Question

Hi am stuck on this question its partly implicit differentiation partly integration.

Find dy/dx given that y² =x(3-x)². State the values of x for which dy/dx is: (a) zero (b) infinite.

Sketch the curve of y² =x(3-x)² and find the volume of the solid formed when the area enclosed by the loop of the curve is rotated through 180 degrees about the x axis.

I HAVE SOLVED PART A, AM STUCK ON B AND THE SKETCH THANKS.

Answer 1

what did you get for part A ?

Answer 2

answer to the first part is ± (3/2√x-(3/2)√x)
answer to the second part is 1

Answer 3

y^2 = x(3-x)^2

differentiate both sides with-respect-to-x implicitly

2yy' = -2x(3-x) + (3-x)^2

y' = [-2x(3-x) + (3-x)^2]/y

Answer 4

for part A, set y' to 0 ans solve x :

[-2x(3-x) + (3-x)^2]/y = 0

(3-x)(-2x+3-x) = 0
(3-x)(3-3x) = 0
x = 3, 1

Answer 5

so the first derivative is 0 at x = 1 and 3, right ?

Answer 6

yes i agree, sorry about the late reply am using a tablet that is not showing the post option hence my slow reply i have to get to the computer and type

Answer 7

Part B: "Sketch the curve of y² =x(3-x)² and find the volume of the solid formed when the area enclosed by the loop of the curve is rotated through 180 degrees about the x axis." To start off: Recognize that y² =x(3-x)² is zero at both x=0 and x=3. Plot the points (0,0) and (3,0) You might want to choose a few other x-values and to calculate the corresponding y-values; in your shoes I'd spring for x values between 0 and 3 and also determine whether or not a negative x-value, such as x=-1, would be permissible. Hint: y² =x(3-x)² has to be zero or greater, since y² can't be negative.

Note: You should make maximum use of your results in Part A: Mark the point at which the slope of the tangent line is zero and that where the slope is infinite. Draw in the actual tangent lines.

This should certainly give you good insights into what this graph and the area it contains looks like.

Share your graph. Then we can progress on to finding the volume of the solid described.

Answer 8

graph
http://www.wolframalpha.com/input/?i=plot+y%5E2+%3D+x%283-x%29%5E2

volume using disc method
\[\pi \int\limits_a^b R^2\]
loop enclosed from x=0 to 3
radius of cross-section circle is the "y" value
\[y = \sqrt{x}(3-x)\]
\[V = \pi \int\limits_0^3 x (3-x)^2 dx\]