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maths question

hi, please could i have help on part c
The ms says we need to find a lower limit, not sure what that means:
question: https://ibb.co/KcfCgDtn
ms: https://ibb.co/Hp93qNdW
thanks!

Reply 1

Original post
by anonymous56754
hi, please could i have help on part c
The ms says we need to find a lower limit, not sure what that means:
question: https://ibb.co/KcfCgDtn
ms: https://ibb.co/Hp93qNdW
thanks!

If you're asking about the minimum value specifically, imagine sliding that normal line down vertically - there are points where the line crosses the curve at 2 points and then it will reach a point where it intersects once. Since you're equating a quadratic to a line, this is the usual "discriminant = 0" problem.

Reply 2

Original post
by davros
If you're asking about the minimum value specifically, imagine sliding that normal line down vertically - there are points where the line crosses the curve at 2 points and then it will reach a point where it intersects once. Since you're equating a quadratic to a line, this is the usual "discriminant = 0" problem.
But when the discriminant=0, isn’t there only one root and the questions asks for two distinct roots so shouldn’t b2-4ac>0? This is shown in the upper limit but I don’t get the lower limit part. Thank you.IMG_1344.jpeg

Reply 3

Original post
by anonymous56754
But when the discriminant=0, isn’t there only one root and the questions asks for two distinct roots so shouldn’t b2-4ac>0? This is shown in the upper limit but I don’t get the lower limit part. Thank you.IMG_1344.jpeg

Youve found the value of k (y intercept) where the line touches the curve. So when k > that value ....

Youd get the same result solving b^2-4ac>0 but its usually less hassle to solve an equation rather than an inequality and interpret it appropriately.

Reply 4

Original post
by mqb2766
Youve found the value of k (y intercept) where the line touches the curve. So when k > that value ....
Youd get the same result solving b^2-4ac>0 but its usually less hassle to solve an equation rather than an inequality and interpret it appropriately.

Sorry, I’m still quite confused. I’ve attached my working below through solving but I don’t get how you can get two k values because there is no quadratic?IMG_1348.jpeg

Reply 5

Original post
by anonymous56754
Sorry, I’m still quite confused. I’ve attached my working below through solving but I don’t get how you can get two k values because there is no quadratic?IMG_1348.jpeg

just getting your pencil out and sliding it up and down the y axis (different values of k) and with a gradient of -1/2 should convince you that there is only one value of k where the pencil touches the curve and it corresponds to the lower value of k. So your working is correct in that k>43/8 corrresponds to the range of k where the l intersects the quadratic. So Im not sure what youre confused about?

You stated in the OP you were only asking about the lower value, but the upper value for k (two values?) is when l intersects with the starting point of the parametric curve, as should be clear from the given diagram and moving your pencil up and down the y-axis with a grad of -1/2. The curve C is a quadratic function but only defined on a restricted domain as should be clear from the earlier parts.

Reply 6

Original post
by mqb2766
just getting your pencil out and sliding it up and down the y axis (different values of k) and with a gradient of -1/2 should convince you that there is only one value of k where the pencil touches the curve and it corresponds to the lower value of k. So your working is correct in that k>43/8 corrresponds to the range of k where the l intersects the quadratic. So Im not sure what youre confused about?
You stated in the OP you were only asking about the lower value, but the upper value for k (two values?) is when l intersects with the starting point of the parametric curve, as should be clear from the given diagram and moving your pencil up and down the y-axis with a grad of -1/2. The curve C is a quadratic function but only defined on a restricted domain as should be clear from the earlier parts.
Wait so my working is correct and you don’t need the lower limit?

Reply 7

Original post
by anonymous56754
Wait so my working is correct and you don’t need the lower limit?

Im a bit confused about what youre asking but k>43/8 is correct (see the ms). You can find it as an equality discriminant=0, to get 43/8 then put the > in. Or solve the discriminant>0 as youve done. Personally I think the ms way is a bit simpler/less error prone but theres not much in it.

Reply 8

Original post
by mqb2766
Im a bit confused about what youre asking but k>43/8 is correct (see the ms). You can find it as an equality discriminant=0, to get 43/8 then put the > in. Or solve the discriminant>0 as youve done. Personally I think the ms way is a bit simpler/less error prone but theres not much in it.

IMG_1359.jpeg Sorry if I’m being unclear, basically the red part I didn’t do but I don’t get why they are doing that, the green part is fine

Reply 9

Original post
by anonymous56754
IMG_1359.jpeg Sorry if I’m being unclear, basically the red part I didn’t do but I don’t get why they are doing that, the green part is fine

I realize its that part, but you get the same ans now k>43/8 from disciminant>0 whereas the ms does discriminant=0 and then replaces the = with a >. So what part is unclear?

Loosely speaking theyve slid their pencil down the y axis from the upper bound where the two intersections start and noted that the two solutions become one when the pencil is tangent to the curve (discriminant equals zero). As far as I can see it, this is the same logic that youve done.

Reply 10

Original post
by mqb2766
I realize its that part, but you get the same ans now k>43/8 from disciminant>0 whereas the ms does discriminant=0 and then replaces the = with a >. So what part is unclear?
Loosely speaking theyve slid their pencil down the y axis from the upper bound where the two intersections start and noted that the two solutions become one when the pencil is tangent to the curve (discriminant equals zero). As far as I can see it, this is the same logic that youve done.

Oh I see, but why did they get 13/2? I only got 43/8 as you can see in my working but it’s only one solution?

Reply 11

Original post
by anonymous56754
Oh I see, but why did they get 13/2? I only got 43/8 as you can see in my working but it’s only one solution?

13/2 is the upper bound for the y intercept which is obained by finding where your pencil intersects with the starting point of C (and one other point which is trivially observed from the sketch). Its got nothing to do with the discriminant / lower bound inequality.

It comes about because your curve is a quadratic only over a restricted domain because of the parametric definition. If it was an "unrestricted" quadratic, there would be no upper bound for the y intercept.
(edited 10 months ago)

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