Author Topic: Mathematics Question Thread (Read 1626101 times) Tweet Share

itssona · « **Reply #2205 on:** July 18, 2017, 05:48:39 pm »

Ohh wait can u correct me on the red ones

Thank u for explaining that part

RuiAce · « **Reply #2206 on:** July 18, 2017, 05:53:55 pm »

According to the graph, the range of $y=x^2$ is $y\ge 0$

Shadowxo · « **Reply #2207 on:** July 18, 2017, 05:55:44 pm »

Quote from: sssona09 on July 18, 2017, 05:48:39 pm

Ohh wait can u correct me on the red ones

Thank u for explaining that part

x² (basic parabola)
Domain: all real x
Range: [0,infinity)

log_ex (basic log)
Domain: All positive x (0,infinity)
Range: All real y

|x| (basic modulus graph)
Domain: All real x
Range: [0,infinity)

x³ (basic cubic)
Domain: All real x
Range: All real y

e^x (basic exponential)
Domain: All real x
Range: All positive y (0,infinity)

itssona · « **Reply #2208 on:** July 18, 2017, 05:58:07 pm »

So for log (x+1) is the domain x>-1 ?

itssona · « **Reply #2209 on:** July 18, 2017, 05:59:08 pm »

Quote from: Shadowxo on July 18, 2017, 05:55:44 pm

x² (basic parabola)
Domain: all real x
Range: [0,infinity)

log_ex (basic log)
Domain: All positive x (0,infinity)
Range: All real y

|x| (basic modulus graph)
Domain: All real x
Range: [0,infinity)

x³ (basic cubic)
Domain: All real x
Range: All real y

e^x (basic exponential)
Domain: All real x
Range: All positive y (0,infinity)

Ah thank u!!

RuiAce · « **Reply #2210 on:** July 18, 2017, 06:03:45 pm »

Quote from: Shadowxo on July 18, 2017, 05:55:44 pm

x² (basic parabola)
Domain: all real x
Range: [0,infinity)

log_ex (basic log)
Domain: All positive x (0,infinity)
Range: All real y

|x| (basic modulus graph)
Domain: All real x
Range: [0,infinity)

x³ (basic cubic)
Domain: All real x
Range: All real y

e^x (basic exponential)
Domain: All real x
Range: All positive y (0,infinity)

They don't use interval notation in the HSC.

Quote from: sssona09 on July 18, 2017, 05:58:07 pm

So for log (x+1) is the domain x>-1 ?

Yes

bigsweetpotato2000 · « **Reply #2211 on:** July 18, 2017, 09:26:05 pm »

What is the best way of understanding Motion ?

I'm still rote-learning it and I just can't seem to find the logic in it!

Unsatisfied,

Bigsweetpotato Farm

RuiAce · « **Reply #2212 on:** July 18, 2017, 09:48:19 pm »

Quote from: bigsweetpotato2000 on July 18, 2017, 09:26:05 pm

What is the best way of understanding Motion ?

I'm still rote-learning it and I just can't seem to find the logic in it!

Unsatisfied,

Bigsweetpotato Farm

$\text{All the logic you need is}\\\ \text{The derivative of the displacement is the velocity}\\ \text{The derivative of the velocity is the acceleration}\\ \text{and you reverse everything for integration}$
$\text{What other logic do you require?}$

bigsweetpotato2000 · « **Reply #2213 on:** July 18, 2017, 09:50:45 pm »

Quote from: RuiAce on July 18, 2017, 09:48:19 pm

$\text{All the logic you need is}\\\ \text{The derivative of the displacement is the velocity}\\ \text{The derivative of the velocity is the acceleration}\\ \text{and you reverse everything for integration}$
$\text{What other logic do you require?}$

Apply to real life situation logic?

RuiAce · « **Reply #2214 on:** July 18, 2017, 10:12:02 pm »

Quote from: bigsweetpotato2000 on July 18, 2017, 09:50:45 pm

Apply to real life situation logic?

Can you please provide an example to supplement your question? It's very ambiguous as to what you want right now.

2U scenarios are pretty much immediate applications of the above results.

bigsweetpotato2000 · « **Reply #2215 on:** July 18, 2017, 10:28:05 pm »

Quote from: RuiAce on July 18, 2017, 10:12:02 pm

Can you please provide an example to supplement your question? It's very ambiguous as to what you want right now.

2U scenarios are pretty much immediate applications of the above results.

Graphing?

RuiAce · « **Reply #2216 on:** July 18, 2017, 10:32:19 pm »

$\text{This goes beyond 'logic'}\\ \text{This combines the applications topic }\textbf{with}\text{ the geometry of the derivative.}$
$\text{We are given the displacement-time graph, i.e. }x=f(t)\\ \text{and we want to describe the velocity and acceleration at }P\\ \textit{using}\text{ our knowledge of the first and second derivatives.}$
$\text{Because all we have is a graph, we need to interpret what the derivatives mean geometrically.}\\ \text{We start with the first derivative, which measures the gradient of the tangent}$
$\text{If you sketch (by hand) a tangent at }P\\ \text{you will find that it slopes downwards.}\\ \text{This is equivalent to saying that }\frac{dx}{dt}<0,\text{ i.e. }v < 0\text{ at }P\\ \text{Hence the velocity is negative.}$
Alternatively, just recognise that it's decreasing without sketching the tangent by inspection.
$\text{Then, we know that the second derivative measures concavity}\\ \text{By just inspection, we see that at }P\text{ the displacement-time graph}\\\text{is concave up. This hints that the second derivative is positive}\\ \text{That is to say, }\frac{d^2x}{dt^2} > 0,\text{ i.e. }a >0\\ \text{So the acceleration is positive}$
$\therefore A$

Shadowxo · « **Reply #2217 on:** July 18, 2017, 10:33:55 pm »

Quote from: bigsweetpotato2000 on July 18, 2017, 10:28:05 pm

Graphing?

v=dx/dt. As the gradient is negative, v is negative at P
a=dv/dt. As the gradient is increasing (becoming less negative) a is positive at P

Edit: Rui beat me to it but hope this helps anyway

bigsweetpotato2000 · « **Reply #2218 on:** July 19, 2017, 12:03:12 am »

Quote from: RuiAce on July 18, 2017, 10:32:19 pm

$\text{This goes beyond 'logic'}\\ \text{This combines the applications topic }\textbf{with}\text{ the geometry of the derivative.}$
$\text{We are given the displacement-time graph, i.e. }x=f(t)\\ \text{and we want to describe the velocity and acceleration at }P\\ \textit{using}\text{ our knowledge of the first and second derivatives.}$
$\text{Because all we have is a graph, we need to interpret what the derivatives mean geometrically.}\\ \text{We start with the first derivative, which measures the gradient of the tangent}$
$\text{If you sketch (by hand) a tangent at }P\\ \text{you will find that it slopes downwards.}\\ \text{This is equivalent to saying that }\frac{dx}{dt}<0,\text{ i.e. }v < 0\text{ at }P\\ \text{Hence the velocity is negative.}$
Alternatively, just recognise that it's decreasing without sketching the tangent by inspection.
$\text{Then, we know that the second derivative measures concavity}\\ \text{By just inspection, we see that at }P\text{ the displacement-time graph}\\\text{is concave up. This hints that the second derivative is positive}\\ \text{That is to say, }\frac{d^2x}{dt^2} > 0,\text{ i.e. }a >0\\ \text{So the acceleration is positive}$
$\therefore B$

Isn't it A?

Thanks btw

bigsweetpotato2000 · « **Reply #2219 on:** July 19, 2017, 12:05:02 am »

Quote from: Shadowxo on July 18, 2017, 10:33:55 pm

v=dx/dt. As the gradient is negative, v is negative at P
a=dv/dt. As the gradient is increasing (becoming less negative) a is positive at P

Edit: Rui beat me to it but hope this helps anyway

Heheh
Repeated trials with consistent results lead to reliability

Many thanks!

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