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###### Need help with this calculus question

Reply 1

1 month ago

Original post by A4M1NA

I have completed part a of this question but I'm stuck on part b. Can anyone help me with this part?

Attach the question

Original post by BankaiGintoki

Attach the question

Attach the question

How do I attach an image?

Original post by A4M1NA

How do I attach an image?

idk ab that, i use a laptop... but if you can type out the question?

Original post by eenie_49

idk ab that, i use a laptop... but if you can type out the question?

idk ab that, i use a laptop... but if you can type out the question?

There’s a diagram on the question

Original post by A4M1NA

There’s a diagram on the question

oh um.... did you get it from a website? is there a link?

Original post by A4M1NA

How do I attach an image?

um try looking at this thread?? https://www.thestudentroom.co.uk/showthread.php?t=1722634

Original post by eenie_49

oh um.... did you get it from a website? is there a link?

https://imagizer.imageshack.com/img924/5522/1YhwO8.jpg

I'm struggling with part b of this question

Original post by mqb2766

if the tangent is at 45 degrees, whats the gradient at those points?

-1

Original post by A4M1NA

-1

So sub it in to the gradient equation and get a relationship for y in terms of x, then sub that into the equation of the ellipse and solve for x (and then y)

Original post by mqb2766

So sub it in to the gradient equation and get a relationship for y in terms of x, then sub that into the equation of the ellipse and solve for x (and then y)

So sub it in to the gradient equation and get a relationship for y in terms of x, then sub that into the equation of the ellipse and solve for x (and then y)

I managed to do part B. For the coordinates of A(-2root10, -root10) and B(2root10, root10)

Original post by A4M1NA

I managed to do part B. For the coordinates of A(-2root10, -root10) and B(2root10, root10)

Not worked it through, but looks plausible and easy enough to verify by subbing back in.

(edited 1 month ago)

Original post by mqb2766

Not worked it through, but looks plausible and easy enough to verify by subbing back in.

Not worked it through, but looks plausible and easy enough to verify by subbing back in.

okay thanks. For part C would I just use the formula for distance between two points

Original post by A4M1NA

okay thanks. For part C would I just use the formula for distance between two points

Looks like it. The "hence ..." and 2 marks is about right for pythagoras.

Original post by mqb2766

Looks like it. The "hence ..." and 2 marks is about right for pythagoras.

Looks like it. The "hence ..." and 2 marks is about right for pythagoras.

Thank youuu

Reply 18

1 month ago

Original post by A4M1NA

I have completed part a of this question but I'm stuck on part b. Can anyone help me with this part?

Since part a) was a show that question, you will need to use that for part b).

Points A and B are the points of inflection. Points of inflection are defined as ‘curve changing from concave to convex, and when second differential is zero’.

So you: differentiate again, set it equal to 0 and find the x coordinates. Proof it is a point of inflection by substituting numbers into the second differential just below and just above the x values. You should get a positive (convex) and negative (concave) values. Than find the y values.

Reply 19

1 month ago

Original post by BankaiGintoki

Since part a) was a show that question, you will need to use that for part b).

Points A and B are the points of inflection. Points of inflection are defined as ‘curve changing from concave to convex, and when second differential is zero’.

So you: differentiate again, set it equal to 0 and find the x coordinates. Proof it is a point of inflection by substituting numbers into the second differential just below and just above the x values. You should get a positive (convex) and negative (concave) values. Than find the y values.

Points A and B are the points of inflection. Points of inflection are defined as ‘curve changing from concave to convex, and when second differential is zero’.

So you: differentiate again, set it equal to 0 and find the x coordinates. Proof it is a point of inflection by substituting numbers into the second differential just below and just above the x values. You should get a positive (convex) and negative (concave) values. Than find the y values.

Just catching up with this thread and have a few observations on your post.

* There's no reason to suppose that points A and B correspond to points of inflection;

* Just looking at the given graph you can see there's no change of convexity near points A and B;

* I don't believe there are any points on the curve where the second derivative with respect to x is zero.

What the question is asking you to do is find points on the curve where the tangential gradient is equal to -1 (corresponding to the 45deg slope shown on the given diagram).

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