Question

8.) Evaluate the line integral $\int_C xe^{2y}ds$, where C is the first quarter of the unit circle oriented in the counter-clockwise direction from (1,0) to (0,1). 9.) Find a potential function f such that $\nabla f = F$ where F = 10.) Using F = $(2xe^{x^2+y^3} + 3x^2)i + (3y^2e^{x^2+y^3} + 2ycos(y^2))j$. the line segment joining (0,2) and (4,0)

          8.) Evaluate the line integral $\int_C xe^{2y}ds$, where C is the first quarter of the unit circle oriented in the counter-clockwise direction from (1,0) to (0,1).
9.) Find a potential function f such that $\nabla f = F$ where F = 
10.) Using F = $(2xe^{x^2+y^3} + 3x^2)i + (3y^2e^{x^2+y^3} + 2ycos(y^2))j$.
the line segment joining (0,2) and (4,0)

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8.) Evaluate the line integral xe^2yds, where C is the first quarter of the unit circle oriented in the counter-clockwise direction from (1,0) to (0,1).
9.) Find a potential function f such that ∇ f = F where F =
10.) Using F = (2xe^x^2+y^3 + 3x^2)i + (3y^2e^x^2+y^3 + 2ycos(y^2))j.
the line segment joining (0,2) and (4,0)

Added by Gina G.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Lauren Shelton

11/17/2022

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Transcript

00:01 So if i'm going from the second derivative to the original, then this is what's known as the anti -derivative, or i'm taking the integral of each derivative.

00:15 Then i'm taking the anti -integral of each to get to the next.

00:17 So i'm going to take the anti -derivative here to get to f -prime of theta.

00:23 So this means i'm going to take the anti -derivative with respect to d -theta, anti -derivative of sign.

00:31 What's the opposite of sign? well, recall that the derivative of sine of x is cosine of x, and the derivative of cosine is negative sine of x.

00:50 So if i'm going backwards, the anti -derivative of sign, technically this would mean the derivative of negative cosine of x is negative negative sign of x or sign of x, which means the anti -derivative of sign is negative cosine.

01:04 So this will be negative cosine of theta.

01:07 Plus antide derivative of cosine, going backwards, is just positive sign.

01:13 And this equals plus a constant c.

01:17 And this is why i'm given f prime of 0 equals 2.

01:22 So this means when i plug in 0 into my function, so negative cosine of 0 plus sine of 0, i don't know what c is yet, equals my output 2.

01:33 What's cosine at 0? well, cosine of 0 is just 1.

01:38 So this would be negative 1.

01:40 Sided 0 is just 0 plus c...

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