First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

leehuan · Jun 16, 2016

Re: MATH1131 help thread

Two day-before-exam questions confusing me that I'm gonna chuck in this thread...

$$Q1: Consider the function $\phi(x,y)=x^2+\cosh{(x-1)}+y^3+y$

$$a) Find the equation of the tangent line to the curve $\phi (x,y)=4$ at the point $P(1,1)$

This part isn't hard with implicit differentiation but how would the "line" y=-x/2 + 3/2 be visualised? I guess, I mean in R³

$b) Find $\nabla \phi (1,1)$ and show that it is normal to the tangent line.$

This has me confused.
_______________

$$Q2: By using a corollary to the Mean Value Theorem, or otherwise, show that $\\ \cosh{x}>1+\frac{x^2}{2}$ for $x>0$

...The answers just have me confused. Like I follow it but don't get how they get to the final result. Mainly because they started on the second derivative and that scared me.

InteGrand · Jun 16, 2016

Re: MATH1131 help thread

leehuan said:
Two day-before-exam questions confusing me that I'm gonna chuck in this thread...

$$Q1: Consider the function $\phi(x,y)=x^2+\cosh{(x-1)}+y^3+y$

$$a) Find the equation of the tangent line to the curve $\phi (x,y)=4$ at the point $P(1,1)$

This part isn't hard with implicit differentiation but how would the "line" y=-x/2 + 3/2 be visualised? I guess, I mean in R³

$b) Find $\nabla \phi (1,1)$ and show that it is normal to the tangent line.$

This has me confused.
_______________

$$Q2: By using a corollary to the Mean Value Theorem, or otherwise, show that $\\ \cosh{x}>1+\frac{x^2}{2}$ for $x>0$

...The answers just have me confused. Like I follow it but don't get how they get to the final result. Mainly because they started on the second derivative and that scared me.

$\noindent 1. a) It's a line in $\mathbb{R}^2$. The curve is a function of two variables equated to a constant, something like how $x^2 + y^2 = 1$ is just a curve in 2D. So the tangent line is tangent to a curve in 2D, so the line is in 2-space.$

$\noindent b) Note $\phi (x,y) = x^2+\cosh{(x-1)}+y^3+y\Rightarrow \nabla \phi \left(x,y\right) = \left(2x+\sinh (x-1), 3y^2 +1\right)$. Sub. in $x=1,y=1$ to get the vector $\vec{v}\equiv \nabla \phi (1,1)$. Then show this is normal to the tangent line by dotting it with the vector $\left(1,m\right)$, where $m$ is the slope of the tangent line.$

$\noindent (Aside: Note that in general if $\vec{n}=(a,b)$ is a normal to a line in $\mathbb{R}^2$, then the line has equation $ax+by=c$, where $c$ is found by subbing in the known point on the line (completely analogous to planes in $\mathbb{R}^3$ with a given normal). Remembering that $\nabla \phi$ is perpendicular to the level sets of $\phi$, it is thus of no surprise that $\nabla \phi (1,1)$ is normal to the tangent line at $(1,1)$.)$

InteGrand · Jun 16, 2016

Re: MATH1131 help thread

leehuan said:
Two day-before-exam questions confusing me that I'm gonna chuck in this thread...

$$Q1: Consider the function $\phi(x,y)=x^2+\cosh{(x-1)}+y^3+y$

$$a) Find the equation of the tangent line to the curve $\phi (x,y)=4$ at the point $P(1,1)$

This part isn't hard with implicit differentiation but how would the "line" y=-x/2 + 3/2 be visualised? I guess, I mean in R³

$b) Find $\nabla \phi (1,1)$ and show that it is normal to the tangent line.$

This has me confused.
_______________

$$Q2: By using a corollary to the Mean Value Theorem, or otherwise, show that $\\ \cosh{x}>1+\frac{x^2}{2}$ for $x>0$

...The answers just have me confused. Like I follow it but don't get how they get to the final result. Mainly because they started on the second derivative and that scared me.

$\noindent 2. We can derive these similarly to the trig. ones. We can do it via either differentiation, or integration. Here it is done via integration. Let $x>0$ be fixed. Recall $\cosh t > 1$ for all $t>0$, so integrating this inequality from $0$ to $x$ gives $\sinh x - \sinh 0 > x$, i.e. $\sinh t > t$, $t>0$. Now integrate from $0$ to $x$ again: $\cosh x - \cosh 0 > \frac{x^2}{2}\Rightarrow \cosh x > 1 + \frac{x^2}{2}$, as $\cosh 0=1$. This completes the proof.$

$\noindent \textbf{Remark.} We can keep going arbitrarily often to generate the following results for any integer $n$: $\cosh x > 1+\frac{x^2}{2!}+\cdots + \frac{x^{2n}}{(2n)!}$ and $\sinh x > x + \frac{x^3}{3!}+\cdots + \frac{x^{2n-1}}{(2n-1)!}$, for $x>0$. If we want results for $x<0$, the cosh inequality is the same as all terms there are even functions of $x$, whilst the sinh inequality just flips (becomes $<$), since all functions are odd.$

leehuan · Jun 16, 2016

Re: MATH1131 help thread

$$Consider the point $A(1,0,-1,2)\in \mathbb R^4$ and the plane $\\ \mathcal P =\left\{\textbf{x} \in \mathbb R^4: \textbf{x} = \begin{pmatrix}2\\1\\1\\0\end{pmatrix} + \lambda \begin{pmatrix}0\\3\\-1\\0\end{pmatrix}+ \mu \begin{pmatrix} 0\\1\\3\\0 \end{pmatrix} \right\}$

$$a) Find a parametric vector equation of the plane which goes through the point $A$ and is parallel to the plane $\mathcal P$

Easy enough. The answer to this is just x=(1,0,-1,2)^T+lambda₁(0,3,-1,0)^T+lambda₂(0,1,3,0)^T

$$b) Find a parametric vector equation of a plane which goes though the point $A$, does not intersect $\mathcal P$ and is not parallel to $\mathcal P$

Bit confused as to how to do this question properly. I randomly guessed this one:

x=(1,0,-1,2)^T+mu₁(0,0,-1,0)^T+mu₂(0,1,3,0)^T

InteGrand · Jun 16, 2016

Re: MATH1131 help thread

leehuan said:
$$Consider the point $A(1,0,-1,2)\in \mathbb R^4$ and the plane $\\ \mathcal P =\left\{\textbf{x} \in \mathbb R^4: \textbf{x} = \begin{pmatrix}2\\1\\1\\0\end{pmatrix} + \lambda \begin{pmatrix}0\\3\\-1\\0\end{pmatrix}+ \mu \begin{pmatrix} 0\\1\\3\\0 \end{pmatrix} \right\}$

$$a) Find a parametric vector equation of the plane which goes through the point $A$ and is parallel to the plane $\mathcal P$

Easy enough. The answer to this is just x=(1,0,-1,2)^T+lambda₁(0,3,-1,0)^T+lambda₂(0,1,3,0)^T

$$b) Find a parametric vector equation of a plane which goes though the point $A$, does not intersect $\mathcal P$ and is not parallel to $\mathcal P$

Bit confused as to how to do this question properly. I randomly guessed this one:

x=(1,0,-1,2)^T+mu₁(0,0,-1,0)^T+mu₂(0,1,3,0)^T

Note that the first component of every vector in plane P is 2. So one way to guarantee we don't intersect with P is to make sure vectors in our new plane never have first component 2. And one way to do this is to make the first component of both direction vectors 0, which is what you did, so that's good. (I assume you can see why this ensures the planes won't intersect.)

For the non-parallelism part, note that the span of the direction vectors for P is like "R^2" with first and last component 0. I.e. It is {(0,x,y,0): x, y are real}. This is essentially because the first and last components of the direction vectors in P are both 0 (which is why their span has this property too), and the middle two components (components 2 and 3), if treated like two-tuples, form a basis for R^2. This means in summary that the span of the two direction vectors in P is precisely all vectors of the form (0,x,y,0).

We want our new plane to NOT have this span (since we want non-parallelism). Unfortunately, the direction vectors you chose have exactly the same span as before (so the two planes ARE parallel)! This is because the direction vectors you chose for your proposed plane have first and last component 0 and (0,-1) and (1,3) (the two middle two-tuples in your proposed plane's direction vectors) still span R^2. (I assume you can see why they do this.)

So instead an easy way to prevent the parallelism is to just make one of your current proposed direction vectors have non-zero fourth component (we want 0 first components to ensure first component in proposed plane never has value 2, thus guaranteeing non-intersection with P). So just change your first direction vector in the proposed plane to (0,0,-1,1) say (many possible choices).

Then with this choice of first direction vector (and same second one, which again could be many things), we have ensured the spans of the two planes are different. (Why? Because second plane has (0,0,-1,1) in the span of its direction vectors, which isn't in the first plane's one since this has 1 in its fourth component.)

leehuan · Jun 17, 2016

Re: MATH1131 help thread

The hardest question on the exam

Prove that if a matrix A and its inverse A^-1 both have only integer terms, then the determinant of the matrix must equal to 1 or -1 (any n by n matrix)

Sent from my iPhone using Tapatalk

InteGrand · Jun 17, 2016

Re: MATH1131 help thread

leehuan said:
The hardest question on the exam

Prove that if a matrix A and its inverse A^-1 both have only integer terms, then the determinant of the matrix must equal to 1 or -1 (any n by n matrix)

Sent from my iPhone using Tapatalk

Note it is clear that det(A) is a non-zero integer. (Non-zero 'cos it's invertible and integer as seen by the fact the determinant expansion contains all integer terms as A has only integer entries; one way to see this via Laplace expansion and an induction on the size of the matrix to show that the minors are all integers too. It'll be similarly easy to see from other definitions of determinants.) Say det(A) = k.

Since det(A^-1) = 1/(det(A)) for any invertible matrix A, we have det(A^-1) = 1/k. But similarly det(A^-1) must be an integer (as all of A^-1's entries are given to be integers), say K. So K = 1/k. Clearly the only way we can get integers k and K satisfying K = 1/k is if k = +/- 1 (‡). In other words, det(A) must be equal to 1 or -1, and the proof is complete.

(‡) Because if k is an integer with magnitude bigger than 1, then 1/k isn't an integer, but we already said 1/k = K is an integer.

leehuan · Jun 17, 2016

Re: MATH1131 help thread

Sorta frustrating because I swear in the exam I was headed in that direction but the I walked backwards

Sent from my iPhone using Tapatalk

leehuan · Jun 17, 2016

Re: MATH1131 help thread

This was the other hard one

$$A random event occurs at one of the $n$ times $t_k, k=1,2,\dots , n$ where $t_1<t_2< \dots < t_n$ and $n>1$. The probability that the event occurs at time $t_k$ is given by $p_k=\alpha \left(r^{k-1}-r^{k}\right)$ for $k=1,2, \dots , n$

$a) Define a suitable random $X$ for this problem, making it clear whether $X$ is a discrete or continuous random variable.$

I let X be the time that said event occurs, which is discrete given how we're considering t_i as a finite set

$\textbf{b}$) Determine the possible values of $r$ and for these valuse, find the $\alpha$ in terms of $r$

InteGrand · Jun 17, 2016

Re: MATH1131 help thread

leehuan said:
This was the other hard one

$$A random event occurs at one of the $n$ times $t_k, k=1,2,\dots , n$ where $t_1<t_2< \dots < t_n$ and $n>1$. The probability that the event occurs at time $t_k$ is given by $p_k=\alpha \left(r^{k-1}-r^{k}\right)$ for $k=1,2, \dots , n$

$a) Define a suitable random $X$ for this problem, making it clear whether $X$ is a discrete or continuous random variable.$

I let X be the time that said event occurs, which is discrete given how we're considering t_i as a finite set

$\textbf{b}$) Determine the possible values of $r$ and for these valuse, find the $\alpha$ in terms of $r$

For b), the pmf has to sum to 1 when summed from k = 1 to n. The sum is telescopic (factoring out alpha). The sum is: alpha*(1- r^n). So alpha = 1/(1 – r^n).

In terms of what values r can be, note p_k = alpha*r^(k-1) * (1-r). The pmf is such that p_k is non-negative for all k and sums to 1. This occurs iff r > 0, r =/= 1. (The reason that r can't be 0 or 1 since then all the probabilities are 0, so the pmf wouldn't sum to 1. Also, r can't be negative, since then the pmf values will alternate in sign (as n > 1) and then we'd have one of the probabilities be negative. Note r can be greater than 1, as then the pmf is still positive always as the alpha is negative and the 1-r is negative. The sum is still 1 of course by the choice of alpha.)

leehuan · Jun 17, 2016

Re: MATH1131 help thread

InteGrand said:
For b), the pmf has to sum to 1 when summed from k = 1 to n. The sum is telescopic (factoring out alpha). The sum is: alpha*(1- r^n). So alpha = 1/(1 – r^n).

In terms of what values r can be, note p_k = alpha*r^(k-1) * (1-r). The pmf is such that p_k is non-negative for all k and sums to 1. This occurs iff 0 < r < 1. (The reason that r can't be 0 or 1 since then all the probabilities are 0, so the pmf wouldn't sum to 1. Also, r can't be negative, since then the pmf values will alternate in sign (as n > 1) and then we'd have one of the probabilities be negative.)

Oh wow I might've been able to scab off some marks then.

leehuan · Jun 26, 2016

Re: MATH1131 help thread

Gonna dump this technicality question here as I don't know where to put it.

$a \le \lim_{n \to \infty}f(n) \le a$
$a < \lim_{n \to \infty}f(n) < a$

With the squeeze theorem, does it matter if the inequality is strict?

InteGrand · Jun 26, 2016

Re: MATH1131 help thread

leehuan said:
Gonna dump this technicality question here as I don't know where to put it.

$a \le \lim_{n \to \infty}f(n) \le a$
$a < \lim_{n \to \infty}f(n) < a$

With the squeeze theorem, does it matter if the inequality is strict?

It doesn't matter. I.e. if x(n) < ƒ(n) < y(n) for all positive integers n (or after some point at least), and x(n) and y(n) both tend to a as n → ∞, then ƒ(n) also tends to a; and this statement holds also if some of those inequality signs were strict.

Paradoxica · Jun 26, 2016

Re: MATH1131 help thread

leehuan said:
Gonna dump this technicality question here as I don't know where to put it.

$a \le \lim_{n \to \infty}f(n) \le a$
$a < \lim_{n \to \infty}f(n) < a$

With the squeeze theorem, does it matter if the inequality is strict?

It's irrelevant for the purposes of your mathematical analysis.

Like, say, 1/x tends to zero as x grows in size, but never is equal to zero for any real value x

However, since you are analysing the limit behaviour it is equal to zero in the limit.

Flop21 · Aug 19, 2016

Re: MATH1131 help thread

What does this mean

(∂^2*z)/∂y∂x

So I've seen the partial derivative sign and the second partial derivative sign/expression, but what does that mean^

Trebla · Aug 19, 2016

Re: MATH1131 help thread

Flop21 said:
What does this mean

(∂^2*z)/∂y∂x

So I've seen the partial derivative sign and the second partial derivative sign/expression, but what does that mean^

It's a second derivative where you partially differentiate wrt x then partially differentiate again wrt y.

Flop21 · Aug 19, 2016

Re: MATH1131 help thread

What do they want here, I don't understand

I always get really stuck on these online tutorials because of weird things I just can't do like that ^

InteGrand · Aug 19, 2016

Re: MATH1131 help thread

Flop21 said:
What do they want here, I don't understand

I always get really stuck on these online tutorials because of weird things I just can't do like that ^

Looks like they want you to give a non-zero solution for alpha, beta, gamma.

Flop21 · Aug 19, 2016

Re: MATH1131 help thread

InteGrand said:
Looks like they want you to give a non-zero solution for alpha, beta, gamma.

and how do I do that? Wouldn't you use back substitution, but I'm confused how since it has infinite solutions doesn't it?

EDIT: Hold up I'm figuring it out, shall report back if still don't understand.

Flop21 · Aug 19, 2016

Re: MATH1131 help thread

Still have no clue

First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Administrator

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)