VCE Methods Question Thread!

[camrenis]

Another one
Determine when

∫(e^2xsin(2x))dx=∫(e^2xcos(2x))dx, given that 0≤x≤2π

There are other ways to do this, but hopefully this is straight forward enough!

Method:  Find the derivatives of both sides with respect to \(x\) using the product rule.  You can then perform two substitutions to obtain the solutions to the indefinite integrals.  These solutions can then be substituted in, and then exponents and \(\sin\) terms will cancel and you'll just retain the \(\cos\) terms which can be simply solved for.

Solution:

Spoiler

Derivative of LHS:
$$
\begin{align}
\frac{d}{dx}(e^{2x}\sin(2x))&=2e^{2x}\sin(2x)+2e^{2x}\cos(2x)
\end{align}
$$
Now take the integral of both sides and hence
$$
\begin{align}
\int2e^{2x}\sin(2x)+2e^{2x}\cos(2x)\;dx&=e^{2x}\sin(2x)\\
\implies \int2e^{2x}\sin(2x)\;dx+\int2e^{2x}\cos(2x)\;dx&=e^{2x}\sin(2x)\quad\quad\quad\quad\quad\quad\mathbf{(1)}
\end{align}
$$
Now do the same to the RHS.
\begin{align}
\frac{d}{dx}(e^{2x}\cos(2x))&=2e^{2x}\cos(2x)-2e^{2x}\sin(2x)\\
\implies \int2e^{2x}\cos(2x)-2e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)\\
\implies \int2e^{2x}\cos(2x)\;dx-\int2e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)\quad\quad\quad\quad\quad\quad\mathbf{(2)}
\end{align}
Using (1) and (2) we can find the solution to both of the integrals.  Lets find the LHS first.  Rearrange (1) in terms of the \(\cos\) term so then
$$
\begin{align}
\int2e^{2x}\cos(2x)\;dx&=e^{2x}\sin(2x)-\int2e^{2x}\sin(2x)\;dx
\end{align}
$$
and now substitute this into (2) and then move all non-integral terms to the RHS so then
$$
\begin{align}
\left(e^{2x}\sin(2x)-\int2e^{2x}\sin(2x)\;dx\right)-\int2e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)\\
\implies -4\int e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)-e^{2x}\sin(2x)\\
\implies \int e^{2x}\sin(2x)\;dx&=\frac{1}{4}e^{2x}\left(\sin(2x)-\cos(2x)\right)\quad\quad\quad\quad\quad\mathbf{(3)}
\end{align}
$$
Now the solution for the RHS can be found the same way, but finding (1) in terms of the \(\sin\) term so that
$$
\begin{align}
\implies \int2e^{2x}\sin(2x)\;dx&=e^{2x}\sin(2x)-\int2e^{2x}\cos(2x)\;dx
\end{align}
$$
Substituting this into (2) and then moving all non-integral terms onto the RHS gives
$$
\begin{align}
\int2e^{2x}\cos(2x)\;dx-\left(e^{2x}\sin(2x)-\int2e^{2x}\cos(2x)\;dx\right)&=e^{2x}\cos(2x)\\
\implies \int4e^{2x}\cos(2x)\;dx&=e^{2x}\cos(2x)+e^{2x}\sin(2x)\\
\implies \int e^{2x}\cos(2x)\;dx&=\frac{1}{4}e^{2x}\left(\sin(2x)+\cos(2x)\right)\quad\quad\quad\quad\quad\mathbf{(4)}
\end{align}
$$
Substitute (3) and (4) back into the original problem so then
$$
\begin{align}
\frac{1}{4}e^{2x}\left(\sin(2x)-\cos(2x)\right)&=\frac{1}{4}e^{2x}\left(\sin(2x)+\cos(2x)\right)
\end{align}
$$
The quarters and exponents simply cancel so that
$$
\begin{align}
\sin(2x)-\cos(2x)&=\sin(2x)+\cos(2x)\\
\implies -\cos(2x)&=\cos(2x)\\
\implies 2\cos(2x)&=0\\
\implies \cos(2x)&=0
\end{align}
$$
This is nice and easy to solve - half the period of a normal cos graph.  Since a normal cos graph has intecepts of 0 at \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\), then this has intercepts of 0 at \(\frac{\pi}{4}\), \(\frac{3\pi}{4}\).  We need the solutions for the next cycle as well as the question wants \([0,2\pi]\) so then the final solution is
$$
\begin{align}
x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}
\end{align}
$$
Happy to try and explain the steps a bit further if you get stuck somewhere!

[captkirk]

There are other ways to do this, but hopefully this is straight forward enough!

Method:  Find the derivatives of both sides with respect to \(x\) using the product rule.  You can then perform two substitutions to obtain the solutions to the indefinite integrals.  These solutions can then be substituted in, and then exponents and \(\sin\) terms will cancel and you'll just retain the \(\cos\) terms which can be simply solved for.

Solution:

Spoiler

Derivative of LHS:
$$
\begin{align}
\frac{d}{dx}(e^{2x}\sin(2x))&=2e^{2x}\sin(2x)+2e^{2x}\cos(2x)
\end{align}
$$
Now take the integral of both sides and hence
$$
\begin{align}
\int2e^{2x}\sin(2x)+2e^{2x}\cos(2x)\;dx&=e^{2x}\sin(2x)\\
\implies \int2e^{2x}\sin(2x)\;dx+\int2e^{2x}\cos(2x)\;dx&=e^{2x}\sin(2x)\quad\quad\quad\quad\quad\quad\mathbf{(1)}
\end{align}
$$
Now do the same to the RHS.
\begin{align}
\frac{d}{dx}(e^{2x}\cos(2x))&=2e^{2x}\cos(2x)-2e^{2x}\sin(2x)\\
\implies \int2e^{2x}\cos(2x)-2e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)\\
\implies \int2e^{2x}\cos(2x)\;dx-\int2e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)\quad\quad\quad\quad\quad\quad\mathbf{(2)}
\end{align}
Using (1) and (2) we can find the solution to both of the integrals.  Lets find the LHS first.  Rearrange (1) in terms of the \(\cos\) term so then
$$
\begin{align}
\int2e^{2x}\cos(2x)\;dx&=e^{2x}\sin(2x)-\int2e^{2x}\sin(2x)\;dx
\end{align}
$$
and now substitute this into (2) and then move all non-integral terms to the RHS so then
$$
\begin{align}
\left(e^{2x}\sin(2x)-\int2e^{2x}\sin(2x)\;dx\right)-\int2e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)\\
\implies -4\int e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)-e^{2x}\sin(2x)\\
\implies \int e^{2x}\sin(2x)\;dx&=\frac{1}{4}e^{2x}\left(\sin(2x)-\cos(2x)\right)\quad\quad\quad\quad\quad\mathbf{(3)}
\end{align}
$$
Now the solution for the RHS can be found the same way, but finding (1) in terms of the \(\sin\) term so that
$$
\begin{align}
\implies \int2e^{2x}\sin(2x)\;dx&=e^{2x}\sin(2x)-\int2e^{2x}\cos(2x)\;dx
\end{align}
$$
Substituting this into (2) and then moving all non-integral terms onto the RHS gives
$$
\begin{align}
\int2e^{2x}\cos(2x)\;dx-\left(e^{2x}\sin(2x)-\int2e^{2x}\cos(2x)\;dx\right)&=e^{2x}\cos(2x)\\
\implies \int4e^{2x}\cos(2x)\;dx&=e^{2x}\cos(2x)+e^{2x}\sin(2x)\\
\implies \int e^{2x}\cos(2x)\;dx&=\frac{1}{4}e^{2x}\left(\sin(2x)+\cos(2x)\right)\quad\quad\quad\quad\quad\mathbf{(4)}
\end{align}
$$
Substitute (3) and (4) back into the original problem so then
$$
\begin{align}
\frac{1}{4}e^{2x}\left(\sin(2x)-\cos(2x)\right)&=\frac{1}{4}e^{2x}\left(\sin(2x)+\cos(2x)\right)
\end{align}
$$
The quarters and exponents simply cancel so that
$$
\begin{align}
\sin(2x)-\cos(2x)&=\sin(2x)+\cos(2x)\\
\implies -\cos(2x)&=\cos(2x)\\
\implies 2\cos(2x)&=0\\
\implies \cos(2x)&=0
\end{align}
$$
This is nice and easy to solve - half the period of a normal cos graph.  Since a normal cos graph has intecepts of 0 at \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\), then this has intercepts of 0 at \(\frac{\pi}{4}\), \(\frac{3\pi}{4}\).  We need the solutions for the next cycle as well as the question wants \([0,2\pi]\) so then the final solution is
$$
\begin{align}
x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}
\end{align}
$$
Happy to try and explain the steps a bit further if you get stuck somewhere!

There are other ways to do this, but hopefully this is straight forward enough!

Method:  Find the derivatives of both sides with respect to \(x\) using the product rule.  You can then perform two substitutions to obtain the solutions to the indefinite integrals.  These solutions can then be substituted in, and then exponents and \(\sin\) terms will cancel and you'll just retain the \(\cos\) terms which can be simply solved for.

Solution:

Spoiler

Derivative of LHS:
$$
\begin{align}
\frac{d}{dx}(e^{2x}\sin(2x))&=2e^{2x}\sin(2x)+2e^{2x}\cos(2x)
\end{align}
$$
Now take the integral of both sides and hence
$$
\begin{align}
\int2e^{2x}\sin(2x)+2e^{2x}\cos(2x)\;dx&=e^{2x}\sin(2x)\\
\implies \int2e^{2x}\sin(2x)\;dx+\int2e^{2x}\cos(2x)\;dx&=e^{2x}\sin(2x)\quad\quad\quad\quad\quad\quad\mathbf{(1)}
\end{align}
$$
Now do the same to the RHS.
\begin{align}
\frac{d}{dx}(e^{2x}\cos(2x))&=2e^{2x}\cos(2x)-2e^{2x}\sin(2x)\\
\implies \int2e^{2x}\cos(2x)-2e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)\\
\implies \int2e^{2x}\cos(2x)\;dx-\int2e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)\quad\quad\quad\quad\quad\quad\mathbf{(2)}
\end{align}
Using (1) and (2) we can find the solution to both of the integrals.  Lets find the LHS first.  Rearrange (1) in terms of the \(\cos\) term so then
$$
\begin{align}
\int2e^{2x}\cos(2x)\;dx&=e^{2x}\sin(2x)-\int2e^{2x}\sin(2x)\;dx
\end{align}
$$
and now substitute this into (2) and then move all non-integral terms to the RHS so then
$$
\begin{align}
\left(e^{2x}\sin(2x)-\int2e^{2x}\sin(2x)\;dx\right)-\int2e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)\\
\implies -4\int e^{2x}\sin(2x)\;dx&=e^{2x}\cos(2x)-e^{2x}\sin(2x)\\
\implies \int e^{2x}\sin(2x)\;dx&=\frac{1}{4}e^{2x}\left(\sin(2x)-\cos(2x)\right)\quad\quad\quad\quad\quad\mathbf{(3)}
\end{align}
$$
Now the solution for the RHS can be found the same way, but finding (1) in terms of the \(\sin\) term so that
$$
\begin{align}
\implies \int2e^{2x}\sin(2x)\;dx&=e^{2x}\sin(2x)-\int2e^{2x}\cos(2x)\;dx
\end{align}
$$
Substituting this into (2) and then moving all non-integral terms onto the RHS gives
$$
\begin{align}
\int2e^{2x}\cos(2x)\;dx-\left(e^{2x}\sin(2x)-\int2e^{2x}\cos(2x)\;dx\right)&=e^{2x}\cos(2x)\\
\implies \int4e^{2x}\cos(2x)\;dx&=e^{2x}\cos(2x)+e^{2x}\sin(2x)\\
\implies \int e^{2x}\cos(2x)\;dx&=\frac{1}{4}e^{2x}\left(\sin(2x)+\cos(2x)\right)\quad\quad\quad\quad\quad\mathbf{(4)}
\end{align}
$$
Substitute (3) and (4) back into the original problem so then
$$
\begin{align}
\frac{1}{4}e^{2x}\left(\sin(2x)-\cos(2x)\right)&=\frac{1}{4}e^{2x}\left(\sin(2x)+\cos(2x)\right)
\end{align}
$$
The quarters and exponents simply cancel so that
$$
\begin{align}
\sin(2x)-\cos(2x)&=\sin(2x)+\cos(2x)\\
\implies -\cos(2x)&=\cos(2x)\\
\implies 2\cos(2x)&=0\\
\implies \cos(2x)&=0
\end{align}
$$
This is nice and easy to solve - half the period of a normal cos graph.  Since a normal cos graph has intecepts of 0 at \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\), then this has intercepts of 0 at \(\frac{\pi}{4}\), \(\frac{3\pi}{4}\).  We need the solutions for the next cycle as well as the question wants \([0,2\pi]\) so then the final solution is
$$
\begin{align}
x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}
\end{align}
$$
Happy to try and explain the steps a bit further if you get stuck somewhere!

Thank you so much for your explanation camrenis 

[lovelyperson]

Noob question: how do I find the mean point and asymptotes of a tan graph?

[Sine]

Noob question: how do I find the mean point and asymptotes of a tan graph?

what do you mean by mean point? VCAA haven't ever used this terminology

consider tan(x) = sin(x)/cos(x)
At a basic level asymptotes are when tan(x) can't exist  - so this occurs when the denominator is = 0  so solve the cos(x) = 0 for the asymptotes. Remember for something like tan(2x+pi) you are solving cos(2x+pi) = 0.

You really only need one asymptote then can work around that using the period of the graph to find the other asymptotes.

[lovelyperson]

what do you mean by mean point? VCAA haven't ever used this terminology

consider tan(x) = sin(x)/cos(x)
At a basic level asymptotes are when tan(x) can't exist  - so this occurs when the denominator is = 0  so solve the cos(x) = 0 for the asymptotes. Remember for something like tan(2x+pi) you are solving cos(2x+pi) = 0.

You really only need one asymptote then can work around that using the period of the graph to find the other asymptotes.

Thank you! Mean point is the point of inflection part (?) of the tan graph, sorry for the vague term - always thought it was an actual term.

Do I even need it to solve/sketch tan graphs?

[Ancora_Imparo]

Yep, point of inflection would be the right term! Once you find the asymptotes, since a tan curve is symmetric, you know that each point of inflection will occur exactly in the middle of two asymptotes.

E.g. y = tan(x)
You know that there is an asymptote at x=pi/4. Since the period of the curve is pi, you know that there will also be asymptotes at x=-pi/4, 3pi/4 etc. So the points of inflection will be exactly in between each pair of asymptotes. E.g. middle of -pi/4 and pi/4 is x=0 and the middle of pi/4 and 3pi/4 is x=pi/2

To sketch a tan curve, you need to determine the asymptotes and the axes intercepts. If a tan curve isn't translated up/down (like y=tan(x)), then each x-intercept will also be a point of inflection, but if it is translated up/down, then you'll need to find the roots first.

[The Waterboy]

Halo! For my attached question part d, iii) based of VCAA's definitions of increasing, we would use inclusive brackets around the extreme values of the domain ay? So [0, 5.21] http://imgur.com/a/3qAn1

[LPadlan]

3a with step by step solution pls

[MisterNeo]

(Image removed from quote.) 3a with step by step solution pls

Step1: Write the equation.
Step 2: Rearrange it to make it easier.
Step 3: Differentiate it by bringing 1/2 to the front, and getting rid of the x from 4x.
Step 4: Sub in x from the point (0,1) to find the gradient of the tangent at that point.
Step 5: Simplify to get 4.5 (e^0 is 1)

[LPadlan]

hello how did you get e^1/2x?

[keltingmeith]

hello how did you get e^1/2x?

It's in the question you asked for help with. Did you maybe mean to ask us a different question?

[Aaron]

It's in the question you asked for help with. Did you maybe mean to ask us a different question?

I think they are referring to the working out provided by MisterNeo.. 1st & 2nd lines (how the working out went from: x/2 -> 1/2 x)

[MisterNeo]

hello how did you get e^1/2x?

Why does...

This is because is "half of x", which can also be written as
Does this clear things up?
Differentiating an exponential requires the power to be differentiated too. Making the x into that form makes it easier to bring the half down to the front.

[clarke54321]

Could I please have some help with part c for the first one and part biii and c for the second?

Thanks very much!

[Abbey McMahon]

HELP PLEASE:

For a methods question we are given a table of values:
x   |   y
0   |   10
7   |   6.35
14   | 2.7
21   |   7.05
28   |   11.4

And the graph which passes through these points is in the form y = ax + b + dcos(nx)
We need to find the values of a,b,d&n

First thing:
I created 5 equations and tried to solve them simultaneously on CAS to find the values, the equations were:
10 = b + d
6.35 = 7a + b +dcos(7n)
2.7 = 14a + b +dcos(14n)
7.05 = 21a + b +dcos(21n)
11.4 = 28a + b +dcos(28n)

CAS did not give me any of the values.

So what am I doing wrong?? I have never seen a question in the form y = ax + b +dcos(nx) before and I am seriously stuck!!!!!!