# Optimisation Grade 12: Minimum Length

TLDRThe transcript describes a calculus problem aimed at finding the minimum length of a segment PQ between a straight line and a hyperbola. The presenter suggests using the first derivative to find the minimum, setting it to zero. They demonstrate the process by substituting the equations of the line and hyperbola into the length formula, simplifying, and differentiating to find critical points. After solving the resulting equation, they determine that the minimum length occurs at x = -1, resulting in a positive length of 8, excluding the negative solution as it doesn't represent a valid length.

###### Takeaways

- π The problem is to find the minimum length of PQ, which is approached using calculus by setting the first derivative equal to zero.
- π Point P is given as (-2, 6) and point Q as (-2, 2), and the vertical distance between them is calculated as 4 units.
- π The length of PQ is represented by the difference in the y-values of the top and bottom curves, which are a straight line and a hyperbola respectively.
- π The equation of the top curve (straight line) is used to replace the y-value in the length formula for PQ.
- π The bottom curve is a hyperbola given by "y = 4/x + 4", which is also included in the length formula for PQ.
- π The simplified length formula for PQ becomes "-4x - 4/x", which is then differentiated to find the minimum length.
- π The first derivative of the length formula is "-4 + 4/x^2", which is set to zero to find critical points.
- π’ By solving the equation "x^2 = 1", two potential solutions for x are found: "x = 1" and "x = -1".
- β The solution "x = 1" results in a negative length for PQ, which is not applicable, so it is discarded.
- β The solution "x = -1" gives a positive length for PQ, which is the minimum length of 8 units, and is thus the correct answer.
- π The process involves understanding the calculus of finding minimums and maximums by setting the first derivative to zero and evaluating the results.

###### Q & A

### What mathematical concept is suggested for finding the minimum or maximum values in the given problem?

-The mathematical concept suggested for finding the minimum or maximum values is calculus, specifically by setting the first derivative equal to zero.

### What are the coordinates of points P and Q mentioned in the transcript?

-Point P has coordinates (-2, 6) and point Q has coordinates (-2, 2).

### How is the vertical length between points P and Q calculated?

-The vertical length between points P and Q is calculated by subtracting the y-values of the points, which in this case is 6 - 2, resulting in 4.

### What is the equation used to replace the y-top value in the length formula?

-The equation used to replace the y-top value is the equation of the straight line, which is not explicitly given in the transcript but is implied to be a linear function.

### What is the equation of the hyperbola given for y-bottom?

-The equation of the hyperbola for y-bottom is 4/(x + 4).

### What is the simplified expression for the length of PQ in terms of x?

-The simplified expression for the length of PQ is -4(x - 4)/x.

### What is the process to find the first derivative of the length of PQ?

-The process involves rewriting the expression for the length of PQ as a function of x, and then differentiating it with respect to x to find the first derivative.

### What is the first derivative of the length of PQ after simplification?

-The first derivative of the length of PQ after simplification is -4 + 4x^(-2).

### How is the minimum or maximum length found using the first derivative?

-The minimum or maximum length is found by setting the first derivative equal to zero and solving for x.

### What are the two possible values of x obtained by setting the first derivative to zero?

-The two possible values of x obtained are x = 1 and x = -1.

### Which of the two values of x, 1 or -1, gives a valid length for PQ, and what is that length?

-The value x = -1 gives a valid length for PQ, which is 8, as the length cannot be negative.

###### Outlines

##### π Calculus Application in Finding Minimum Length

This paragraph introduces a calculus-based approach to finding the minimum length of a segment PQ between two curves. The speaker begins by suggesting that minimum or maximum problems are often tackled with calculus, specifically by setting the first derivative to zero. The example given involves a vertical distance calculation between points P and Q with coordinates (-2,6) and (-2,2), respectively, which is straightforwardly 4 units. The speaker then transitions to a more complex scenario where P and Q are defined by a straight line and a hyperbola, respectively. The length PQ is expressed in terms of x, and the first derivative of this expression is taken to find critical points. The process involves simplifying the expression for PQ, differentiating it, and solving for when the derivative equals zero to find potential minimum and maximum values. The speaker emphasizes the importance of checking these critical points to determine which one corresponds to the minimum length.

###### Mindmap

###### Keywords

##### π‘Calculus

##### π‘First Derivative

##### π‘Minimum Length

##### π‘Coordinates

##### π‘Vertical Length

##### π‘Straight Line Equation

##### π‘Hyperbola

##### π‘Simplification

##### π‘Extrema

##### π‘Roots

##### π‘Contextual Application

###### Highlights

The question involves finding the minimum length using calculus.

The vertical length between points P and Q is calculated as the difference in their y-coordinates.

For points P(-2, 6) and Q(-2, 2), the vertical length is 4 units.

The length of PQ can be represented as the difference between the y-values of the top and bottom curves.

The top curve is a straight line, and the bottom curve is a hyperbola given by 4/x + 4.

The length of PQ is simplified to -4x - 4/x.

To find the minimum, take the first derivative of the length function with respect to x.

The first derivative is simplified to -4 + 4x^(-2).

Setting the first derivative equal to zero to find critical points.

Solving for x gives x^2 = 1, which leads to x = Β±1.

Both solutions must be checked to determine which one corresponds to the minimum length.

For x = 1, the length of PQ is negative, which is not applicable.

For x = -1, the length of PQ is positive 8, which is a valid solution.

The minimum length of PQ is 8 units when x = -1.

The process involves calculus to find the minimum length between two curves.

The method simplifies the problem by using the difference in y-values and applying derivatives.

The solution requires checking both critical points to identify the minimum length.

###### Transcripts

## Browse More Related Video

Maxima , minima and point of inflection

Optimization Example: Minimizing Surface Area Given a Fixed Volume

AP CALCULUS AB 2022 Exam Full Solution FRQ#3(c,d)

Finding Arc Length Using an Integral

Second Derivative Test to find relative maximums and minimums (extrema)

Physics 15 Torque (4 of 27) Bar Held Up by Friction

5.0 / 5 (0 votes)

Thanks for rating: