(a) Find the derivative of y = x\(^2\) (1 + x)\(^{\frac{3}{2}}\) with respect to x.
(b) The centre of a circle lies on the line 2y - x = 3. If the circle passes through P(2,3) and Q(6,7), find its equation.
(b) The centre of a circle lies on the line 2y - x = 3. If the circle passes through P(2,3) and Q(6,7), find its equation.
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Correct Answer: Option
Explanation:
y = x\(^2\)(1 + x)\(^{\frac{3}{2}}\)
Let u = x\(^2\)
v = (1 + x)\(^{\frac{3}{2}}\)
\(\frac{du}{dx} 2x\)
\(\frac{dv}{dx} = \frac{3}{2}\)(1 + x)\(^\frac{1}{2}\)
\(\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}\)
= (1 + x)\(^{\frac{3}{2}}\)(2x) + x\(^2\)\(\frac{3}{2}(1 + x)^{\frac{1}{2}}\)
= 2x(1 + x)\(^{\frac{3}{2}}\) + \(\frac{3x^2(1 + x)^{\frac{1}{2}}}{2}\)
(b)
2y - x = 3
when x = 0, y = \(\frac{3}{2}\)
when y = 0, x = -3
h = -3, k = \(\frac{3}{2}\)
Using
(x - h)\(^2\) + (y - k)\(^2\) = r\(^2\)
(x + 3)\(^2\) + (y - \(\frac{3}{2}\))\(^2\) = r\(^2\)
r(2, 3), x = 2, y = 3
(2 + 3)\(^2\) + (3 + \(\frac{3}{2}\))\(^2\) = r\(^2\)
\(\frac{109}{4}\) = r\(^2\)
The equation will give
(x + 3)\(^2\) + (y - \(\frac{3}{2}\))\(^2\)
= \(\frac{109}{4}\)
y = x\(^2\)(1 + x)\(^{\frac{3}{2}}\)
Let u = x\(^2\)
v = (1 + x)\(^{\frac{3}{2}}\)
\(\frac{du}{dx} 2x\)
\(\frac{dv}{dx} = \frac{3}{2}\)(1 + x)\(^\frac{1}{2}\)
\(\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}\)
= (1 + x)\(^{\frac{3}{2}}\)(2x) + x\(^2\)\(\frac{3}{2}(1 + x)^{\frac{1}{2}}\)
= 2x(1 + x)\(^{\frac{3}{2}}\) + \(\frac{3x^2(1 + x)^{\frac{1}{2}}}{2}\)
(b)
2y - x = 3
when x = 0, y = \(\frac{3}{2}\)
when y = 0, x = -3
h = -3, k = \(\frac{3}{2}\)
Using
(x - h)\(^2\) + (y - k)\(^2\) = r\(^2\)
(x + 3)\(^2\) + (y - \(\frac{3}{2}\))\(^2\) = r\(^2\)
r(2, 3), x = 2, y = 3
(2 + 3)\(^2\) + (3 + \(\frac{3}{2}\))\(^2\) = r\(^2\)
\(\frac{109}{4}\) = r\(^2\)
The equation will give
(x + 3)\(^2\) + (y - \(\frac{3}{2}\))\(^2\)
= \(\frac{109}{4}\)