Anonymous
Anonymous asked in Education & ReferenceHomework Help · 1 month ago

# Need some help with finding the X coordinate?

Hi, I need some help with this question. I have found the answer to part A of the question which is b=-3 though not sure how to approach part (b) Relevance

My thoughts are, since it divides the area into two equal parts, we must have integral from 0 to t of bx(x - 2) - mx dx = integral from 0 to t of mx + integral from t to 2 of bx(x - 2).

Where t is the x-coordinate of point A.

I'll assume you're correct about b = -3.

Then, int_0^t -3x^2 + (6 - m)x dx = mx^{2}/2 (t to 0) + int_t^{2} -3x^{2} + 6x dx

I.e. -x^3 + (6 - m)/2 x^2 (t to 0) = mt^{2}/2 + -x^{3} + 3x^{2} (2 to t)

-t^{3} + (6 - m)/2 t^{2} = mt^{2}/2 - 8 + 12 - (-t^{3} + 3t^2)

-t^{3} + (6 - m)(1/2)t^{2} = m(1/2)t^{2} + 4 + t^{3} - 3t^{2}

-t^{3} + (6 - m)(1/2)t^{2} = m(1/2)t^{2} + 4 + t^{3} - 6(1/2)t^{2}

-t^{3} + (6 - m)(1/2)t^{2} = 4 - (6 - m)(1/2)t^{2} + t^{3}

0 = 2t^{3} - (6 - m)t^{2} + 4

It is known that the height of t is -3t(t - 2), thus the slope m is (-3t^{2} + 6t - 0)/(t - 0) = -3t + 6.

Substituting yields:0 = 2t^{3} - (6 - (-3t + 6))t^{2} + 4

0 = 2t^{3} - (3t)t^{2} + 4

-4 = -t^{3}

t = (4)^{1/3}

I.e. cubrt(4).

• Login to reply the answers
• You'd get better help if you posted this in "mathematics" too.  We aren't as bright here.  :)

• Login to reply the answers