R50 l3m f50mhz vswr a 218 b 128 c 1 d none prove how i know...

R=50Ω, L=3m, f=50MHz, VSWR=? (a) 2.18 (b) 1.28 (c) 1 (d) None Prove how. I know that the known variables are missing, but just want to ask what is the use of R, L, and F in finding VSWR?

Transcript

00:01 And this question, which of the following is the equation of the tangent line to this particular curve at x equal to minus 1? so the tangent line is y is equal to 2 minus x square times x cube.

00:14 For the tangent line, first we need to find the coordinate.

00:17 So x is already given to minus 1 as minus 1.

00:21 So what is the corresponding y coordinate? that will be 2 minus minus 1 whole square whole cube.

00:26 So that will be 2 minus 1 because minus 1 square is 1 and 2 minus 1 is 1.

00:30 So that will become one cube, which is just one.

00:33 So that will be minus one one.

00:36 And then we find the derivative.

00:38 So dy over dx will be three comes down.

00:42 We have two minus x square whole square.

00:44 And then inside is d over dx of two minus x square.

00:49 So that will become three times two minus x square whole square.

00:53 And the derivative of this following function is going to be zero minus two x using the chain rule or power chain rule.

01:01 So 0 minus 2x is minus 2x and 3 times minus 2x is minus 6x.

01:05 So that will be minus 6x times 2 minus x square whole square.

01:09 That is the value of dy over dx.

01:12 So the value of dy over dx at x equal to minus 1 because that's what we have to find the slope.

01:18 So dy over dx at x equal to minus 1 is going to be negative 6 times minus 1.

01:24 And then 2 minus 1 is 1...

Question

R=50Ω, L=3m, f=50MHz, VSWR=? (a) 2.18 (b) 1.28 (c) 1 (d) None Prove how. I know that the known variables are missing, but just want to ask what is the use of R, L, and F in finding VSWR?

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