Consider the curve given by x^2+4y^2=7+3xy
a) Show that dy/dx=(3y-2x)/(8y-3x)
b) Show that there is a point P with x-coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P.
c) Find the value of d^2y/dx^2 (second derivative) at the point P found in part b). Does the curve have a local maximum, a local minimum, or neither at the point P? Justify your answer..
The only part that I did was a). I don't really know how to find Point P for b) and c)
a)
2 x dx + 8 y dy = 3x dy + 3y dx
dy(3x-8y)=dx(2x-3y)
b)
Where is the slope = 0?
where 2x = 3y
if x = 3
9 + 4 y^2 = 7 + 9 y
4 y^2 - 9 y + 2 = 0
(4y-1)(y-2) = 0
y = 1/4 or y = 2
check both of those to find which works
(3,2)
3*2 - 2*3 = 0
yes, (3,2) works
(3,1/4)
3/4 -6 = 0 no way so (3,2) is it
c)
dy/dx=(3y-2x)/(8y-3x)
d^2y/dx^2 = [(8y-3x)(-2)-(3y-2x)(-3)] / (8y-3x)^2
= [(16-9)(-2)]/(16-9)^2 = -2/7
that is negative, so a maximum of the function
Thank you so much! It makes perfect sense and when I calculated it, it matched up.
I'm confused as to how to do part C), I don't remember how to find the second derivative of an implicit differentiation problem!
Well, don't worry! I'm here to help you with parts b) and c). Let's get started!
b) To find the point P with x-coordinate 3 at which the tangent line to the curve is horizontal, we need to find the corresponding y-coordinate. We can do this by using the equation of the curve.
Start with the given curve:
x^2 + 4y^2 = 7 + 3xy
Take the derivative of both sides with respect to x:
2x + 8y(dy/dx) = 0
To find the point P where the tangent line is horizontal (meaning dy/dx = 0), we substitute x = 3 into the equation above:
2(3) + 8y(dy/dx) = 0
Solving for dy/dx:
6 + 8y(dy/dx) = 0
8y(dy/dx) = -6
dy/dx = -6/(8y)
dy/dx = -3/(4y)
Now we have the expression for dy/dx. To find the y-coordinate of P, we substitute x = 3 and dy/dx = 0 into the original equation:
3^2 + 4y^2 = 7 + 3(3)y
9 + 4y^2 = 7 + 9y
4y^2 - 9y + 2 = 0
Solving this quadratic equation, we find that y = 2/4 or y = 1/2. So the point P is (3, 1/2).
c) To find the second derivative (d^2y/dx^2) at the point P (3, 1/2), we need to differentiate the expression for dy/dx with respect to x. Let's go!
Start with the expression for dy/dx:
dy/dx = -3/(4y)
Differentiate both sides with respect to x:
(d^2y/dx^2)(1) = -3(4(dy/dx))/(4y)^2
Since we already know that dy/dx = 0 at P, we substitute 0 for dy/dx:
(d^2y/dx^2)(1) = -3(4(0))/(4(1/2))^2
(d^2y/dx^2) = 0
So at the point P (3, 1/2), the second derivative is 0. This means that the curve doesn't have a local maximum or minimum at that point. It could be a saddle point or an inflection point.
I hope this helps! Let me know if you have any other questions.
To solve parts b) and c), let's start by rewriting the given equation of the curve:
x^2 + 4y^2 = 7 + 3xy
a) To find dy/dx, we need to implicitly differentiate the equation with respect to x.
Differentiating both sides of the equation with respect to x:
2x + 8yy' = 3xdy/dx + 3y - 3xy'
Rearranging the terms and factoring out dy/dx:
8yy' + 3xy' = 3xy - 2x
Combining like terms:
y'(8y + 3x) = 3xy - 2x
Dividing both sides by (8y + 3x) to solve for dy/dx:
dy/dx = (3xy - 2x)/(8y + 3x)
So, we have shown that dy/dx = (3y - 2x)/(8y - 3x).
b) To find the point P with x-coordinate 3 at which the tangent line is horizontal, we need to find the y-coordinate of P.
In a horizontal tangent line, dy/dx = 0. So, we set dy/dx = 0 and solve for y.
(3y - 2x)/(8y - 3x) = 0
To make the numerator zero, we set 3y - 2x = 0.
Substituting x = 3:
3y - 2(3) = 0
3y - 6 = 0
3y = 6
y = 2
Therefore, the point P with x-coordinate 3 is P(3, 2).
c) To find the second derivative, d^2y/dx^2, we need to take the derivative of dy/dx with respect to x.
Differentiating dy/dx = (3y - 2x)/(8y - 3x) with respect to x:
d^2y/dx^2 = [d/dx((3y - 2x)/(8y - 3x))] / (8y - 3x) - [(3y - 2x)d/dx(8y - 3x)] / (8y - 3x)^2
Simplifying further:
d^2y/dx^2 = [(2(8y - 3x) - (3y - 2x)(-3))/(8y - 3x)^2] / (8y - 3x)
d^2y/dx^2 = [(16y - 6x + 9y - 6x)/(8y - 3x)^2] / (8y - 3x)
d^2y/dx^2 = (25y - 12x) / (8y - 3x)^2
To find the value of d^2y/dx^2 at P(3, 2), substitute x = 3 and y = 2 into the expression:
d^2y/dx^2 = (25(2) - 12(3)) / (8(2) - 3(3))^2
d^2y/dx^2 = (-21) / (16 - 9)^2
d^2y/dx^2 = (-21) / 49
Since d^2y/dx^2 is negative at P(3, 2), the curve does not have a local maximum or minimum at this point.