# Thread: Crazy AP style derivative problems

1. ## Crazy AP style derivative problems

Big homework assignment this week, and luckily i was able to do most of the problems. However, there are these 3 that i just can't go right.

1. Equation is $\displaystyle 2(y^3) + 6(x^2)y- 12(x^2) + 6y = 1$

a. Show that the derivative of the above equation is equal to $\displaystyle (4x -2xy)/(x^2 + y^2 + 1)$

b. Write an equation of each horizontal tangent line to the curve.

c. The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates of point P.

2. Equation is $\displaystyle x^2 + 4(y^2) = 7 = 3xy$

a. Show that the derivative of the above equation is equal to $\displaystyle (3y - 2x)/(8y - 3x)$

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 the 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.

3. Equation is $\displaystyle x(y^2) - (x^3)y = 6$

a. Show that the derivative of the above euation is equal to $\displaystyle (3(x^2)y - y^2)/(2xy - x^3)$

b. Find all the points on the curve whose x coordinate is 1, and write an equation for the tangent line at each of these points.

c. Find the x coordinate of each point on the curve where the tangent line is vertical.

Not easy, i know! Hopefully someone can help me out with some of these. Thanks in advance!!!!

2. ## I will give you the derivatives of each

Originally Posted by Blue Griffin
Big homework assignment this week, and luckily i was able to do most of the problems. However, there are these 3 that i just can't go right.

1. Equation is $\displaystyle 2(y^3) + 6(x^2)y- 12(x^2) + 6y = 1$

a. Show that the derivative of the above equation is equal to $\displaystyle (4x -2xy)/(x^2 + y^2 + 1)$

b. Write an equation of each horizontal tangent line to the curve.

c. The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates of point P.

2. Equation is $\displaystyle x^2 + 4(y^2) = 7 = 3xy$

a. Show that the derivative of the above equation is equal to $\displaystyle (3y - 2x)/(8y - 3x)$

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 the 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.

3. Equation is $\displaystyle x(y^2) - (x^3)y = 6$

a. Show that the derivative of the above euation is equal to $\displaystyle (3(x^2)y - Y^2)/(2xy - x^3)$

b. Find all the points on the curve whose x coordinate is 1, and write an equation for the tangent line at each of these points.

c. Find the x coordinate of each point on the curve where the tangent line is vertical.

Not easy, i know! Hopefully someone can help me out with some of these. Thanks in advance!!!!
$\displaystyle 2y^3+6x^2y-12x^2+6y=1$...then$\displaystyle 6y^2\cdot{y'}+6x^2\cdot{y'}+12xy-24x+6y'=0$...now solve for $\displaystyle y'$....$\displaystyle x^2+4y^2-7=3xy$...then $\displaystyle 2x+8y\cdot{y'}=3x\cdot{y'}+3y$..once again solve for $\displaystyle y'$....$\displaystyle xy^2-x^3y=6$...then $\displaystyle 2xy\cdot{y'}+y^2-x^3\cdot{y'}+3x^2y=0$....and solve for $\displaystyle y'$...I am sure you can do the rest

3. Originally Posted by Blue Griffin
Big homework assignment this week, and luckily i was able to do most of the problems. However, there are these 3 that i just can't go right.

1. Equation is $\displaystyle 2(y^3) + 6(x^2)y- 12(x^2) + 6y = 1$

a. Show that the derivative of the above equation is equal to $\displaystyle (4x -2xy)/(x^2 + y^2 + 1)$

b. Write an equation of each horizontal tangent line to the curve.

c. The line through the origin with slope -1 is tangent to the curve at point P. Find the x and y coordinates of point P.

2. Equation is $\displaystyle x^2 + 4(y^2) = 7 = 3xy$

a. Show that the derivative of the above equation is equal to $\displaystyle (3y - 2x)/(8y - 3x)$

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 the 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.

3. Equation is $\displaystyle x(y^2) - (x^3)y = 6$

a. Show that the derivative of the above euation is equal to $\displaystyle (3(x^2)y - y^2)/(2xy - x^3)$

b. Find all the points on the curve whose x coordinate is 1, and write an equation for the tangent line at each of these points.

c. Find the x coordinate of each point on the curve where the tangent line is vertical.

Not easy, i know! Hopefully someone can help me out with some of these. Thanks in advance!!!!
They are all simialr so lets do 3.

$\displaystyle x(y^2) - (x^3)y = 6$

using implicit differentation we get

$\displaystyle \underbrace{1 \cdot y^2+x(2y\frac{dy}{dx})}_{Product Rule}-(\underbrace{3x^2y+x^3\frac{dy}{dx}}_{Product Rule})=0$

$\displaystyle (2xy-x^3)\frac{dy}{dx}=3x^2y-y^2 \iff \frac{dy}{dx}= \frac{3x^2y-y^2}{2xy-x^3}$

for b) eval at x=1

$\displaystyle 1(y^2)-(1^3)y=6 \iff y^2-y-6=0 \iff (y-3)(y+2)=0$

so we get the points (1,3) and (1,-2)

for c) set the derivative equal to zero.

$\displaystyle 0=\frac{3x^2y-y^2}{2xy-x^3} \iff 0=3x^2y-y^2$

$\displaystyle 0=y(3x^2-y)$

so $\displaystyle y=0 \mbox{ or } y=3x^2$

y=0 is not on the curve (why?)

Now we can plug $\displaystyle y=3x^2$ into the original equation.

$\displaystyle x(y^2) - (x^3)y = 6 \iff x(3x^2)^2-x^3(3x^2)=6$

$\displaystyle 9x^5-3x^5=6 \iff x^5=1 \iff x=1$

Try a few of the others and see what you can do.

Good luck.

4. Ok, your advice helped a lot, and i was able to figure out 2. It was easier than i thought....

Need a bit of help with the rest of one, though. I don't really know where to start with b or c.

5. Originally Posted by Blue Griffin
Ok, your advice helped a lot, and i was able to figure out 2. It was easier than i thought....

Need a bit of help with the rest of one, though. I don't really know where to start with b or c.

Horizontal tangents occur when the derivative is zero

so set the derivative equal to zero and solve for the points

you will need to use the similar proceedure that I used in 3 to find the points on the graph.

set the derivative equal to -1 and see what you can do from there.

Good luck.