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secondary 4 | A Maths
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kathy2
Kathy2

secondary 4 chevron_right A Maths chevron_right Singapore

Thanks

Date Posted: 4 months ago
Views: 41
Eric Nicholas K
Eric Nicholas K
4 months ago
Now

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Eric Nicholas K
Eric Nicholas K's answer
3767 answers (Tutor Details)
1st
Q1, same idea as the previous question
kathy2
Kathy2
3 months ago
Can you check for me why the normal part my book answer is different from you
Eric Nicholas K
Eric Nicholas K
3 months ago
Kathy, I made a mistake in the copying, carrying forward the error. I miswrote the x = 1, y = -3 as (1, 3). I will redo this one later. Just finished teaching my student techniques on trigo identity solving.
kathy2
Kathy2
3 months ago
Ok. Thanks
Eric Nicholas K
Eric Nicholas K
3 months ago
That means for the tangent and normal, 1, -3 should be used instead of 1, 3.
kathy2
Kathy2
3 months ago
Are u around?
Eric Nicholas K
Eric Nicholas K
3 months ago
I will help you once ready
Eric Nicholas K
Eric Nicholas K
3 months ago
Kathy, my phone ran into the same trouble again. I will upload my answers by scanning shortly and then answer questions from online.
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Eric Nicholas K
Eric Nicholas K's answer
3767 answers (Tutor Details)
This method of finding equations of tangents extends to curves which are not polynomials, such as graphs of ln x and so on.
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Eric Nicholas K
Eric Nicholas K's answer
3767 answers (Tutor Details)
Here
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Eric Nicholas K
Eric Nicholas K's answer
3767 answers (Tutor Details)
Q1, corrected version
kathy2
Kathy2
3 months ago
I thought u will change all, haha.
I try myself
Eric Nicholas K
Eric Nicholas K
3 months ago
Which one would you like me to change? I will help you later once I am ready.
Eric Nicholas K
Eric Nicholas K
3 months ago
The techniques to form equations of tangent and normal to a curve at a specific point are as follows.

1. Obtain the coordinates required for the tangent. If the question says “find the equation...at x = 2”, we need to find the value of y based on the equation of the curve for the coordinates.

2. Obtain dy/dx of the curve for any general point and then sub in the x-value of the coordinate to find the value of dy/dx at that x-value. This is the gradient of the tangent. For the normal, the new value is -1 divided by this number.

3. Now that you have the coordinates and the gradient, you can then form the equation using your y = mx + c.
kathy2
Kathy2
3 months ago
Qns 2 and qns 3
Eric Nicholas K
Eric Nicholas K
3 months ago
Ok will redo them once ready, I am about to finish a question from another person and then doing other things first. Writing up Q2 and Q3 at 3 pm.
kathy2
Kathy2
3 months ago
Ok. Thanks
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Eric Nicholas K
Eric Nicholas K's answer
3767 answers (Tutor Details)
Q2, Q3, redone
kathy2
Kathy2
3 months ago
Thank you. I post a new chapter
Eric Nicholas K
Eric Nicholas K
3 months ago
Ok wait