Diagram shows the straight line $\displaystyle y = -x + k$ touching the curve $\displaystyle y = 3x - x^2$ at point $\displaystyle A$ .
a)Find the value of $\displaystyle k$
b)the coordinates of point $\displaystyle A$ .
Hi!
The point A has the x-coordinate a.
You do have the straight line $\displaystyle y_1 = -x+k$and the curve $\displaystyle 3x-x^2 $.
you know the straight line touches the curve at point A with the x-coordinate a
So that means at x = a (point A) the straight line has the same increase at point A as the curve.
That means
$\displaystyle y_1 ' = y_2 '$
$\displaystyle \frac{dy_1}{dx}= \frac{dy_2}{dx}$
more precisely:
$\displaystyle \frac{d}{dx}(-x+k) = \frac{d}{dx}(3x-x^2) $
$\displaystyle -1 = \frac{d}{dx}(3x-x^2) $
$\displaystyle -1 = 3-2x $
if the question want the area of shaded region
at the curve $\displaystyle y = 3x-x^2$ the $\displaystyle x-intercept $ is
$\displaystyle 0=3x - x^2$
$\displaystyle x(3 - x)=0$
$\displaystyle x = 3$
hence, $\displaystyle y=-x^2 + 3x$
$\displaystyle \int_{2}^{3} {y} \ dx$
$\displaystyle \int_{2}^{3} (-x^2 + 3x) dx$
$\displaystyle \left[ \frac{-x^3}{3} + \frac {3x^2}{2} \right]_{2}^{3}$
$\displaystyle \left[\frac{-(3)^3}{3} + \frac{3(3)^2}{2} \right] - \left[\frac{-(2)^3}{3} + \frac{3(2)^2}{2} \right]$
$\displaystyle \left[-9 + \frac{27}{2} \right] - \left[\frac{-8}{3} + 6 \right]$
$\displaystyle =\frac{7}{6} unit^2$
am i right?
No! It is A=(2,2) and k =4.
As you know we shall solve
3 - 2a = -1
solving for a :
-2a = -1 -3
-2a = -4
a = 2
now substitute a = 2 in -a + k = 3a - a^2 (as mr fantastic was saying)
-2+k = 3*2 - 2^2
-2+k = 6-4
-2+k = 2
k = 2+2 = 4
So the straight line is y = -x+4
now substitute x = 2 and you have
y = -2+4 =2
So A = (2,2)
Yes, you are! GJ
Ok..I very understand sir..Thanks!
lastly,if the question want the volume generated,in term of $\displaystyle \pi$,
when the region bounded by the curve $\displaystyle y=3x - x^2$ and the $\displaystyle x-axis$ is revolved through $\displaystyle 360^0$ about the $\displaystyle x - axis$
i get,
$\displaystyle =\int_{2}^{3} \pi{y^2} dx$
$\displaystyle =\int_{2}^{3} \pi(3x-x^2)^2 dx$
$\displaystyle =\pi\int_{2}^{3} (9x^2-6x^3+x^4) dx$
$\displaystyle =\pi \left[\frac{9x^3}{3}-\frac{6x^4}{4}+\frac{x^5}{5} \right]_{2}^{3}$
$\displaystyle =\pi\left[\frac{9(3)^3}{3}-\frac{6(3)^4}{4}+\frac{(3)^5}{5} \right] - \left[\frac{9(2)^3}{3}-\frac{6(2)^4}{4}+\frac{(2)^5}{5} \right]$
$\displaystyle =\pi\left(81-\frac{243}{2} + \frac{243}{5} \right) - \left(24-24 + \frac{32}{5}\right)$
$\displaystyle =\frac{17}{10} \pi unit^3$
right?