Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 4 | A Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
need help with this qn, pls explain too
Date Posted: 3 years ago
Views: 535
Functions work like this: not every x-value will have a corresponding y value on your graph. Hence, you will need to find the range of values for x so that there is a corresponding value of y.
What you can do is either:
- plot the graph of this equation, so that you can see the overall shape and see the limits of x from your sketch. Plug in x-values to get corresponding y-values (I recommend using whole numbers to do this).
or:
- adjust the function so that it is in terms of y.
(i) The clue is √9 - 2x^2. Remember that negative numbers cannot have a square root. Hence, 9 - 2x^2 ≥ 0.
Solve the inequality, and you should get x ≥-3/√2 and x ≤ 3/√2.
Hence, your range is -3/√2 ≤ x ≤ 3/√2.
(ii) You can differentiate both sides and attempt to work your way to this.
Here's how I did it:
y = x^2 √9 - 2x^2
y^2 = x^4 (9 - 2x^2)
= 9x^4 - 2x^6 (I removed the square root first)
(Differentiate with respect to y on both sides)
2y dy/dx = 36 x^3 - 12 x^5
y dy/dx = 18 x^3 - 6 x^5
y dy/dx + 6 x^5 - 18 x^3 = 0
Hence,
y dy/dx + 6 x^3 (x^2 - 3) = 0 (shown)
What you can do is either:
- plot the graph of this equation, so that you can see the overall shape and see the limits of x from your sketch. Plug in x-values to get corresponding y-values (I recommend using whole numbers to do this).
or:
- adjust the function so that it is in terms of y.
(i) The clue is √9 - 2x^2. Remember that negative numbers cannot have a square root. Hence, 9 - 2x^2 ≥ 0.
Solve the inequality, and you should get x ≥-3/√2 and x ≤ 3/√2.
Hence, your range is -3/√2 ≤ x ≤ 3/√2.
(ii) You can differentiate both sides and attempt to work your way to this.
Here's how I did it:
y = x^2 √9 - 2x^2
y^2 = x^4 (9 - 2x^2)
= 9x^4 - 2x^6 (I removed the square root first)
(Differentiate with respect to y on both sides)
2y dy/dx = 36 x^3 - 12 x^5
y dy/dx = 18 x^3 - 6 x^5
y dy/dx + 6 x^5 - 18 x^3 = 0
Hence,
y dy/dx + 6 x^3 (x^2 - 3) = 0 (shown)
y = x^2 √9 - 2x^2
y^2 = x^4 (9 - 2x^2)
= 9x^4 - 2x^6 (I removed the square root first)
(Differentiate with respect to y on both sides)
2y dy/dx = 36 x^3 - 12 x^5
-----------------------------------
1. I don't think differentiating functions implicitly is taught in the O Levels at the moment.
2. Isn't it differentiate with respect to x on both sides? (you wrote "with respect to y")
y^2 = x^4 (9 - 2x^2)
= 9x^4 - 2x^6 (I removed the square root first)
(Differentiate with respect to y on both sides)
2y dy/dx = 36 x^3 - 12 x^5
-----------------------------------
1. I don't think differentiating functions implicitly is taught in the O Levels at the moment.
2. Isn't it differentiate with respect to x on both sides? (you wrote "with respect to y")
LockB, to see this question, follow Marcus' steps to see what it means by the validity (the curve does not exist at some values of x).
See 1 Answer
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
The validity is simply due to that square root stuff.
For the second part, we just sub in and inspect the elements accordingly. You will only learn a different method of obtaining the equation in part ii by a method called implicit differentiation, taught in the A Levels if you are pursuing studies in the A Levels.
For the second part, we just sub in and inspect the elements accordingly. You will only learn a different method of obtaining the equation in part ii by a method called implicit differentiation, taught in the A Levels if you are pursuing studies in the A Levels.
Date Posted:
3 years ago
Alternative to ii) :
y = x²√(9 - 2x²)
y = √(x⁴(9 - 2x²))
y = √(9x⁴ - 2x⁶) = (9x⁴ - 2x⁶)¹/²
dy/dx = ½ (9x⁴ - 2x⁶)-¹/² (36x³ - 12x⁵)
dy/dx = (18x³ - 6x⁵) / √(9x⁴ - 2x⁶)
Multiply by √(9x⁴ - 2x⁶) on both sides,
√(9x⁴ - 2x⁶) dy/dx = 18x³ - 6x⁵
y dy/dx = 18x³ - 6x⁵
[Since y = √(9x⁴ - 2x⁶), rewrite the one on the LHS as y]
y dy/dx + 6x⁵ - 18x³ = 0
y dy/dx + 6x³(x² - 3) = 0
(Shown)
y = x²√(9 - 2x²)
y = √(x⁴(9 - 2x²))
y = √(9x⁴ - 2x⁶) = (9x⁴ - 2x⁶)¹/²
dy/dx = ½ (9x⁴ - 2x⁶)-¹/² (36x³ - 12x⁵)
dy/dx = (18x³ - 6x⁵) / √(9x⁴ - 2x⁶)
Multiply by √(9x⁴ - 2x⁶) on both sides,
√(9x⁴ - 2x⁶) dy/dx = 18x³ - 6x⁵
y dy/dx = 18x³ - 6x⁵
[Since y = √(9x⁴ - 2x⁶), rewrite the one on the LHS as y]
y dy/dx + 6x⁵ - 18x³ = 0
y dy/dx + 6x³(x² - 3) = 0
(Shown)
LockB :
The goal of this question is is for students to realise that multiplying by y on LHS is the same as multiplying by the square root term on the RHS, eventually reaching the desired equation by bringing terms over to the other side.
Inspection is merely verifying and only leads to the conclusion that 'oh, so its correct and it fits' but misses out the above skill of realising that there needs to be a multiplication by y from dy/dx to get y dy/dx.
The goal of this question is is for students to realise that multiplying by y on LHS is the same as multiplying by the square root term on the RHS, eventually reaching the desired equation by bringing terms over to the other side.
Inspection is merely verifying and only leads to the conclusion that 'oh, so its correct and it fits' but misses out the above skill of realising that there needs to be a multiplication by y from dy/dx to get y dy/dx.
thx :)
360 Tutoring Program