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Problem 21 Hard Difficulty

Evaluate the line integral $\int_{C} \mathbf{F} \cdot d \mathbf{r},$ where $C$ is given by the vector function $\mathbf{r}(t) .$
$$\mathbf{F}(x, y, z)=\sin x \mathbf{i}+\cos y \mathbf{j}+x z \mathbf{k}$$
$$\mathbf{r}(t)=t^{3} \mathbf{i}-t^{2} \mathbf{j}+t \mathbf{k}, \quad 0 \leqslant t \leqslant 1$$

Answer

$\int_{C} \mathbf{F} \cdot d \mathbf{r}=\frac{6}{5}-\cos (1)-\sin (1)$

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Video Transcript

All right, so in this question, were asked to determine the lining to grow. And we're given that big for the force, or the field is sine X. I have must coincide like you had a sexy key. Have. And we're also given that the position vector is two cubed. I have minus teas. Where'd you have plasticky As Ticos from Syria. So now we want to write in terms of tea. So instead of X, we're gonna plug in T Cube instead of why you were gonna put in negative. Do you square that instead of Z we're gonna looking So, um, sign of X becomes sign cube. Uh, co sign of, uh, why becomes co sign of Negativity Square. And then, uh, it's xz becomes cute times. All right, so, uh, there is something we can simplify. So, for example, we know that co sign is an even function. So co sign of negativity. Swear to just coast side of the square. The negative groups anti cube times t is t to the power for So now we can write f in terms of t a scientific cubed co scientist squared and T to the power for and now, Artie, we can read Are the institute common negativity squared comity and our prime of team. We just take the derivative of our with respect. So cute becomes pretty squared. Um, negative t swear. Becomes negative to t and he becomes one. That's our primary. All right, so now the line into growth is just We can write it as into growth of a to be of FFR 15 about product with our primary teeth. And, uh, our upper limb are lower limit is zero in an upper limit is one just like its specified above. Ah, now we find a dot product of these two vectors. Uh, and this is simple. So we just multiply her first term here by the first term second term here by the second term here, their term here by the searcher. And then we get this integral right over here. All right, so now we have three terms, and two of the Sturm's involves science co site. But what we notice is that we have sine of t Q And on the outside, we have three squared. So we realized that we can use U substitution. So if we let you be cute than use just three t squared and for the co signs swear Reckon that w beauty square than the CW's Just to t again, we're gonna use U substitution in both cases. All right, so now we have the integral from 0 to 1 of signs in new and we have the integral like also from 0 to 1 of coastline wdw If you pay attention, he's the limits. Although so the limits for the 1st 1 the imprint of Lord limit of the integral remains from 01 Why? Because we want to convert. T of t is equal to zero. We get you is equal to zero t equals one again. You is gonna be one cube of one. Same thing for a W. He is zero. We get w w Z equals zero squared or zero and when he is one, we have believes equal to one squared. But I know for the last term we just keep it in terms. All right, So the integral of sine you the users negative co sign on the integral coastline W's just signed w on the integral of t to the power for just a to the power five divided. Okay, now we were going to subtract. So we have our upper lord living. We're gonna subtract the upper limit from the living. So we get negative. Co sign you Bus co sign of zero but co signing 00 minus sign of one minus Sign of zero. Oh, sorry. Co Sign of zero. It is one by apologize. Sign of zero is zero. And then here, this zero to the power of five. Divided by five, which is zero. So you have negative co sign zero A sorry, negative coastline one minus sign one. And then we have 1/2 and one. So 1/2 plus 1/5. No, no one have. Sorry, it's one plus one. So we have one plus one, which is just 6/5. And then we have negative course. Sign one minus sign. So our final answer is just negative. Go side one. Minus sign. One plus 6 50

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