rate of change??

Question

In a figure, the graph of y=1/x^2.
line l is tangent to graph at point P, with coordinates (w, 1/w^2), where w>0. Point Q has coordinates (w, 0). Line l crosses xaxis at point R with coordinates (k,0) . ((line l intersects the x axis and forms triangle PQR.))
a. find value of k when w=3.
b. for all w>0, find k in terms of w.
c. suppose that w is increasint at constant rate of 7 unites per second. when the cordinate w=5, what is the rate of change of k with respect to time?
d. suppose that w is still increasing at the constant of 7 units per second. when w=5, what is the rate of change of the area of the triangle PQR with respect to time? determine the area is increasing or decreasing at this instant.

im confused...i dnt kno where to start!!!!
PLEASE HELP!!! THANKS!

well for a, i got k to equal 27/6? is this correct?
i did what you guys did for the first part, and solved for the x intercept point K, and got that answer...hmm..

Answer 1

a. y = 1/x^2
dy/dx = -2/x^3
so, at P, x = w so dy/dw = -2/w^3
Equation of tangent at P is given by:
y = mx + b
= -2/w^3 . x + b
The tangent passes through P (w, 1/w^2)
So 1/w^2 = -2/w^3 . w + b
= -2/w^2
Thus b = 3/w^2
So tangent is y = -2/w^3 x + 3/w^2
Tangent crosses x-axis at (k, 0)
ie -2/w^3 . k + 3/w^2 = 0
So 2/w^3 . k = 3/w^2
Thus k = 3w/2 = 9/2 when k = 3

b. k = 3w/2 as shown in a above

c. dk/dt =3/2 . dw/dt
So when dw/dt = 7, dk/dt = 7.3/2 = 21/2 units/sec

d. ΔPQR is a right angled triangle (PQ is perp to QR)
Therefore Area ΔPQR = A = ½PQ.QR
Length PQ = 1/w^2 (same x-coordinates so length = difference in y coordiantes)
Length PR = (3w/2 - w) (same y-coordinates so length = difference in x coordiantes)
=w/2.
So A = ½ .1/w^2 . w/2
=1/(4w)
dA/dt = -1/(4w^2) . dw/dt
When w = 5
dA/dt = -1/100 . 7
= -7/100 units^2/sec and this means that the area is decreasing at this instant as dA/dt < 0

Answer 2

First, get the equation of the tangent line.

Use the point-slope form,

y - y0 = m(x-x0)

where m is the slope, and (x0,y0) is any point on the line.

In this case, (x0,y0) = (w,1/w^2), the point of tangency. m is the slope of the tangent line, which is given by the derivative at x=w.

y' = -2/x^3, so at w, m = -2/w^3.

Therefore, the equation of the tangent line is

y - 1/w^2 = (-2/w^3)(x - w)
so
y = -2x/w^3 + 3/w^2,
after simplifying.

Next, you need to find the x-intercept of this line. Set y=0 and x=k, and solve for k. You have the equation

2k/w^3 = 3/w^2

and solving for k gives you k=3w^3/(2w^2) = 3w/2.

That takes care of parts a and b.

c. You have dw/dt = 7, and you want dk/dt. Differentiate the equation k = 3w/2 with respect to t and plug in dw/dt = 7.
d. The triangle has base k-w and height w, so the area is A = w(k-w)/2. Plug in k = 3w/2, then differentiate this equation with respect to t. Plug in w=5 and dw/dt = 7 to find dA/dt. If it's positive, it's increasing.

Answer 3

To get started, figure the derivative at point (w, 1/w^2). This will give you the slope. Then figure the x intercept. (This is point R). That will give you k.