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[True/False] See attached image_ (True / False] Consider the model delined by Yi Bo+81T,+u;. To find the parameter estimates- set the derivative of RSS t0 zero and ...

Question

[True/False] See attached image_ (True / False] Consider the model delined by Yi Bo+81T,+u;. To find the parameter estimates- set the derivative of RSS t0 zero and solve for bp and b1, S0 it mnust be that ORSS 2NBu + 2(81T, Vi) 06TrueFalse

[True/False] See attached image_ (True / False] Consider the model delined by Yi Bo+81T,+u;. To find the parameter estimates- set the derivative of RSS t0 zero and solve for bp and b1, S0 it mnust be that ORSS 2NBu + 2(81T, Vi) 06 True False



Answers

True-False Determine whether the statement is true or false. Explain your answer. If $C$ is the graph of a smooth vector-valued function $\mathbf{r}(t)$ in 2 -space, then the unit tangent vector $\mathbf{T}(t)$ to $C$ is orthogonal to $\mathbf{r}(t)$ and points in the direction of increasing parameter.

Okay, tour faults. To find the derivative of a vector function R equals R of T find the derivative of each of its components. Well, that is true. Once you have that in your i j and K, you can individually take the derivative of each.

Hey, we have a true or false question here. If a vector function R equals RFP is differentiable at T. Some not. And where tisa not is between A and B. And point. Piece of not is on the curb traced out by R equals R. M. T corresponding to T equals C. Okay, there is a lot of stuff going on here. But the main thing here it says that the tangent line to the curve that contains P. At T. Some not has the direction of our. So you have to remember your tangent line is all about your derivative. When they say that the direction of our well you have a vector function and that vector is always from the origin. So if you're describing your vector are that would be what I've just drawn. Where your tangent line is the line that is tangent. So um the main thing here is uh it does contain the point but the direction is wrong. So this is false.

In this problem, we will cover local linearity. So to solve this problem, we have to refer back to the tangent plane approximation, which I have written here in green and we see that we will have to double check to make sure that this tangent plane approximation is correct. So we have to find the partial derivatives With respect to X&Y at the .01, and also the value of the function at 01. So we'll begin by finding f. 01. And if we plug that in, we will get that. This is just one Now to find the partial derivative of respective X. And that's just taking the derivative of the function holding the variable. Why fixed? So keeping Y in front we have E. To the X. Squared. And we also want to multiply by the derivative of the exponents itself. So two X. And that gives us two x. Y. Each of the X square. If we were to plug in the .01 into this partial derivative, we would get that. The whole thing is zero. And now we move on lastly to the partial derivative with respect to Y. And that's just take more function, the derivative of the function holding the X. Variable fixed now. And when we take the derivative with respect to why we're just going to get E. To the X. Square. And if we were to plug in the .01 into this partial derivatives, we would get one because each of the zero is just one. So now using our information, we can write out our approximation. So we have F X. Y is approximately equal. So one plus. Since the partial derivative respect the X. Is zero, there's not gonna be any X terms in this equation. So we can move right along to the Y. And we see that the partial derivative is one. So we'll just have why my just one here and of course the positive one and negative one cancels out. So this can be simplified to F. Of X. Y approximately equal. So why? That means that the statement given and blue is true.

For this problem we are asked if the following statement is true or false. That being that for a functions that equals F of X. Y. The partial derivative Fx of Excellent. Why not? Is the directional derivative at X? Not? Why not? In the direction of I had. So we can turn to the definition of our directional derivative. Well, we know that it's going to be the dot product of the gradient of F with our um with the vector that we're taking, the direction of which in this case it's the gradient of F with a hat. And that means that it's just going to be the partial derivative with respect to X. And are evaluated in this case at X. Not Why not started with I. Hat. Which means that it's just going to be that fxx not Why not? I had. So the one thing and we need to be careful of here, is that actually uh I take that back. The only thing that we can really say here is that we see that this is definitely true.


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