Need help, URGENT maths question: (1 Viewer)

1008 · Jun 3, 2016

InteGrand said:
Well it's because the notations are both referring to the partial derivative of ƒ wrt its first designated variable. As seanieg89 said:

.

Thanks so much InteGrand for clearing that up. Yeah I knew that x and u are dummy variables, but I couldn't translate my understanding in my previous answer, I guess substituting u and v indeed make the notation a little less chaotic lol... How would you do this one?

$\\\text{(i) a) Find }\int_{1}^{2} x^{-2}dx\text{ by using Riemann sums. Use a uniform partition but set }x^*_i=\sqrt{x_{i-1}x_i},\text{ and use }\\\frac{1}{\alpha(\alpha +1)} = \frac{1}{\alpha}-\frac{1}{\alpha+1}\\\text{b) Find }\int_{1}^{b}x^{-1/2}dx\text{ by using Riemann sums. Use a uniform partition but set }\\x^*_i=\left((\sqrt{x_{i-1}}+\sqrt{x_i})/2\right)^2\\$

InteGrand · Jun 3, 2016

1008 said:
Thanks so much InteGrand for clearing that up. Yeah I knew that x and u are dummy variables, but I couldn't translate my understanding in my previous answer, I guess substituting u and v indeed make the notation a little less chaotic lol... How would you do this one?

$\\\text{(i) a) Find }\int_{1}^{2} x^{-2}dx\text{ by using Riemann sums. Use a uniform partition but set }x^*_i=\sqrt{x_{i-1}x_i},\text{ and use }\\\frac{1}{\alpha(\alpha +1)} = \frac{1}{\alpha}-\frac{1}{\alpha+1}\\\text{b) Find }\int_{1}^{b}x^{-1/2}dx\text{ by using Riemann sums. Use a uniform partition but set }\\x^*_i=\left((\sqrt{x_{i-1}}+\sqrt{x_i})/2\right)^2\\$

What have you gotten up to so far? The first part of doing part (a) (just setting up the Riemann sum) should be rather routine. What did you get for the Riemann sum?

1008 · Jun 6, 2016

Alright, I've got another question:

$\\\text{1) Let }A\text{ be a }5\times 4\text{ matrix with real entries and let }\textbf{b }\text{ be the second column of }A.\text{ It is }\\\text{given that }\textbf{Ax = 0}\text{ has the general solution }\\\textbf{x}=\mu(1,2,2,1)^T,\mu\in\mathbb{R} \\\text{Find the following explicity}\\\text{a) one explicit solution to }x_p\text{to }\textbf{Ax}_p=\textbf{b};\\\text{b) the general solution to }\textbf{Ax = b}\\\text{c) Give a geometric interpretation of the general solution}\\\\\\\text{2) Consider the surface }S\text{ defined by }z=x^2+2xy^2+y^4,\text{ and the point }P=(2,1,9)\text{ on it.}\\\text{a) Find the equation of the tangent plane to }S \text{ at }P.\\\text{b) Find the equation of the normal }S\text{ at }P.$

For the second question, in the answers they implicitly differentiate the equation wrt x and y and then the normal is the vector: $\begin{pmatrix} \frac{\partial x}{\partial z} \\\frac{\partial y}{\partial z}\\-1\end{pmatrix}$ . I just wanna know why they always use -1 for the third element of this vector, and is there an alternate way of doing this question? And can we pick any number instead of -1?

EDIT: Sorry for being cold and not replying to your previous post IG, I was heading to uni when you replied, and I had work after. And then I've been busy with exams prep D: I'm still working on those q's, but I'm finishing off other subjects first

InteGrand · Jun 6, 2016

1008 said:
Alright, I've got another question:

$\\\text{1) Let }A\text{ be a }5\times 4\text{ matrix with real entries and let }\textbf{b }\text{ be the second column of }A.\text{ It is }\\\text{given that }\textbf{Ax = 0}\text{ has the general solution }\\\textbf{x}=\mu(1,2,2,1)^T,\mu\in\mathbb{R} \\\text{Find the following explicity}\\\text{a) one explicit solution to }x_p\text{to }\textbf{Ax}_p=\textbf{b};\\\text{b) the general solution to }\textbf{Ax = b}\\\text{c) Give a geometric interpretation of the general solution}\\\\\\\text{2) Consider the surface }S\text{ defined by }z=x^2+2xy^2+y^4,\text{ and the point }P=(2,1,9)\text{ on it.}\\\text{a) Find the equation of the tangent plane to }S \text{ at }P.\\\text{b) Find the equation of the normal }S\text{ at }P.$

For the second question, in the answers they implicitly differentiate the equation wrt x and y and then the normal is the vector: $\begin{pmatrix} \frac{\partial x}{\partial z} \\\frac{\partial y}{\partial z}\\-1\end{pmatrix}$ . I just wanna know why they always use -1 for the third element of this vector, and is there an alternate way of doing this question? And can we pick any number instead of -1?

EDIT: Sorry for being cold and not replying to your previous post IG, I was heading to uni when you replied, and I had work after. And then I've been busy with exams prep D: I'm still working on those q's, but I'm finishing off other subjects first

No worries.

$\noindent 1) a) A particular solution is $\vec{x}_p = (0,1,0,0) = \vec{e}_2$. In fact in general, for ANY $m\times n$ matrix $A$, we have $A \vec{e}_j = \text{ column } j\text{ of }A$, where $\vec{e}_j$ is the $j$th standard basis vector for $\mathbb{C}^n$ or $\mathbb{R}^n$, for $j=1,\ldots ,n$ ($n$ being the no. of \emph{columns}$).$

$\noindent b) The general solution is $\vec{x} = \vec{x}_p + \vec{x}_h$, where $\vec{x}_h$ is the \emph{homogeneous solution}. This is always the case for ANY matrix equation $A \vec{x} =\vec{b}$ (and also generalises to any such equation involving a linear operator, e.g. linear differential equations on the space of smooth functions). So here, general solution is $\vec{x} = (0,1,0,0) + \mu (1,2,2,1)$, $\mu \in \mathbb{R}$.$

$\noindent c) Geometrically, the solution set is a line through $(0,1,0,0)$ parallel to $(1,2,2,1)$ in $\mathbb{R}^4$.$

InteGrand · Jun 6, 2016

1008 said:
Alright, I've got another question:

$\\\text{1) Let }A\text{ be a }5\times 4\text{ matrix with real entries and let }\textbf{b }\text{ be the second column of }A.\text{ It is }\\\text{given that }\textbf{Ax = 0}\text{ has the general solution }\\\textbf{x}=\mu(1,2,2,1)^T,\mu\in\mathbb{R} \\\text{Find the following explicity}\\\text{a) one explicit solution to }x_p\text{to }\textbf{Ax}_p=\textbf{b};\\\text{b) the general solution to }\textbf{Ax = b}\\\text{c) Give a geometric interpretation of the general solution}\\\\\\\text{2) Consider the surface }S\text{ defined by }z=x^2+2xy^2+y^4,\text{ and the point }P=(2,1,9)\text{ on it.}\\\text{a) Find the equation of the tangent plane to }S \text{ at }P.\\\text{b) Find the equation of the normal }S\text{ at }P.$

For the second question, in the answers they implicitly differentiate the equation wrt x and y and then the normal is the vector: $\begin{pmatrix} \frac{\partial x}{\partial z} \\\frac{\partial y}{\partial z}\\-1\end{pmatrix}$ . I just wanna know why they always use -1 for the third element of this vector, and is there an alternate way of doing this question? And can we pick any number instead of -1?

EDIT: Sorry for being cold and not replying to your previous post IG, I was heading to uni when you replied, and I had work after. And then I've been busy with exams prep D: I'm still working on those q's, but I'm finishing off other subjects first

$\noindent 2) I think you've mixed up the normal vector formula (or typoed, since you just put the $z$'s in the `denominators' instead of `numerators'). A normal to the graph of $z=f\left(x,y\right)$ at a given point $\vec{a}$ is $\vec{N} = \begin{bmatrix} - \frac{\partial f}{\partial x} \\ -\frac{\partial f}{\partial y}\\1\end{bmatrix}\equiv \begin{bmatrix} - f_x \\ -f_y\\1\end{bmatrix} $, where the partials are evaluated at the point $\vec{a}$ (and will be throughout this post). (Or equivalently, we can use $\begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y}\\-1\end{bmatrix}$.)$

$\noindent One way to see why $ \begin{bmatrix} -f_x \\ -f_y\\1 \end{bmatrix}$ is a normal to the surface is to note that for a smooth surface, the following are tangent vectors: $ \begin{bmatrix} 1 \\ 0\\ f_x \end{bmatrix}$ and $ \begin{bmatrix}0 \\ 1\\ f_y \end{bmatrix}$. Then taking the cross product of these vectors gives us a normal, since those two vectors are parallel to the tangent plane. If you cross these, you should get $\begin{bmatrix} - f_x \\ - f_y\\1\end{bmatrix}$. So no, we can't just pick a random number for the third component, because there's a reason why it's that (came from the cross product). Plus, changing the last element of the vector would generally change the direction of the vector, so it would no longer be a normal if we did that.$

leehuan · Jun 6, 2016

InteGrand said:
No worries.

$\noindent 1) a) A particular solution is $\vec{x}_p = (0,1,0,0) = \vec{e}_2$. In fact in general, for ANY $m\times n$ matrix $A$, we have $A \vec{e}_j = \text{ column } j\text{ of }A$, where $\vec{e}_j$ is the $j$th standard basis vector for $\mathbb{C}^n$ or $\mathbb{R}^n$, for $j=1,\ldots ,n$ ($n$ being the no. of \emph{columns}$).$

$\noindent b) The general solution is $\vec{x} = \vec{x}_p + \vec{x}_h$, where $\vec{x}_h$ is the \emph{homogeneous solution}. This is always the case for ANY matrix equation $A \vec{x} =\vec{b}$ (and also generalises to any such equation involving a linear operator, e.g. linear differential equations on the space of smooth functions). So here, general solution is $\vec{x} = (0,1,0,0) + \mu (1,2,2,1)$, $\mu \in \mathbb{R}$.$

$\noindent c) Geometrically, the solution set is a line through $(0,1,0,0)$ parallel to $(1,2,2,1)$ in $\mathbb{R}^4$.$

Holy crap. This took me one whole semester to finally understand what is going on.
_______

Also @1008

I'm pretty sure Ian has the proof of why the normal vector is IG's correction in his lecture slides.

Need help, URGENT maths question: (1 Viewer)

1008

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InteGrand

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1008

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InteGrand

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InteGrand

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leehuan

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