Find All Points On The Graph Of $y^3 - 27y = X^2 - 90$ At Which The Tangent Line Is Vertical. (Order Your Answers From Smallest To Largest $x$, Then From Smallest To Largest \$y$[/tex\].)$\begin{array}{l} (x, Y)

by ADMIN 217 views

Introduction

In this article, we will explore the problem of finding all points on the graph of the given equation at which the tangent line is vertical. This involves finding the derivative of the given function and setting it equal to zero to determine the critical points. We will then use these critical points to find the corresponding values of x and y.

Understanding the Problem

The given equation is $y^3 - 27y = x^2 - 90$. To find the points at which the tangent line is vertical, we need to find the derivative of this function with respect to x. This will give us the slope of the tangent line at each point on the graph.

Finding the Derivative

To find the derivative of the given function, we can use the chain rule and the power rule. The derivative of $y^3$ is $3y^2$, and the derivative of $-27y$ is $-27$. The derivative of $x^2$ is $2x$, and the derivative of $-90$ is $0$. Therefore, the derivative of the given function is:

ddx(y3−27y)=ddx(x2−90)\frac{d}{dx}(y^3 - 27y) = \frac{d}{dx}(x^2 - 90)

3y2dydx−27dydx=2x3y^2 \frac{dy}{dx} - 27 \frac{dy}{dx} = 2x

Setting the Derivative Equal to Zero

To find the points at which the tangent line is vertical, we need to set the derivative equal to zero and solve for x and y. This will give us the critical points on the graph.

3y2dydx−27dydx=2x3y^2 \frac{dy}{dx} - 27 \frac{dy}{dx} = 2x

dydx(3y2−27)=2x\frac{dy}{dx}(3y^2 - 27) = 2x

dydx=2x3y2−27\frac{dy}{dx} = \frac{2x}{3y^2 - 27}

Setting the derivative equal to zero, we get:

2x3y2−27=0\frac{2x}{3y^2 - 27} = 0

This implies that:

2x=02x = 0

x=0x = 0

Solving for y

Now that we have found the value of x, we can substitute it into the original equation to solve for y.

y3−27y=x2−90y^3 - 27y = x^2 - 90

y3−27y=02−90y^3 - 27y = 0^2 - 90

y3−27y=−90y^3 - 27y = -90

y(y2−27)=−90y(y^2 - 27) = -90

y(y+33)(y−33)=−90y(y + 3\sqrt{3})(y - 3\sqrt{3}) = -90

Finding the Critical Points

To find the critical points, we need to find the values of y that satisfy the equation.

y(y+33)(y−33)=−90y(y + 3\sqrt{3})(y - 3\sqrt{3}) = -90

We can see that y = 0 is a solution. To find the other solutions, we can set each factor equal to zero and solve for y.

y=0y = 0

y+33=0y + 3\sqrt{3} = 0

y−33=0y - 3\sqrt{3} = 0

Solving for y, we get:

y=0y = 0

y=−33y = -3\sqrt{3}

y=33y = 3\sqrt{3}

Finding the Corresponding Values of x

Now that we have found the values of y, we can substitute them into the original equation to find the corresponding values of x.

y3−27y=x2−90y^3 - 27y = x^2 - 90

03−27(0)=x2−900^3 - 27(0) = x^2 - 90

0=x2−900 = x^2 - 90

x2=90x^2 = 90

x=±90x = \pm \sqrt{90}

x=±310x = \pm 3\sqrt{10}

Finding the Corresponding Values of y

Now that we have found the values of x, we can substitute them into the original equation to find the corresponding values of y.

y3−27y=x2−90y^3 - 27y = x^2 - 90

(−33)3−27(−33)=x2−90(-3\sqrt{3})^3 - 27(-3\sqrt{3}) = x^2 - 90

−813+813=x2−90-81\sqrt{3} + 81\sqrt{3} = x^2 - 90

0=x2−900 = x^2 - 90

x2=90x^2 = 90

x=±310x = \pm 3\sqrt{10}

y=−33y = -3\sqrt{3}

y=33y = 3\sqrt{3}

Conclusion

In this article, we have found the points on the graph of the given equation at which the tangent line is vertical. We have used the derivative of the function to find the critical points, and then used these critical points to find the corresponding values of x and y. The points at which the tangent line is vertical are:

(0,0),(0,−33),(0,33),(−310,−33),(310,−33),(−310,33),(310,33)(0, 0), (0, -3\sqrt{3}), (0, 3\sqrt{3}), (-3\sqrt{10}, -3\sqrt{3}), (3\sqrt{10}, -3\sqrt{3}), (-3\sqrt{10}, 3\sqrt{3}), (3\sqrt{10}, 3\sqrt{3})

These points are ordered from smallest to largest x, and then from smallest to largest y.

Discussion

The problem of finding the points at which the tangent line is vertical is an important one in mathematics. It involves finding the derivative of a function and setting it equal to zero to determine the critical points. The critical points are then used to find the corresponding values of x and y. This problem is a classic example of how calculus can be used to solve real-world problems.

References

  • [1] Calculus, 3rd edition, by Michael Spivak
  • [2] Differential Equations and Dynamical Systems, 3rd edition, by Lawrence Perko
  • [3] Calculus: Early Transcendentals, 7th edition, by James Stewart

Introduction

In our previous article, we explored the problem of finding all points on the graph of the given equation at which the tangent line is vertical. We used the derivative of the function to find the critical points, and then used these critical points to find the corresponding values of x and y. In this article, we will answer some of the most frequently asked questions about this problem.

Q: What is the significance of finding the points at which the tangent line is vertical?

A: Finding the points at which the tangent line is vertical is an important problem in mathematics because it involves finding the critical points of a function. Critical points are points on the graph of a function where the derivative is zero or undefined. These points are important because they can help us understand the behavior of the function.

Q: How do we find the derivative of the given function?

A: To find the derivative of the given function, we can use the chain rule and the power rule. The derivative of $y^3$ is $3y^2$, and the derivative of $-27y$ is $-27$. The derivative of $x^2$ is $2x$, and the derivative of $-90$ is $0$. Therefore, the derivative of the given function is:

ddx(y3−27y)=ddx(x2−90)\frac{d}{dx}(y^3 - 27y) = \frac{d}{dx}(x^2 - 90)

3y2dydx−27dydx=2x3y^2 \frac{dy}{dx} - 27 \frac{dy}{dx} = 2x

Q: How do we set the derivative equal to zero and solve for x and y?

A: To find the points at which the tangent line is vertical, we need to set the derivative equal to zero and solve for x and y. This will give us the critical points on the graph.

3y2dydx−27dydx=2x3y^2 \frac{dy}{dx} - 27 \frac{dy}{dx} = 2x

dydx(3y2−27)=2x\frac{dy}{dx}(3y^2 - 27) = 2x

dydx=2x3y2−27\frac{dy}{dx} = \frac{2x}{3y^2 - 27}

Setting the derivative equal to zero, we get:

2x3y2−27=0\frac{2x}{3y^2 - 27} = 0

This implies that:

2x=02x = 0

x=0x = 0

Q: How do we solve for y?

A: Now that we have found the value of x, we can substitute it into the original equation to solve for y.

y3−27y=x2−90y^3 - 27y = x^2 - 90

y3−27y=02−90y^3 - 27y = 0^2 - 90

y3−27y=−90y^3 - 27y = -90

y(y2−27)=−90y(y^2 - 27) = -90

y(y+33)(y−33)=−90y(y + 3\sqrt{3})(y - 3\sqrt{3}) = -90

Q: What are the critical points on the graph?

A: The critical points on the graph are the points at which the tangent line is vertical. These points are:

(0,0),(0,−33),(0,33),(−310,−33),(310,−33),(−310,33),(310,33)(0, 0), (0, -3\sqrt{3}), (0, 3\sqrt{3}), (-3\sqrt{10}, -3\sqrt{3}), (3\sqrt{10}, -3\sqrt{3}), (-3\sqrt{10}, 3\sqrt{3}), (3\sqrt{10}, 3\sqrt{3})

Q: How do we order the critical points?

A: The critical points are ordered from smallest to largest x, and then from smallest to largest y.

Q: What is the significance of the critical points?

A: The critical points are important because they can help us understand the behavior of the function. They can also help us identify the points at which the tangent line is vertical.

Q: How do we use the critical points to find the corresponding values of x and y?

A: We can use the critical points to find the corresponding values of x and y by substituting the values of x into the original equation and solving for y.

Q: What are the corresponding values of x and y?

A: The corresponding values of x and y are:

(0,0),(0,−33),(0,33),(−310,−33),(310,−33),(−310,33),(310,33)(0, 0), (0, -3\sqrt{3}), (0, 3\sqrt{3}), (-3\sqrt{10}, -3\sqrt{3}), (3\sqrt{10}, -3\sqrt{3}), (-3\sqrt{10}, 3\sqrt{3}), (3\sqrt{10}, 3\sqrt{3})

Conclusion

In this article, we have answered some of the most frequently asked questions about finding the points on the graph of $y^3 - 27y = x^2 - 90$ at which the tangent line is vertical. We have used the derivative of the function to find the critical points, and then used these critical points to find the corresponding values of x and y. The critical points are important because they can help us understand the behavior of the function and identify the points at which the tangent line is vertical.

Discussion

The problem of finding the points at which the tangent line is vertical is an important one in mathematics. It involves finding the derivative of a function and setting it equal to zero to determine the critical points. The critical points are then used to find the corresponding values of x and y. This problem is a classic example of how calculus can be used to solve real-world problems.

References

  • [1] Calculus, 3rd edition, by Michael Spivak
  • [2] Differential Equations and Dynamical Systems, 3rd edition, by Lawrence Perko
  • [3] Calculus: Early Transcendentals, 7th edition, by James Stewart