area and tangent to graph

[tsh44]

Hello,

the problem that is giving me some trouble:

f (x) = A (36-x^2) for 0 < A< 1 and let R be the region in the first quadrant bounded by the y-axis, the graph of f, and teh graph of g (x) = 36 - (x^2)

Find the area of R in terms of A
find, in erms of A, the equation of the line L, tangent to the graph of f at the point 6,0
Determine the value of A such that the line L divides the region R into two parts with equal areas

I believe the answer for the first part is the integral from 0 to 6 of (36-x^2) - (A(36-x^2)) dx. I believe this is how far you can go to put into terms of A.

For the seocnd question I know a lower A would make the slope less steep but I am not really sure what to do at all.

Also stuck on getting started with part c
thank you for any help

 

[skeeter]

Hello,

the problem that is giving me some trouble:

f (x) = A (36-x^2) for 0 < A< 1 and let R be the region in the first quadrant bounded by the y-axis, the graph of f, and teh graph of g (x) = 36 - (x^2)

Find the area of R in terms of A
find, in erms of A, the equation of the line L, tangent to the graph of f at the point 6,0
Determine the value of A such that the line L divides the region R into two parts with equal areas

I believe the answer for the first part is the integral from 0 to 6 of (36-x^2) - (A(36-x^2)) dx. I believe this is how far you can go to put into terms of A.

you can go farther ... find the value of the indefinite integral.

For the seocnd question I know a lower A would make the slope less steep but I am not really sure what to do at all.

once you find the area between g and f from part (a), the area between L and f will be 1/2 of that value. the line L will be defined by the equation y = f'(6)(x - 6)

Also stuck on getting started with part c
thank you for any help

 

[galactus]

As Skeeter and yourself pointed out, find \(\displaystyle \L\\\int_{0}^{6}\left[36-x^{2}-A(36-x^{2})\right]dx\)....[1]

The equation of your line at (6,0) in terms of A:

f'(x)=-2Ax, \(\displaystyle y=-12A(x-6)\)

The area between g(x) and this line must equal 1/2 the area from [1]:

\(\displaystyle \L\\\text{A2}=\int_{0}^{6}\left[36-x^{2}-(-12A(x-6))\right]dx\)=1/2 the result from [1].

Solve for A.

Also, the area between the line and f(x) must be the same.

\(\displaystyle \L\\\text{A1}=\int_{0}^{6}\left[-12A(x-6)-A(36-x^{2})\right]dx\)