4條Differentiation應用題..急!2小時內解決

(1) 9x2 + 16y2 = 52
18x + 32y(dy/dx) = 0
dy/dx = -9x/16y
Slope of 9x - 8y - 2 = 0 is 9/8, so
-9x/16y = 9/8
x = -2y
Sub back to the original equation:
36y2 + 16y2 = 52
y2 = 1
y = 1 or -1
x = -2 or 2
So for (-2, 1), equation by point-sklope form is:
(y - 1)/(x + 2) = 9/8
8y - 8 = 9x + 18
9x - 8y + 26 = 0
For (-1, 2), equation by point-sklope form is:
(y - 2)/(x + 1) = 9/8
8y - 16 = 9x + 9
9x - 8y + 25 = 0
(2) dy/dx = 3x2
Slope of x + 3y + 4 = 0 is -1/3
So tangents' slope required = 3
Hence x = 1 or -1, then y = 6 or 4
For (1, 6), equation is:
(y - 6)/(x - 1) = 3
y - 6 = 3x - 3
3x - y + 3 = 0
For (-1, 4), equation is:
(y - 4)/(x + 1) = 3
y - 4 = 3x + 3
3x - y + 7 = 0
(3) Let (h, k) be the point on the curve on which the tangent to the curve at it passes through (3, -3).
Then dy/dx = -4x/9y = -4h/9k
So,
(k + 3)/(h - 3) = -4h/9k
9k2 + 27k = 12h - 4h2
4h2 + 9k2 = 12h - 27k
36 = 12h - 27k
12 = 4h - 9k
h = (12 - 9k)/4
Sub this relation into the curve equation:
(12 - 9k)2/4 + 9k2 = 36
(4 - 3k)2/4 + k2 = 4
(4 - 3k)2 + 4k2 = 16
25k2 - 24k = 0
k = 0 or 24/25
h = 3 or 21/25
So the points are (3, 0) and (21/25, 24/25) with equations:
At (3, 0): x = 3
At (21/25, 24/25):
(y - 24/25)/(x - 21/25) = (-3 - 24/25)/(3 - 21/25) = -11/6
6y - 144/25 = 231/25 - 11x
11x + 6y - 15 = 0
(4) The curve cuts the y-axis at (0, 2) and (0, -2)
Taking implicit differentiation for the equation:
2x + 5x(dy/dx) + 5y + 2y(dy/dx) = 0
dy/dx = -(2x + 5y)/(5x + 2y)
At (0, 2), dy/dx = -5/2
Equation of tangent:
(y - 2)/x = -5/2
2y - 4 = -5x
5x + 2y - 4 = 0
Equation of normal:
(y - 2)/x = 2/5
5y - 10 = 2x
2x - 5y + 10 = 0
At (0, -2), dy/dx = -5/2
Equation of tangent:
(y + 2)/x = -5/2
2y + 4 = -5x
5x + 2y + 4 = 0
Equation of normal:
(y + 2)/x = 2/5
5y + 10 = 2x
2x - 5y - 10 = 0

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