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Reply 20
Avatar for Zaphod77
Zaphod77
1
Ok. I have attached my working, I hope you can understand it! The graph is a positive quartic, as none of the x's in brackets are negative, and you look at when y<0 because that is where the equation is less than 0 - you look at the inequality. If the equation I got was on the other side, you would get the same critical values but the graph would be negative and you would be looking at when y>0 instead. Does this all make sense? :smile:
Inequalities.docx22.4KB
Reply 21
Avatar for Joshmeid
Joshmeid
5
Original post by smith50
Thanks here is the question i just don't know how to find the critical values FP2 INEQU.PNG
Thanks,
Smith


You have factorised incorrectly, best way to attack these are to find all the common factors between both terms straight away then simplify the rest.

See Zaphod77's workings, (although you shouldn't post full workings!), if you are still confused.
Reply 22
Avatar for smith50
smith50
OP
6
Original post by Zaphod77
Ok. I have attached my working, I hope you can understand it! The graph is a positive quartic, as none of the x's in brackets are negative, and you look at when y<0 because that is where the equation is less than 0 - you look at the inequality. If the equation I got was on the other side, you would get the same critical values but the graph would be negative and you would be looking at when y>0 instead. Does this all make sense? :smile:


Hey i don't seem to follow what you did :confused: Immensely confused .
Could you help me out as if you haven't enough :wink:
Thanks,
Smith
Reply 23
Avatar for Zaphod77
Zaphod77
1
Original post by smith50
Hey i don't seem to follow what you did :confused: Immensely confused .
Could you help me out as if you haven't enough :wink:
Thanks,
Smith


Ok, I can try! So first of all I did what you did - I multiplied both side by both the bottoms of the fractions squared. I then moved one side over to the other and took out common factors (x and (x+3)). I also expanded the other brackets. Can you see where I did that?
Reply 24
Avatar for smith50
smith50
OP
6
Original post by Zaphod77
Ok, I can try! So first of all I did what you did - I multiplied both side by both the bottoms of the fractions squared. I then moved one side over to the other and took out common factors (x and (x+3)). I also expanded the other brackets. Can you see where I did that?


For the second bracket what did you get i seem to get 2x^2+x+12 :confused:
Smith
Reply 25
Avatar for Zaphod77
Zaphod77
1
Original post by smith50
For the second bracket what did you get i seem to get 2x^2+x+12 :confused:
Smith


You've got confused when taking out the factors. Take them out separately if you like - take out the x, what are you left with (without expanding the rest)?
Reply 26
Avatar for smith50
smith50
OP
6
Original post by Zaphod77
You've got confused when taking out the factors. Take them out separately if you like - take out the x, what are you left with (without expanding the rest)?


Oops :biggrin: thankyou for this i get this now so do you always square the brackets?
Thanks soo much,
Smith
Reply 27
Avatar for Zaphod77
Zaphod77
1
Original post by smith50
Oops :biggrin: thankyou for this i get this now so do you always square the brackets?
Thanks soo much,
Smith

I'm glad you get it! Yes, it's the only way of ensuring they're positive - if you multiply both sides by something negative the inequality sign has to change, so in order to make sure the thing you're multiplying with is positive you have to square them.
Reply 28
Avatar for smith50
smith50
OP
6
Original post by Zaphod77
I'm glad you get it! Yes, it's the only way of ensuring they're positive - if you multiply both sides by something negative the inequality sign has to change, so in order to make sure the thing you're multiplying with is positive you have to square them.


Hi could you please help me with this question please :smile:
FP2.PNG
Thanks,
Smith
Reply 29
Avatar for Zaphod77
Zaphod77
1
Original post by smith50
Hi could you please help me with this question please :smile:
FP2.PNG
Thanks,
Smith


Ok, what have you done so far?
Reply 30
Avatar for Joshmeid
Joshmeid
5
Original post by smith50
Hi could you please help me with this question please :smile:
FP2.PNG
Thanks,
Smith


Use the chain rule on dydx\frac{dy}{dx} to include t.
Find out dtdx\frac{dt}{dx}.
Substitute it into dydx\frac{dy}{dx}.
Differentiate dydx\frac{dy}{dx} to get d2ydx2\frac{d^{2}y}{dx^{2}} and simplify.

If you struggle with any of these steps message back.
(edited 10 years ago)
Reply 31
Avatar for smith50
smith50
OP
6
Original post by Zaphod77
Ok, what have you done so far?


Thanks :smile: this is what i have done so far
dx/dt = e^t then dy/dx=dy/dt x dt/dx so dy/dx = e^-tdy/dt it's after this point where i get stuck.

Smith
Reply 32
Avatar for smith50
smith50
OP
6
Original post by Joshmeid
Use the chain rule on dydx\frac{dy}{dx} to include t.
Find out dtdx\frac{dt}{dx}.
Substitute it into dydx\frac{dy}{dx}.
Differentiate dydx\frac{dy}{dx} to get d2ydx2\frac{d^{2}y}{dx^{2}} and simplify.

If you struggle with any of these steps message back.


Hi i understand you have to use the chain rule but it is when i am doing the second differentiation.
Smith
Reply 33
Avatar for Joshmeid
Joshmeid
5
Original post by smith50
Thanks :smile: this is what i have done so far
Smith


Remember x=etx = e^{t} so find t in terms of x, substitute into e−te^{-t} then differentiate using product.
(edited 10 years ago)
Reply 34
Avatar for Zaphod77
Zaphod77
1
Original post by smith50
Thanks :smile: this is what i have done so far
Smith


Ok, that's good! So from there you need to get d^2y/dx^2. So you need to differentiate - you'll have to use the chain rule, as they are both functions.
Reply 35
Avatar for smith50
smith50
OP
6
Original post by Zaphod77
Ok, that's good! So from there you need to get d^2y/dx^2. So you need to differentiate - you'll have to use the chain rule, as they are both functions.


when i do i get d^2y/dx^2 = e^-t d^2y/dt^2 - e^-t dy/dt? I don't know where i am going wrong :confused:
Smith
(edited 10 years ago)
Reply 36
Avatar for Joshmeid
Joshmeid
5
Original post by smith50
when i do i get d^2y/dx^2 = e^-t d^2y/dt^2 - e^-t dy/dt?
Smith


ddx(e−t)≠−e−t\frac{d}{dx}(e^{-t}) \not= -e^{-t} and ddx(dydt)≠d2ydt2\frac{d}{dx}(\frac{dy}{dt}) \not= \frac{d^{2}y}{dt^{2}}.

t is a function.

That's like saying ddx(y2+x3+5x)=2y+3x2+5\frac{d}{dx}(y^{2} + x^{3} + 5x) = 2y + 3x^{2} + 5 when infact ddx(y2+x3+5x)=2ydydx+3x2+5\frac{d}{dx}(y^{2} + x^{3} + 5x) = 2y\frac{dy}{dx} + 3x^{2} + 5, you should know implicit differentiation from C4.
(edited 10 years ago)
Reply 37
Avatar for Smaug123
Smaug123
15
Original post by smith50
when i do i get d^2y/dx^2 = e^-t d^2y/dt^2 - e^-t dy/dt? I don't know where i am going wrong :confused:
Smith

It's easier to see if you write t as t(x) rather than just t.
Reply 38
Avatar for smith50
smith50
OP
6
Original post by Joshmeid
ddx(e−t)≠−e−t\frac{d}{dx}(e^{-t}) \not= -e^{-t} and ddx(dydt)≠d2ydt2\frac{d}{dx}(\frac{dy}{dt}) \not= \frac{d^{2}y}{dt^{2}}.

t is a function.

That's like saying ddx(y2+x3+5x)=2y+3x2+5\frac{d}{dx}(y^{2} + x^{3} + 5x) = 2y + 3x^{2} + 5 when infact ddx(y2+x3+5x)=2ydydx+3x2+5\frac{d}{dx}(y^{2} + x^{3} + 5x) = 2y\frac{dy}{dx} + 3x^{2} + 5, you should know implicit differentiation from C4.

I am quiet unsure about this question could you post some solutions please
Thanks,
Smith
Reply 39
Avatar for Joshmeid
Joshmeid
5
Original post by smith50
I am quiet unsure about this question could you post some solutions please
Thanks,
Smith


No, but I will try to explain :smile:

Firstly:

dxdt=et\frac{dx}{dt} = e^{t} as you came to.

We can futher simplify this though because we know x=etx = e^{t}

So dxdt=x\frac{dx}{dt} = x which is helpful later on.

By the chain rule we know dydx=dydt⋅dtdx\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} and so dydx=dydt⋅1x\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{1}{x}.

Now we must find the second derivative:

d2ydx2=ddx(dydx)\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(\frac{dy}{dx})

So we differentiate the RHS of dydx\frac{dy}{dx} again using the product rule.

d2ydx2=1x⋅ddx(dydt)+dydt⋅ddx(1x)\frac{d^{2}y}{dx^{2}} = \frac{1}{x} \cdot \frac{d}{dx}(\frac{dy}{dt}) + \frac{dy}{dt} \cdot \frac{d}{dx} (\frac{1}{x})

Now here is where most people get confused, how to find ddx(dydt)\frac{d}{dx}(\frac{dy}{dt}).

One way to go about understanding this is by using a substitution of u=dydtu = \frac{dy}{dt}, then ddx(dydt)\frac{d}{dx}(\frac{dy}{dt}) turns into dudx\frac{du}{dx} and then by the chain rule, dudx=dudt⋅dtdx\frac{du}{dx} = \frac{du}{dt} \cdot \frac{dt}{dx}, we can then substitute u back into this: ddx(dydt)=ddt(dydt)⋅dtdx=d2ydt2\frac{d}{dx}(\frac{dy}{dt}) = \frac{d}{dt}(\frac{dy}{dt}) \cdot \frac{dt}{dx} = \frac{d^{2}y}{dt^{2}}.

So with that information try to find d2ydx2\frac{d^{2}y}{dx^{2}}. If you are confused at any step reply with it in a quote.
(edited 10 years ago)

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