3U Maths Question Thread

[jamonwindeyer]

Hey guys,

I just need a hand with parts C and D for the question attached below (with answers attachment). For part C can you explain to me if there is an equation we should arrive to or just do all by the calculator? (I let 5t = 20ln(t+1)). And for Part D I'm just not sure how to do it.

Thanks

Hey! So for C, there is no nice way to solve that equation, so calculator work is all you have!

For D, the distance between them is given by (this gives a positive number when Henry is behind Thomas):



For maximum distance, we need to maximise this function. We can just differentiate!



And you can prove that's a maximum really easily! So it happens when \(t=3\)

[RuiAce]

yes there was, thank you so much for your help.

Alright assuming the typo, here's one example of a proof. You can probably find some other ones floating around in various textbooks.
\[ \text{Let }\boxed{\theta = \sin^{-1}x}\\ \text{and note that due to the range of inverse sine,}\\ -\frac\pi2 \leq \theta \leq \frac\pi2. \]
\[ \text{Then } \boxed{\sin \theta = x}\\ \text{but we note that }\cos \left( \frac\pi2 - \theta\right) = \sin x. \]
(This is the complementary angle identity from Year 11, but of course it was taught with \(90^\circ\) instead.)
\[ \text{Note that the range restriction can be rearranged into}\\ 0\leq \frac\pi2 - \theta \leq \pi.\]
\[ \text{As a consequence, we do not need to do any ASTC analysis}\\ \text{and conclude immediately that }\boxed{\frac\pi2 - \theta = \cos^{-1}x}.\\ \text{Subbing back }x=\sin^{-1}\theta\text{ gives}\\ \boxed{ \frac\pi2 - \sin^{-1}x = \cos^{-1}x }\text{ which rearranges to what we wish to prove.}\]

[georgebanis]

Hey! So for C, there is no nice way to solve that equation, so calculator work is all you have!

For D, the distance between them is given by (this gives a positive number when Henry is behind Thomas):



For maximum distance, we need to maximise this function. We can just differentiate!



And you can prove that's a maximum really easily! So it happens when \(t=3\)

Thanks for the explanation Jamon!

[annabeljxde]

Hey!

I've encountered many inverse trig function questions where they ask me to find the domain and range of a function within the function. I'm a little bit confused as to how to work these questions out. For example, this question:

If anyone could lend me a hand, I'd really appreciate it!!

[RuiAce]

Hey!

I've encountered many inverse trig function questions where they ask me to find the domain and range of a function within the function. I'm a little bit confused as to how to work these questions out. For example, this question:

If anyone could lend me a hand, I'd really appreciate it!!

\[ \text{Since }h(x) = \tan^{-1}x\text{ has natural domain all real }x,\\ \text{we only need to consider any domain restrictions on }g(x)=\sqrt{1-x^2}. \]
Which is of course, just a semi-circle with radius 1, so we know its domain is \( -1\leq x \leq 1 \). This will then be the domain of \( h(g(x)) = \tan^{-1} \sqrt{1-x^2} \).
\[ \text{Now, the }\textbf{range}\text{ of the inner function }g(x)\\ \text{will be treated as a }\textbf{domain}\text{ restriction}\\ \textbf{only}\text{ for the outer function }h(x). \]
\[\text{Here, }\sqrt{1-x^2}\text{ has range }0\leq y \leq 1.\\ \text{So we need to consider the range of }\tan^{-1}x\\ \textbf{when}\text{ we restrict the range of }\tan^{-1}x\text{ to }0\leq x \leq 1. \]
\[ \text{Since }h(x) = \tan^{-1}x\text{ is monotonic increasing (just look at its graph)}\\ \text{we can basically infer it from the graph by focusing on its endpoints.}\\ \text{We know that }\tan^{-1}0 = 0\text{ and }\tan^{-1}1 = \frac\pi4\\ \text{so therefore the range will be }\boxed{0\leq y \leq \frac\pi4}. \]

[annabeljxde]

Thank you for the solution to my previous question!

I need help with this question.... Any help will be greatly appreciated!

[RuiAce]

Thank you for the solution to my previous question!

I need help with this question.... Any help will be greatly appreciated!

\[ \text{If we set }\sin^{-1}x = \alpha\text{ we get }\boxed{\sin \alpha = \frac{x}{1}}.\\ \text{So we can draw a right-angled triangle with}\\ \text{opp side equal to }\alpha\text{ and hypotenuse equal to 1}. \]
\[ \text{Then using adjacent/hypotenuse}\\ \boxed{\cos\alpha = \sqrt{1-x^2}}. \]

Alternatively, if you want to do it properly..

\[ \text{Let }\sin^{-1}x = \alpha\\ \text{and note consequently that }-\frac\pi2 \leq \alpha \leq \frac\pi2\\ \text{from the range of the inverse sine function.} \]
\[ \text{Then }x = \sin \alpha.\\ \text{Since }\cos^2 x = 1-\sin^2 x\text{ we then have }\cos^2 x = 1-x^2\\ \text{and hence }\boxed{\cos x = \sqrt{1-x^2}}. \]
\[ \text{However }-\frac\pi2 \leq \alpha \leq \frac\pi2\text{ means that }\\ \alpha\text{ is in the first or fourth quadrants,}\\ \text{both of which we see that }\cos\alpha\text{ is positive.}\]
\[ \text{Therefore, taking the positive root, we have}\\ \boxed{\cos^{-1}\alpha = \sqrt{1-x^2}} \]

\[ \text{Similarly }\cos \beta = \sqrt{1-y^2}\\ \text{and thus}\\ \begin{align*} \sin (\alpha+\beta) &= \sin\alpha\cos\beta+\cos\alpha\sin\beta\\ &= x\sqrt{1-y^2} + y\sqrt{1-x^2}. \end{align*} \]
Take \(\sin^{-1}\) of both sides of that equation, sub back in for \(\alpha\) and \(\beta\) on the LHS, and you're done. The next bit should be easy - just sub those fractions in.

[annabeljxde]

\[ \text{If we set }\sin^{-1}x = \alpha\text{ we get }\boxed{\sin \alpha = \frac{x}{1}}.\\ \text{So we can draw a right-angled triangle with}\\ \text{opp side equal to }\alpha\text{ and hypotenuse equal to 1}. \]
\[ \text{Then using adjacent/hypotenuse}\\ \boxed{\cos\alpha = \sqrt{1-x^2}}. \]

Alternatively, if you want to do it properly..

\[ \text{Let }\sin^{-1}x = \alpha\\ \text{and note consequently that }-\frac\pi2 \leq \alpha \leq \frac\pi2\\ \text{from the range of the inverse sine function.} \]
\[ \text{Then }x = \sin \alpha.\\ \text{Since }\cos^2 x = 1-\sin^2 x\text{ we then have }\cos^2 x = 1-x^2\\ \text{and hence }\boxed{\cos x = \sqrt{1-x^2}}. \]
\[ \text{However }-\frac\pi2 \leq \alpha \leq \frac\pi2\text{ means that }\\ \alpha\text{ is in the first or fourth quadrants,}\\ \text{both of which we see that }\cos\alpha\text{ is positive.}\]
\[ \text{Therefore, taking the positive root, we have}\\ \boxed{\cos^{-1}\alpha = \sqrt{1-x^2}} \]

\[ \text{Similarly }\cos \beta = \sqrt{1-y^2}\\ \text{and thus}\\ \begin{align*} \sin (\alpha+\beta) &= \sin\alpha\cos\beta+\cos\alpha\sin\beta\\ &= x\sqrt{1-y^2} + y\sqrt{1-x^2}. \end{align*} \]
Take \(\sin^{-1}\) of both sides of that equation, sub back in for \(\alpha\) and \(\beta\) on the LHS, and you're done. The next bit should be easy - just sub those fractions in.

Thank you so much for the quick reply! This helped me a lot!! 

[saloni.aphale]

Can someone please help me with the question i have attatched below, THANK YOU !!

[DrDusk]

Can someone please help me with the question i have attatched below, THANK YOU !!

                                                                             

'scuse me, just making an entrance into the math forums :c

[annabeljxde]

I think I need a refresher of Perms and Combs... I still struggle with it as it's my worst performing topic for 3U I know this question is probably not too difficult, I just need some help;'')

[fun_jirachi]

Hey there!

This one is a more intuitive solution than you'd think
There are 8! or 40320 ways for those people to go through the door. Now, John either goes before or after Barbara, so it makes sense that the number of ways John goes in after Barbara is just half that, or 20160 ways.

Hope this helps

[Minivasili]

hey guys!!! i really need some help in my maths assesment for ext 1 which is on inverse functions.


the question is:
 use the theorem [g'(x) = 1/f'(g(x))] to find the gradient of the tangent to the curve y=f^-1(x) at the point (4,1)         
[in the previous question it is given that f(x) = x^3+3x]

thank you for your help

[Opengangs]

hey guys!!! i really need some help in my maths assesment for ext 1 which is on inverse functions.


the question is:
 use the theorem [g'(x) = 1/f'(g(x))] to find the gradient of the tangent to the curve y=f^-1(x) at the point (4,1)         
[in the previous question it is given that f(x) = x^3+3x]

thank you for your help

Let \( g(x) \) be \( f^{-1}(x) \). Then using the theorem, we have:
\[ \begin{align*} \left(f^{-1}(x) \right)’ = \frac{1}{f’\left(f^{-1}(x) \right)} \end{align*}. \]

Now at the point \( x = 1 \), we have \( f(1) = 4 \Rightarrow 1 = f^{-1}(4) \).

Next, we take the derivative of \( f(x) \) which gives us \( f’(x) = 3x^2 + 3 \). So at 1, we have \( f’\left( f^{-1}(4) \right) = f’(1) = 6.\)

Putting this all together finally gives you:
\[ \left(f^{-1}(4)\right)’ = \frac{1}{6}. \]

[Minivasili]

Hey, thank you so much for your help but I am still confused as to why we are using x=1 when the point is (4,1) so shouldn't X be 4. I'm sorry


Thanks again for your help