FER - Mathematics#
Contribution(s)
Dr. Fei is a content contribuitor for this section. Her notes are listed as first reading activity below.
Readings#
Videos#
Example 1#
A line goes through the point (4, −6) and is perpendicular to the line 𝑦 = 4𝑥 + 10. What is the equation of the line?
(A) 𝑦 = 𝑚𝑥 − 20
(B) 𝑦 = − 1/4 x - 5
(C) y= 1/5𝑥 + 5
(D) y= 1/4 + 5
Solving a Perpendicular Line Problem
Let’s solve the following problem step by step.
A line passes through the point (4, −6) and is perpendicular to the line \(y = 4x + 10\). We want to find the equation of the new line.
Step 1: Identify the slope of the given line#
The equation is already in slope–intercept form:
\(y = mx + b\)
where \(m\) is the slope.
From \(y = 4x + 10\), the slope is:
\(m = 4\)
Step 2: Determine the slope of a perpendicular line#
Perpendicular lines have slopes that are negative reciprocals.
That means:
Flip the fraction
Change the sign
Since 4 can be written as \(4/1\), the negative reciprocal is:
\(m = -\frac{1}{4}\)
So the slope of the line we want is \(-1/4\).
Step 3: Use the point–slope formula#
We know:
slope \(m = -\frac{1}{4}\)
point \((4, -6)\)
Use the point–slope formula:
\(y - y_1 = m(x - x_1)\)
Substitute the known values:
\(y - (-6) = -\frac{1}{4}(x - 4)\)
This simplifies to:
\(y + 6 = -\frac{1}{4}(x - 4)\)
Step 4: Convert to slope–intercept form#
Distribute the slope:
\(y + 6 = -\frac{1}{4}x + 1\)
Now subtract 6 from both sides:
\(y = -\frac{1}{4}x + 1 - 6\)
\(y = -\frac{1}{4}x - 5\)
Step 5: Choose the correct answer#
The equation of the line is:
\(y = -\frac{1}{4}x - 5\)
This matches choice B.
Final Answer:
B) \(y = -\frac{1}{4}x - 5\)
Quick recap#
Identify the slope of the given line → 4
Find the negative reciprocal → −1/4
Use point–slope form with \((4, −6)\)
Convert to slope–intercept form
Match with answer choices
Understanding perpendicular slopes and using point–slope form makes problems like this straightforward and systematic.
Example 2#
The vertical angle to the top of a flagpole from point A on the ground is observed to be 37°11′. The observer walks 17 m directly away from point A and the flagpole to point B and finds the new angle to be 25°43′. What is the approximate height of the flagpole?
(A) 10 m
(B) 22 m
(C) 82 m
(D) 300 m
Finding the Height of a Flagpole Using Angles of Elevation
Let’s solve this trigonometry problem step by step.
An observer at point A measures the angle to the top of a flagpole as \(37^\circ 11'\). The observer then walks 17 meters directly away to point B and measures a new angle of \(25^\circ 43'\).
We want to find the height of the flagpole.
Step 1: Visualize the geometry#
Imagine:
a vertical flagpole
point A on the ground
point B, 17 m farther away
two lines of sight from A and B to the top of the pole
Each line of sight creates a right triangle.
The height of the pole is the opposite side of each triangle.
The figure below illustrates these relationships:
*
|\
h | \
| \
| \
| \
| \
| \
| \
+--------+-------- ground
pole A B
Angle at A = 37°11′ Angle at B = 25°43′ AB = 17 m
Caption: Geometry used to determine flagpole height from two angles of elevation measured at points A and B.
Alt-Text: The diagram shows a vertical flagpole on level ground. Point A is on the ground near the pole, and point B is 17 meters farther away. Lines from A and B to the top of the pole form angles of elevation of 37°11′ and 25°43′, respectively.
Step 2: Convert angle measurements to decimal degrees#
This makes calculator work easier.
\(37^\circ 11' = 37 + \frac{11}{60} \approx 37.18^\circ\)
\(25^\circ 43' = 25 + \frac{43}{60} \approx 25.72^\circ\)
Step 3: Define variables#
Let:
\(h\) = height of the flagpole
\(x\) = horizontal distance from point A to the pole
Since point B is 17 m farther away, its distance is:
\(x + 17\)
Step 4: Write tangent relationships#
Using \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\):
From point A:
\(\tan(37.18^\circ) = \frac{h}{x}\)
From point B:
\(\tan(25.72^\circ) = \frac{h}{x+17}\)
Step 5: Express \(h\) from each equation#
\(h = x \tan(37.18^\circ)\)
\(h = (x+17)\tan(25.72^\circ)\)
Since both equal \(h\), set them equal:
\(x \tan(37.18^\circ) = (x+17)\tan(25.72^\circ)\)
Step 6: Substitute tangent values#
\(\tan(37.18^\circ) \approx 0.758\)
\(\tan(25.72^\circ) \approx 0.482\)
So:
\(0.758x = 0.482(x + 17)\)
Step 7: Solve for \(x\)#
\(0.758x = 0.482x + 8.194\)
\(0.276x = 8.194\)
\(x \approx 29.7 \text{ m}\)
Step 8: Solve for height#
\(h = x \tan(37.18^\circ)\)
\(h \approx 29.7(0.758)\)
\(h \approx 22.5 \text{ m}\)
Step 9: Choose the closest answer#
The height is approximately 22 meters.
Correct answer: (B) 22 m
Final Answer
(B) 22 m
Quick recap#
Convert angles to decimal form
Let \(x\) be the distance from point A
Write tangent equations for both positions
Set equations equal and solve for \(x\)
Compute height using tangent
Select the closest answer
Example 3#
What is the maximum value of the following function on the interval 𝑥 ≤ 0?
\(𝑦 = 2𝑥^3 + 12𝑥^2− 30𝑥 + 10 \)
(A) -210
(B) -36
(C) -5
(D) 210
Maximizing a Polynomial on the Interval \(x \le 0\)
Let’s solve this calculus problem carefully and systematically.
We are asked:
What is the maximum value of the function on the interval \(x \le 0\)?
The function is:
\(y = 2x^3 + 12x^2 - 30x + 10\)
And the answer choices are:
(A) \(-210\) (B) \(-36\) (C) \(-5\) (D) \(210\)
Step 1: Understand the interval#
The interval is all real numbers less than or equal to zero, meaning:
\(x \le 0\)
So we are searching for the biggest output value the function can produce anywhere to the left of 0, including at 0.
This is important because it tells us two things:
We must check any critical points that occur at \(x \le 0\).
We must also check the endpoint at \(x = 0\).
And we should think about what happens as \(x \to -\infty\).
Step 2: Find critical points using the derivative#
To find maxima and minima, start by differentiating.
\(y = 2x^3 + 12x^2 - 30x + 10\)
Differentiate term by term:
derivative of \(2x^3\) is \(6x^2\)
derivative of \(12x^2\) is \(24x\)
derivative of \(-30x\) is \(-30\)
derivative of \(10\) is \(0\)
So:
\(y' = 6x^2 + 24x - 30\)
Now factor out 6:
\(y' = 6(x^2 + 4x - 5)\)
Factor the quadratic:
\(x^2 + 4x - 5 = (x+5)(x-1)\)
So:
\(y' = 6(x+5)(x-1)\)
Critical points occur where \(y' = 0\), so set it equal to zero:
\(6(x+5)(x-1) = 0\)
This gives:
\(x = -5\) or \(x = 1\)
Step 3: Keep only the critical points in the interval \(x \le 0\)#
Our interval is \(x \le 0\).
\(x = -5\) is allowed.
\(x = 1\) is not allowed.
So the only critical point we need to check is:
\(x = -5\)
Step 4: Check the endpoint and the behavior at negative infinity#
Because our interval includes \(x = 0\), we must also check:
\(x = 0\)
And because the interval extends forever to the left, we should check what happens as \(x \to -\infty\).
Notice the leading term is \(2x^3\).
As \(x \to -\infty\), \(x^3 \to -\infty\), so:
\(2x^3 \to -\infty\)
That means the function goes down without bound to the left, so it cannot have a maximum out at negative infinity. The maximum must happen at either:
the critical point \(x=-5\), or
the endpoint \(x=0\).
Step 5: Evaluate the function at \(x = -5\) and \(x = 0\)#
First evaluate at \(x = -5\):
\(y(-5) = 2(-5)^3 + 12(-5)^2 - 30(-5) + 10\)
Compute each piece:
\((-5)^3 = -125\), so \(2(-125) = -250\) \((-5)^2 = 25\), so \(12(25) = 300\) \(-30(-5) = +150\) Then \(+10\)
Now add them:
\(y(-5) = -250 + 300 + 150 + 10 = 210\)
So:
\(y(-5) = 210\)
Now evaluate at \(x = 0\):
\(y(0) = 2(0)^3 + 12(0)^2 - 30(0) + 10 = 10\)
So:
\(y(0) = 10\)
Step 6: Identify the maximum value on \(x \le 0\)#
We compare the candidate values:
At \(x=-5\), the function is \(210\)
At \(x=0\), the function is \(10\)
The larger value is:
\(210\)
So the maximum value on the interval \(x \le 0\) is:
\(210\)
That matches answer choice (D).
Final Answer
(D) \(210\)
Example 4#
If \(f(x,y) = x^2 y^3 + x y^4 + \sin x + \cos^2 x + \sin^3 y\), find the partial derivative \(\dfrac{\partial f}{\partial x}\).
(A) \((2x+y)y^3 + 3 sin^2 y cos y\)
(B) \((4x - 3y^2)xy^2 + 3 sin^2 y cos y\)
(C) \((3x + 4y^2)xy + 3 sin^2 y cos y\)
(D) \(2x y^3 + y^4 + \cos x - 2\sin x \cos x\)
Step 1: Differentiate term-by-term with respect to \(x\)#
When taking a partial derivative with respect to \(x\), treat \(y\) as a constant.
Term 1: \(x^2 y^3\)
\(y^3\) is constant, so:
\(\frac{\partial}{\partial x}(x^2 y^3) = 2x y^3\)
Term 2: \(x y^4\)
\(y^4\) is constant:
\(\frac{\partial}{\partial x}(x y^4) = y^4\)
Term 3: \(\sin x\)
\(\frac{\partial}{\partial x}(\sin x) = \cos x\)
Term 4: \(\cos^2 x\)
Rewrite as \((\cos x)^2\) and use the chain rule:
\(\frac{\partial}{\partial x}(\cos^2 x) = 2\cos x(-\sin x)\)
\(= -2\sin x \cos x\)
Term 5: \(\sin^3 y\)
This depends only on \(y\), so:
\(\frac{\partial}{\partial x}(\sin^3 y) = 0\)
Step 2: Combine the results#
\(\frac{\partial f}{\partial x} = 2x y^3 + y^4 + \cos x - 2\sin x \cos x\)
Step 3: Optional trig simplification#
Since \(2\sin x \cos x = \sin(2x)\), we can also write:
\(\frac{\partial f}{\partial x} = 2x y^3 + y^4 + \cos x - \sin(2x)\)
Either form is correct.
Final Answer
\(\boxed{\frac{\partial f}{\partial x} = 2x y^3 + y^4 + \cos x - 2\sin x \cos x}\)
(or equivalently)
\(\boxed{2x y^3 + y^4 + \cos x - \sin(2x)}\)
Example 5#
Select all the root(s) in \(f(x)=x^2−4x−3\).
A. \(4+\sqrt{3}\)
B. \(2+\sqrt{7}\)
C. \(2−\sqrt{7}\)
D. \(4−\sqrt{3}\)
Work it out#
To find the roots of an equation, first set \(f(x)\) to 0.
\(x^2−4x−3=0\)
For simple equations, you can factor them by hand, but this equation does not factor easily. 3 can be factored into 1 and 3. However, one must be negative and one positive for their multiple to be -3. There is no way to sum those to get -4. Instead simply use the quadratic formula (a.k.a. the quadratic equation) if the equation can’t be factored easily or if you are unsure. A quadratic equation is of the form:
\(ax^2+bx+c=0\)
So we have \(a=1\), \(b=−4\), and \(c=−3\). The quadratic formula is:
\(x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
Substitute and solve as
\(x= \frac{-(-4) \pm \sqrt{(-4)^2-4(1)(-3)}}{2(1)} = 2 \pm \sqrt{7}\)
Choose B and C
Section End#
Mathematics and Probability Quiz Preview
ans=2*(-5)**3 + 12*(-5)**2 - 30*(-5) +10
ans
210