I just need the 3rd part!!!
(2 points) Let y be defined implicitly by the equation ln(2y)=5xy. Use implicit differentiation to find the first derivative of y with respect to x.
dy/dx= 5y^2/(1-5xy) (I ALREADY GOT THIS)
Use implicit differentiation to find the second derivative of y with respect to x.
d^2ydx^2= y^3(250xy-75)/((5xy-1)^3) (ALREADY GOT THIS)
Note: Your answer should only involve the variables x and y. You should simplify your answer as much as possible before entering it into WeBWorK.
Find the point on the curve where d^2ydx^2=0.
d2ydx2=0 at the point (x,y)=
. Include the parentheses in your answer.
Well, you want the 2nd derivative = 0
so
y^3(250xy-75)/((5xy-1)^3) = 0
y^3(250xy-75) = 0
y^3 = 0, but that would make ln(2y) as undefined, so y ≠ 0
or
250xy-75 = 0
xy = -75/250 = -3/10
but ln(2y)=5xy
xy = ln(2y)/5
ln(2y)/5 = -3/10
ln(2y) = - 3/2
2y = e^(-3/2)
y = .111565
xy = ln(2y)/5
x = ln(2y)/(5y) = -2.689
Oh, finding where the second derivative equals zero? That's like trying to find a clown with no sense of humor – it's a bit tricky, but we'll make it work!
So, at the point where d^2y/dx^2 = 0, it means that the second derivative equals zero. In other words, the clown took a break from juggling and forgot how to make the roller coaster go up or down.
Let's find that point!
Setting the second derivative equal to zero: d^2y/dx^2 = y^3(250xy-75)/((5xy-1)^3) = 0
Now, to solve for the point (x,y), you'll need to find the values of x and y that satisfy this equation. It's like asking the clown for the secret to making people laugh – it's a mystery!
Be patient, my friend, and with a little detective work, you'll uncover the coordinates of that point!
To find the point on the curve where d^2y/dx^2 = 0, we can set the expression for d^2y/dx^2 equal to zero and solve for (x, y).
From the previous step, we have the expression for d^2y/dx^2:
d^2y/dx^2 = y^3(250xy - 75)/((5xy - 1)^3)
Setting this expression equal to zero:
y^3(250xy - 75)/((5xy - 1)^3) = 0
To solve this equation, we can set each factor equal to zero individually:
1) y^3 = 0
This implies that y = 0.
2) 250xy - 75 = 0
Dividing both sides by 25:
10xy - 3 = 0
10xy = 3
xy = 3/10
Thus, we have two possible points where d^2y/dx^2 = 0:
1) (x, y) = (3/10, 0)
2) (x, y) = (3/10, 0)
Note that in both cases, the value of y is zero.
Therefore, the point on the curve where d^2y/dx^2 = 0 is (x, y) = (3/10, 0).
To find the point on the curve where the second derivative of y with respect to x is equal to zero, we need to solve the equation d^2y/dx^2 = 0.
The second derivative of y with respect to x is given by:
d^2y/dx^2 = y^3(250xy - 75) / ((5xy - 1)^3)
Setting this equal to zero, we have:
y^3(250xy - 75) / ((5xy - 1)^3) = 0
To find the point where this equation is satisfied, we can set each term equal to zero individually.
First, setting y^3 = 0, we obtain:
y = 0
Next, setting (250xy - 75) = 0, we solve for x:
250xy = 75
xy = 75/250
xy = 3/10
x = (3/10y), where y is NOT equal to zero
Therefore, the point (x, y) where d^2y/dx^2 = 0 is given by:
(x, y) = ((3/10y), 0), where y ≠ 0
Note: This point is a critical point on the curve, indicating a possible point of inflection.