Maths P1.. HELP needed!!

[ashutoshamatya]

10. The equation of a curve is y = (5x+4)^0.5

(ii)a point with coordinate (x,y) moves along the curve in such a way that the rate of increase of x has the constant value 0.03 units per second. Find the rate of increase of y at the instant when x=1

(iii)Find the area enclosed by the curve, the x-axis, the y-axis and the line x=1.

 

[ks136]

Let me try:
(i) y=(5x+4)^0.5
dy/dx= 0.5 (5x+4)^(-0.5) X 5
dy/dx= 2.5 (5x+4)^(-0.5)
therefore...as it is given in the question that rate of change of x=0.03
R.O.C of y/ROC of x = 5/6 (replacing x by 1 as question has asked for ROC when x=1)
R0C of y= 5/6 X 0.03
R0C of y= 0.025

I AM NOT SURE THAT IT IS THE RIGHT ANSWER...PLEASE LET ME KNOW....

(ii)│⌠

0⌠1 [(5x+4)^0.5] dx= │2/15 (5x+4)^1.5│
and then put the limits 0-1 and u'll get the answer...


I U EXPERIENCE ANY OTHER PROBLEMS...OR IF I HAD DONE ANY MISTAKE IN ANSWERING UR QUESTION THEN DO LET ME KNOW

 

[Virtuoso]

two ways to solve the problems...
first:
find "x" in terms of "y" then differentiate it by using chain rule.... keep the value of dx\dt.. and you'll get the answer

secod:
directly start differentiating with respect to "t". use chain rule to differentate the right side and ull recieve the answer....

 

[leosco1995]

For Q1, just differentiate y and then use the chain rule. dx/dt is 0.03.

y = (5x+4)^0.5
dy/dx = 0.5 *(5x+4)^-0.5 * (5)
dy/dx = 2.5(5x+4)^-0.5

dy/dt = dy/dx * dx/dt
dy/dt = 2.5(5(1)+4)^-0.5 * 0.03
dy/dt = 2.5(9)^-0.5 * 0.03
dy/dt = 0.025