Differentiation AS maths

Watch
#1
Can anyone help me with these three questions that I'm stuck on?
1
4 years ago
#2
(Original post by aliiceconnor)
Can anyone help me with these three questions that I'm stuck on?
What have you tried?
1
#3
(Original post by SeanFM)
What have you tried?
For 8, I have found the first derivative and the equation of the tangent. I tried solving the curve and tangent as simultaneous equations but that didn't work.
For 9, I substituted the given values for t into the equation.
And for 10, I found the first derivative of the curve and then tried to solve the equations simultaneously.

My teacher doesn't explain things very well so I'm left confused and I'm just trying a load of things that don't work basically.
0
4 years ago
#4
(Original post by aliiceconnor)
For 8, I have found the first derivative and the equation of the tangent. I tried solving the curve and tangent as simultaneous equations but that didn't work.
For 9, I substituted the given values for t into the equation.
And for 10, I found the first derivative of the curve and then tried to solve the equations simultaneously.

My teacher doesn't explain things very well so I'm left confused and I'm just trying a load of things that don't work basically.
For 8 you don't need to find the equation of the tangent you just need two simultaneous equations.

You've probably done most of the work for it by finding the equation, but what is f(-2) and f'(-2) and how does this help?

For 9 that wouldn't work but I think everyone does that the first time they come across a question like that. You need to find a function for the rate of change before plugging in t.

For 10, if two lines are parallel then what can you say about their gradients?
1
#5
(Original post by SeanFM)
For 8 you don't need to find the equation of the tangent you just need two simultaneous equations.

You've probably done most of the work for it by finding the equation, but what is f(-2) and f'(-2) and how does this help?

For 9 that wouldn't work but I think everyone does that the first time they come across a question like that. You need to find a function for the rate of change before plugging in t.

For 10, if two lines are parallel then what can you say about their gradients?
I found f(-2)=4a+2b-a^2 and f'(-2)=-4a+b-2a and I tried to solve them simultaneously to get -a^2-2a+3a but I don't know where that gets me haha

For 9, I have no idea how to find the rate.

And for 10, I know all the gradients are 3 but I don't know where to go from there
0
4 years ago
#6
(Original post by aliiceconnor)
I found f(-2)=4a+2b-a^2 and f'(-2)=-4a+b-2a and I tried to solve them simultaneously to get -a^2-2a+3a but I don't know where that gets me haha

For 9, I have no idea how to find the rate.

And for 10, I know all the gradients are 3 but I don't know where to go from there
For your working for 9, check for a typo or a correction for f(-2), it's not 2b but... Also when differentiating f(x), you don't differentiate a^2 - think of it as a number eg 9 (then a=3 but that's not the point) then differentiating a^2 gives 0 because there's no x term, or it's a^2 * x^0.

(by 9 I see I mean 8..)

The rate is just f'(x)

So the gradient of the normal is also 3 (if you can see why) which is the normal... so..
0
#7
(Original post by SeanFM)
For your working for 9, check for a typo or a correction for f(-2), it's not 2b but... Also when differentiating f(x), you don't differentiate a^2 - think of it as a number eg 9 (then a=3 but that's not the point) then differentiating a^2 gives 0 because there's no x term, or it's a^2 * x^0.

(by 9 I see I mean 8..)

The rate is just f'(x)

So the gradient of the normal is also 3 (if you can see why) which is the normal... so..
Yeah it's -2b so -a^2-2a-b so does 2a=b?? But then how do I find what the values are

For 10, I know it's y=3x+k I just don't know how to find k
0
#8
(Original post by aliiceconnor)
Yeah it's -2b so -a^2-2a-b so does 2a=b?? But then how do I find what the values are

For 10, I know it's y=3x+k I just don't know how to find k
Also I've got 9 now! So thank you very much!!!
0
4 years ago
#9
(Original post by aliiceconnor)
Yeah it's -2b so -a^2-2a-b so does 2a=b?? But then how do I find what the values are

For 10, I know it's y=3x+k I just don't know how to find k
But you also know what the values are, for 8, eg -2a + b = ... = f'(-2).
1
#10
(Original post by SeanFM)
But you also know what the values are, for 8, eg -2a + b = ... = f'(-2).
I don't understand, sorry
0
4 years ago
#11
For example you've put x = 2 into y = ax^2 + bx - a^2 to give you (as you've writtne) 4a -2b - a^2 but what does that equal to? It is correct to put that in, but it is not complete.

(Original post by aliiceconnor)
I don't understand, sorry
0
#12
(Original post by SeanFM)
For example you've put x = 2 into y = ax^2 + bx - a^2 to give you (as you've writtne) 4a -2b - a^2 but what does that equal to? It is correct to put that in, but it is not complete.
-13 but then there's two values to find and I don't know how to do that
0
4 years ago
#13
(Original post by aliiceconnor)
-13 but then there's two values to find and I don't know how to do that
You also know what f'(-2) is so that gives you two equations with two unknowns, hint hint.
0
#14
(Original post by SeanFM)
You also know what f'(-2) is so that gives you two equations with two unknowns, hint hint.
-13=4a-2b-a^2
0=-4a+b-2a x2

-13=4a-2b-a^2
0=-8a+2b-4a

0=-a^2+4a-2b-13
0=-12a+2b +

0=-a^2+4a-13-12a

0=-a^2-8a-13

I don't think this is right but it's what I've done
0
4 years ago
#15
(Original post by aliiceconnor)
-13=4a-2b-a^2
0=-4a+b-2a x2

-13=4a-2b-a^2
0=-8a+2b-4a

0=-a^2+4a-2b-13
0=-12a+2b +

0=-a^2+4a-13-12a

0=-a^2-8a-13

I don't think this is right but it's what I've done
Good attempt, good show of working. Your first equation is correct but your second one isn't.

As I stated in a previous post, you don't differentiate a^2 with respect to x to give 2a. Think of any number without an x next to it (eg a^2) as with an x^0 next to it (which is a 1, so you are multiplying by 1..) and differentiate that and see what happens. It is confusing because it may seem like a is a variable but it is a constant.

Also, f'(-2) is not equal to 0. It is equal to..
0
#16
(Original post by SeanFM)
Good attempt, good show of working. Your first equation is correct but your second one isn't.

As I stated in a previous post, you don't differentiate a^2 with respect to x to give 2a. Think of any number without an x next to it (eg a^2) as with an x^0 next to it (which is a 1, so you are multiplying by 1..) and differentiate that and see what happens. It is confusing because it may seem like a is a variable but it is a constant.

Also, f'(-2) is not equal to 0. It is equal to..
I don't know what f'(-2) is equal to
1
4 years ago
#17
(Original post by aliiceconnor)
I don't know what f'(-2) is equal to
Remember that f'(x) is the gradient function, for a given point of x it finds the value of the gradient.
0
4 years ago
#18
(Original post by SeanFM)
Remember that f'(x) is the gradient function, for a given point of x it finds the value of the gradient.
Man like SeanFM

btw is this C2 or C3?
0
#19
(Original post by SeanFM)
Remember that f'(x) is the gradient function, for a given point of x it finds the value of the gradient.
So f'(-2)=-4

-4=-4a+b-2a
-8=-8a+2b-4a

4a-2b+13=0
-12a+2b+8=0

4a+13-12a+8=0
-8a+21=0
0
#20
(Original post by Epistemolog y)
Man like SeanFM

btw is this C2 or C3?
It's C1
0
X

new posts
Back
to top
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Poll

Join the discussion

Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (132)
14.78%
I'm not sure (40)
4.48%
No, I'm going to stick it out for now (268)
30.01%
I have already dropped out (22)
2.46%
I'm not a current university student (431)
48.26%