Cambridge Prelim MX1 Textbook Marathon/Q&A (4 Viewers)

Blitz_N7 · Dec 9, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Question 14b, Chapter 13A of Year 11 Maths 3U Cambridge.
- Show that both functions are solutions of the differential equation y'' − y = 0.
Functions are coshx=(e^x+e^-x)/2 and sinhx=(e^x-e^-x)/2.

InteGrand · Dec 9, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Blitz_N7 said:
Question 14b, Chapter 13A of Year 11 Maths 3U Cambridge.
- Show that both functions are solutions of the differential equation y'' − y = 0.
Functions are coshx=(e^x+e^-x)/2 and sinhx=(e^x+e^-x)/2.

$Let $y_1 = \cosh x = \frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}$ and $y_2 = \sinh x = \frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}$.$

$Then$

$\begin{align*}y_1 ^\prime &= \frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2} = y_2 \quad (\text{using rules for differentiating exponentials}) \\ \Rightarrow y_1 '' = y_2 ^\prime &= \frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2} \quad (\text{again using rules for differentiating exponentials})\\ &= y_1. \end{align*}$

$\noindent So $y_1$ satisfies $y_1 '' = y_1$. Also, from above, $y_2 ^\prime = y_1 \Rightarrow y_2 '' = y_1 ^\prime = y_2$ (as we showed also that $y_1 ^\prime = y_2$. So $y_2$ also satisfies $y_2 '' = y_2$. So both functions satisfy $y'' = y \Longleftrightarrow y'' - y = 0$, as required.$

$\noindent \textbf{Aside.} These functions $\cosh$ and $\sinh$ are known as \textsl{hyperbolic functions} and are well-known as having the property that they are each the derivative of the other.$

Drongoski · Dec 9, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Test for y = cosh x as well as for y = sinh x.

In each case y" = y so that y"- y = 0.

Blitz_N7 · Dec 9, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
$Let $y_1 = \cosh x = \frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}$ and $y_2 = \sinh x = \frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}$.$

$Then$

$\begin{align*}y_1 ^\prime &= \frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2} = y_2 \quad (\text{using rules for differentiating exponentials}) \\ \Rightarrow y_1 '' = y_2 ^\prime &= \frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2} \quad (\text{again using rules for differentiating exponentials})\\ &= y_1. \end{align*}$

$\noindent So $y_1$ satisfies $y_1 '' = y_1$. Also, from above, $y_2 ^\prime = y_1 \Rightarrow y_2 '' = y_1 ^\prime = y_2$ (as we showed also that $y_1 ^\prime = y_2$. So $y_2$ also satisfies $y_2 '' = y_2$. So both functions satisfy $y'' = y \Longleftrightarrow y'' - y = 0$, as required.$

$\noindent \textbf{Aside.} These functions $\cosh$ and $\sinh$ are known as \textsl{hyperbolic functions} and are well-known as having the property that they are each the derivative of the other.$

Thank you.

appleibeats · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Consider the curve y = e^x

i) Find the equation of the tangent to the curve at the point ( k, e^k)

So i got y = xe^k - ke^k + e^k

Now part ii)

Show that the line y = x + 1 is tangent to the curve

How do I do this?

Is it simultaneous equations with

y' = e^x
and
y = x + 1

And Part iii

Hence with the aid of a diagram, show that e^x - x - 1 = 0 has only 1 solution

leehuan · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

This is one way of tackling ii

$\\ $If any arbitrary tangent has equation $y=x{ e }^{ k }-k{ e }^{ k }+{ e }^{ k } \\ $Then if $y=x+1$ is tangent it can be argued that$ \\ x{ e }^{ k }-k{ e }^{ k }+{ e }^{ k }=x+1 \\ $for ONE SPECIFIC VALUE of $k$

$\\ $We then equate coefficients on $x$ and the constant term: \\ Coefficient of $x$ : ${ e }^{ k }=1\Rightarrow k=0 \\ $Constant terms: $ -k{ e }^{ k }+{ e }^{ k }=1 \\ $Note that $k=0$ satisfies this equation as well. \\ Hence, clearly when $k=0$ we do have that this line is a tangent.$

For iii:

$\\ $Sketch the curve $y={e}^{x}$ and $y=x+1$, clearly indicating the tangent to be at $(0,1)$. \\ From the graph, we observe that the line $y=x+1$ will never intersect the curve $y={e}^{x}$ again (this occurs due to the concavity of the exponential function). \\ Hence, there is only one unique solution to the equation ${e}^{x}=x+1$, which by consequence means there is only one unique solution to ${e}^{x}-x-1=0$

leehuan · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Alternative, if we want to use the derivative:

$\\ $From $ \frac { dy }{ dx } ={ e }^{ x }$ we note that the line $y=x+1$ has a gradient of 1. \\ Hence, to find the required value for $x$, we can equate $ \\ \frac { dy }{ dx } =1 \Rightarrow { e }^{ x }=1 \Rightarrow x=0. \\$

$\\ $Naturally, if we substitute $x=0$ into the original equation we have the point $(0,1)$. From here, we can use the information: \\ ${x}_{1}=0, \, {y}_{1}=1, \, m=1 $ with the point-gradient formula to verify that $y=x+1$ is tangent.$

appleibeats · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I have sketched the curve y = ln |x|

And Found the tangent at

x = 1 to be y = x - 1
x = -1 to be y = x + 1

Now I need to determine the angle with which the two tangents from the curve at the points x = +- 1 meet

So I thought I use the formula

tanE = |(m1 - m2)/(1 + m1m2)|

But using that with m1 = 1 and m2 = 1

gets you tanE = 0

That can't be right.

InteGrand · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
I have sketched the curve y = ln |x|

And Found the tangent at

x = 1 to be y = x - 1
x = -1 to be y = x + 1

Now I need to determine the angle with which the two tangents from the curve at the points x = +- 1 meet

So I thought I use the formula

tanE = |(m1 - m2)/(1 + m1m2)|

But using that with m1 = 1 and m2 = 1

gets you tanE = 0

That can't be right.

$You got the wrong slope for the tangent at $x=-1$. The tangent is downward sloping there, so you should have got a negative gradient for it.$

leehuan · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

So just adding to what InteGrand said:

For negative x, we treat |x| as -x

So we have y=ln(-x)
dy/dx=1/x regardless, using the chain rule.

When x=-1
dy/dx=1/(-1) = -1

appleibeats · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

But the same problem still arises, using m1 = 1 and m2 = -1

The numerator is 2 now but the denomiator becomes 0

Right?

leehuan · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

If tan(theta) is undefined, theta is pi/2 (90deg)
----
Should've noticed that m1m2=-1 --> lines are perpendicular

appleibeats · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Now I need to find the area of the triangle ABC, where A, B represent the points on the curve at x = +- 1 at C represents the point of intersection fo the tangents from these points

So using 1/2 . lenght . width

I know the width is 1 + 1 = 2

But the length would be the x cordinate of the point of intersection of the two tangents

so with y = x - 1 and y = -x - 1

Equating

2x = 0
x = 0

sub into y = ln |x|

y = ln0

but you can't have that

Where did I go wrong??

InteGrand · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Now I need to find the area of the triangle ABC, where A, B represent the points on the curve at x = +- 1 at C represents the point of intersection fo the tangents from these points

So using 1/2 . lenght . width

I know the width is 1 + 1 = 2

But the length would be the x cordinate of the point of intersection of the two tangents

so with y = x - 1 and y = -x - 1

Equating

2x = 0
x = 0

sub into y = ln |x|

y = ln0

but you can't have that

Where did I go wrong??

$It's not $y = \ln |x|$ you should be subbing in to; it's the equation of a tangent. So subbing $x=0$ into the equation of either tangent gives $y=-1$. So the intersection point is $(0,-1)$ and the height is thus $1$. So the area is $\frac{1}{2}\times 2\times 1 = 1$ square unit.$

appleibeats · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

If your find the the volume and its equal to : pi Integral from -1 to 0 of (x + 1)^2 dx

is that just equal to {(x+1)^3}/ 3

or is it necessary to to expand the brackets first then find the integral of it.

I did both methods

The first gives 1/3

and the second gives - 1/3

Why is that. Shouldn't they give the same answer. And also what is the correct method ??

InteGrand · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
If your find the the volume and its equal to : pi Integral from -1 to 0 of (x + 1)^2 dx

is that just equal to {(x+1)^3}/ 3

or is it necessary to to expand the brackets first then find the integral of it.

I did both methods

The first gives 1/3

and the second gives - 1/3

Why is that. Shouldn't they give the same answer. And also what is the correct method ??

Both methods are valid, so you must have made a substitution error or something in the one where you got a negative answer (the answer should be positive).

appleibeats · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I have solved the equation 2sin^2x = 1 + cos2x for 0 less than or equal to x less than or equal to 2pi

Solutions are x = pi/4 , 3pi/4, 5pi/4 and 7pi/4

Now how do I show that if 0 less that or equal to x less that or equal to pi/4 then cos2x > or equal to 2sin^2x - 1

leehuan · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

The expression is equivalent to 2sin^2(x) - 1 - cos(2x) = 0

Therefore, consider the function f(x) = 2sin^2(x) - 1 - cos(2x) for 0≤x≤pi/4

Use double angles to rewrite this:

f(x) = (1 - cos(2x)) - 1 - cos(2x)
f(x) = -2cos(2x)

One way of proceeding from here is just to draw the graph. But if one is to stick to algebraic/calculus-based methods:

f'(x)=4sin(2x)

Note that for 0≤x≤pi/4, f'(x) is positive or zero (f'(0)=0). This means that f(x) is INCREASING.

Because f(pi/4)=0
We know that f(x) increases until it's y-value reaches ZERO.

This means that f(x) is negative or zero. Because how else can you increase to 0?

Hence,
f(x)≤0

So
2sin^2(x) - 1 - cos(2x) ≤ 0
cos(2x) ≥ 2sin^2(x) - 1

InteGrand · Dec 12, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
I have solved the equation 2sin^2x = 1 + cos2x for 0 less than or equal to x less than or equal to 2pi

Solutions are x = pi/4 , 3pi/4, 5pi/4 and 7pi/4

Now how do I show that if 0 less that or equal to x less that or equal to pi/4 then cos2x > or equal to 2sin^2x - 1

$Let $f(x) = \cos (2x) -\left(2\sin ^2 x -1\right)$, so the inequation is $f(x) \geq 0$. Show that this holds at $x=0$ by substituting $x=0$ into the inequation. Then since the first (non-negative) root is at $x =\frac{\pi}{4}$ and the function $f$ is continuous, we have $f(x)\geq 0$ for all $x \in \left[0,\frac{\pi}{4}\right]$ $\Big{(}$with equality if and only if $x=\frac{\pi}{4}$ in this interval$\Big{)}$, as required.$

Blitz_N7 · Dec 14, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Question 9, Chapter 13C of Year 11 Maths 3U Cambridge.
$$Differentiate {xe^x}$. $ Hence find$: \int_0^{2} { xe^x } dx }$

I found the first part being dy/dx=(x+1)e^x but the I have no idea how to do the integration from there :/

Cambridge Prelim MX1 Textbook Marathon/Q&A (4 Viewers)

Member

Well-Known Member

Well-Known Member

Member

Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)