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Differentiation AS maths

IMG_3908.JPGIMG_3909.JPGCan anyone help me with these three questions that I'm stuck on?
(edited 7 years ago)
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Original post by aliiceconnor
IMG_3908.JPGIMG_3909.JPGCan anyone help me with these three questions that I'm stuck on?


What have you tried? :h:
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Original post by SeanFM
What have you tried? :h:

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.
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 :smile: 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? :smile:
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Original post by SeanFM
For 8 you don't need to find the equation of the tangent :smile: 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? :smile:


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
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) :smile:

So the gradient of the normal is also 3 (if you can see why) which is the normal... so..
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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) :smile:

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
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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!!!
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).
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Original post by SeanFM
But you also know what the values are, for 8, eg -2a + b = ... = f'(-2).


I don't understand, sorry
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
(edited 7 years ago)
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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
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 :smile: so that gives you two equations with two unknowns, hint hint.
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Original post by SeanFM
You also know what f'(-2) is :smile: 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
(edited 7 years ago)
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..
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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
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.
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?
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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
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Original post by Epistemolog y
Man like SeanFM

btw is this C2 or C3?


It's C1

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