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Numerade Educator



Problem 29 Hard Difficulty

For the vector field $\mathbf{F}$ and curve $C$, complete the following:
a. Determine the points (if any) along the curve C at which the vector field $\mathbf{F}$ is tangent to $C$.
b. Determine the points (if any) along the curve C at which the vector field $\mathbf{F}$ is normal to $C$
c. Sketch $C$ and a few representative vectors of $\mathbf{F}$ on $C$.

$$\mathbf{F}=\langle x, y\rangle ; C=\{(x, y) ; x=1\}$$


a. (1,1)
b. (1,0)
c. graph


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Video Transcript

This is a vector field in a specific her. This question asks us to find the points where f is tangent to the curb points where f is normal to the curb. And then we want to sketch this curve with some sample F defectors on it. Our vector field is extreme A y our curve is a set of point X come a Y in the plane, such that actually is equal one. This would just be the line of points where X is 1/4 1 It would be a vertical line, but before we can get into that, we have two parameter right curve. So this would just be see of tea is equal 21 because X is always equal. One comedy y is equal to teach. This makes sense because X doesn't very so we're gonna have a constant there. But then why Berries? And so this tea would go from negative infinity to positive infinity here because it is going all the way up and all the way down. So let's find our set of points where f is tangent to the curve. So to do that, we need to take the dot product of our vector f and then our tangent back there, See and see if it is plus or minus one, as that would indicate these vectors of parallel meaning they would be the same, at least in the same direction. So we have to take see prime of tea, just the derivative of those components of respective team in order to get the tangent vector. So it'll be zero comma one and then we'll take the dot product our vector X comma y well, after a present in terms of tea by the equalities here. So we have X equals one y equals one comity started with the vector zero comma one equals plus or minus one. So during the stock product, we would have t is equal to plus or minus one So the values t equals won t was negative. One will give rise to these points. What are these points going to be going to look at this key for them. So we have one comma one and then one common negative one. So these are our points. Where are vector field F is tangent to the curve. For part B, We defined points where it is normal to the curb. So we would said Composed CFT because we put T and instead of x y started with C Prime of team equal to zero. Do we need to find these? Well, we already found the dot product. It's just going to be t so we're finding the points T equals zero where f is normal. This would just be the point. One common zero. So this is the only point where f is normal to the curtain. So lastly, let a sketch the curve and some that there's on it. The curve is just the line. X equals. One will say it's this line here. So let's pick the point one common zero Where is this doctor going? Well, our vector is going to be X comma. Why? The same as a point. So we go this way this makes sense and performs a right angle. And this is the 0.1 times zero, we have the normal vector. So now we'll go to one comma one and one common negative one. Well, the one comma one point goes

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