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junior college 1 | H2 Maths
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Hi can anyone help in qn 1 please. Thank you
Date Posted: 2 years ago
Views: 269
Alternatively, for part iii, we wish to find the point in which is tangent is vertical. This can be done without having to sketch the graph.
If we look at this question from a different point of view, with x being the vertical axis and y being the horizontal axis, then the said "tangent" will be horizontal (because we have rotated the axes by 90 degrees).
In other words, we solve for dx/dy = 0.
y² = 9 (1 - x)
y²/9 = 1 - x
x = 1 - y²/9
Differentiating,
dx/dy = -2y/9
We wish for our original tangent to be vertical (and our "rotated tangent" to be horizontal), so dx/dy = 0.
-2y/9 = 0
y = 0
3 cos t = 0
cos t = 0
t = π/2
If we look at this question from a different point of view, with x being the vertical axis and y being the horizontal axis, then the said "tangent" will be horizontal (because we have rotated the axes by 90 degrees).
In other words, we solve for dx/dy = 0.
y² = 9 (1 - x)
y²/9 = 1 - x
x = 1 - y²/9
Differentiating,
dx/dy = -2y/9
We wish for our original tangent to be vertical (and our "rotated tangent" to be horizontal), so dx/dy = 0.
-2y/9 = 0
y = 0
3 cos t = 0
cos t = 0
t = π/2
See 1 Answer
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Like this maybe?
Date Posted:
2 years ago
Thank you
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