{2} - \frac{4\times9}{100} = \frac{64 - 36}{100} = \frac{28}{100} = \frac{7}{25}$