AP Calculus AB

Question

Particle X moves along the positive x-axis so that its position at the time t (is greater than or equal to 0) is given by x(t)=(5t^3)-(9t^2)+7. A second particle, Y, moves along the positive y-axis so that its position at time t is given by y(t)=7t+3. At any time t, when it is greater than or equal to zero, the origin and the positions of the particles X and Y are the vertices of a triangle in the first quadrant. Find the rate of change of the area of the triangle at time=1

Answer 1

Given X = 5t³  - 9t²  +7

at t= 0, X = 7

 

Y = 7t + 3 

 at t= 0 Y = 3

 

Vertices of triangle at  t= 0 are (0,0) , (7, 0) and ( 0 ,3)

 

Area at t = 0 is given by 1/2 ( 7x3) = 21/2

 Positions of vertices at time t are

(0,0) , ( X , 0) and (Y,0 ) where values of X and Y are as given above

Area at t = 1/2 (5t³ -9t² +7 )( 7t +3) = 1/2 ( 35t⁴ - 63t³ +49t² +15t³ - 27 t² +21) = 1/2( 35 t⁴ -48t³ -27t² +49t +21)

Diffentiating wrt t dA/dt = 1/2(140t³- 144 t² -54 t +49) From this rate of change of area at ant time may be found

 

1/2 ( 140 t³ - 144t2 -54t +49)