Writing the specter valued function, parametric lee. So if we write X equals eat the two T and Y equals E. And let's get this into a form that is a bit more familiar to us. This question is asking us to sketch a plane curve for the vector equation R. So this is saying that the rate of change at this point Is equivalent to going over two and up three. And then Our prime of one Is given by 2:03. But here this sits in the top half so we can go ahead and write Are one right there. Here, we still would have gotten one and the y component will have the negative one giving us the bottom half of the curve. And you see if we had plugged in T equals negative one. Asks us to sketch the position vector and the tangent vector for the given value of T. And part B is asking us to find the derivative of this function, which we do by just taking the derivative of each individual component, which will be to T three T squared. So here is what this plane curve looks like, It's curving slightly. And we will get two cubed or eight 345678. If we try X equals four, which is the next number, that kind of square roots nicely.
#Mumble penguin plus
If we try x equals one and we will get y equals plus or -1. So now for instance if we try applauding X equals zero equals 0. And so each value of X is going to give us two different values of why on the right hand side. Because when we take the square root from the denominator that will give us a plus or minus and then when you cube that plus or minus, that uh positive or negative sign will be retained when you cubit. We should also make sure that we write plus or minus, X equals or plus or minus X to the three halves. And so when we substitute in X equals C squared, we get y equals X to the three halves, notably. And you can see you get that by writing Y equals t cubed which is T to the three halves or rather t squared to the three halves.
And from this we can just plug in Y equals X two the three halves. So we can write out the equations X equals t squared from the X component.Īnd why equals T. All right, so we need to make some substitution to express this vector valued function as a function of X and Y. Okay, so let's go ahead and sketch that here. Here we are asked to sketch the plane curve for the vector valued function R f t equals t squared comets. And so we will draw that here and that is all that the question is asking for. So we are told that it is one common negative too. And then This our prime of negative one here is telling us what the rate of change is at that point. This first one are of negative one is giving us the vector too the point on the curve. We can go ahead and sketch this on the graph here. So when t equals negative one are negative, one is going to be negative three comma the one squared plus one is two. And then finally part C asks us to sketch the position vector R. Component here is just one meaty derivative of the Y component is to t. Heart be is asking us to find the derivative of this vector valued function which is as simple as taking the derivative of each component one at a time. Then we go up by one on each side and continue curving up like that. Okay because there's a plus two inside the parentheses we shift left to negative two and because there's a plus one we shift up to one so the apex of the problem is right there. And so we can go ahead and just write down what that curve looks like. And this gives us a relationship between X&Y for this parametric curve. So we have Y equals X plus two squared Plus one. So we can change this, X equals two plus two to be T equals X plus two.
And from here we're going to solve the system of equations. X equals to my t minus two from the X.Ĭomponent and why equals t squared plus one from the Y component. We can do this by writing out two equations. First we are asked to sketchy playing curb with the given vector equation. So here we are asked to find a few things related to the specter valued function harv t equals t minus two squared plus one.
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