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The Slope of Tangent Lines

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The Slope of Tangent Lines
Calculus is the study of         continuous change        
.

Given a curve y=f(x)y=f\left(x\right)y=f(x), the tangent line at a given point is the line that      most resembles     
 the curve near that point. 

The slope of the tangent line at a given point tells us how quickly the curve is      changing     
 at that point.

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How do we construct a tangent line?
Given 2 points PPP and QQQ on the curve y=f(x)y=f\left(x\right)y=f(x), the      secant line     
 between these 2 points is the straight line that connects the two points.

As we slide point QQQ closer to point P(a,b)P(a,b)P(a,b), the slope of the secant line will be closer to the slope of the tangent line at PPP
m=f(a+h)−f(a)h\boxed{\displaystyle m=\frac{f\left(a+h\right)-f\left(a\right)}{h}}m=hf(a+h)−f(a)​​
Write it Down
The slope of the tangent line to a curve at a point PPP is the slope of the secant line between the point PPP and a point that is        arbitrarily close to P        

 lim⁡h→0f(a+h)−f(a)h\displaystyle \lim_{h\to0}\frac{f\left(a+h\right)-f\left(a\right)}{h}h→0lim​hf(a+h)−f(a)​


Wize Tip
To evaluate the limit, we need to simplify the expression enough so that when we substitute h→0h\to 0h→0 into the expression, we get something that makes sense (i.e. a finite number or ∞\infty∞)

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Example
Given the curve f(x)=x2+1f\left(x\right)=x^2+1f(x)=x2+1,
a) find the slope of the secant line between the points x=0x=0x=0 and x=1x=1x=1
b) find the slope of the secant line between the points  x=0x=0x=0 and x=0.5x=0.5x=0.5
c) find the slope of the secant line between the points x=0x=0x=0 and x=0.1x=0.1x=0.1
d) find the slope of the secant line between the points x=0x=0x=0 and x=0.01x=0.01x=0.01
e) find the slope of the secant line between the points x=0x=0x=0 and x=0.001x=0.001x=0.001
f) use your findings from a) - e) to estimate the slope of the tangent line at the point x=0x=0x=0

a) m=f(1)−f(0)1−0=(12+1)−(02+1)1=11=1\displaystyle m=\frac{f\left(1\right)-f\left(0\right)}{1-0}=\frac{\left(1^2+1\right)-\left(0^2+1\right)}{1}=\frac{1}{1}=1m=1−0f(1)−f(0)​=1(12+1)−(02+1)​=11​=1

b) m=f(0.5)−f(0)0.5−0=(0.52+1)−(02+1)0.5=0.250.5=0.5\displaystyle m=\frac{f\left(0.5\right)-f\left(0\right)}{0.5-0}=\frac{\left(0.5^2+1\right)-\left(0^2+1\right)}{0.5}=\frac{0.25}{0.5}=0.5m=0.5−0f(0.5)−f(0)​=0.5(0.52+1)−(02+1)​=0.50.25​=0.5

c) m=f(0.1)−f(0)0.1−0=(0.12+1)−(02+1)0.1=0.010.1=0.1\displaystyle m=\frac{f\left(0.1\right)-f\left(0\right)}{0.1-0}=\frac{\left(0.1^2+1\right)-\left(0^2+1\right)}{0.1}=\frac{0.01}{0.1}=0.1m=0.1−0f(0.1)−f(0)​=0.1(0.12+1)−(02+1)​=0.10.01​=0.1

d) m=f(0.01)−f(0)0.01−0=(0.012+1)−(02+1)0.01=0.00010.01=0.01\displaystyle m=\frac{f\left(0.01\right)-f\left(0\right)}{0.01-0}=\frac{\left(0.01^2+1\right)-\left(0^2+1\right)}{0.01}=\frac{0.0001}{0.01}=0.01m=0.01−0f(0.01)−f(0)​=0.01(0.012+1)−(02+1)​=0.010.0001​=0.01

e) m=f(0.001)−f(0)0.001−0=(0.0012+1)−(02−1)0.001=0.0000010.001=0.001\displaystyle m=\frac{f\left(0.001\right)-f\left(0\right)}{0.001-0}=\frac{\left(0.001^2+1\right)-\left(0^2-1\right)}{0.001}=\frac{0.000001}{0.001}=0.001m=0.001−0f(0.001)−f(0)​=0.001(0.0012+1)−(02−1)​=0.0010.000001​=0.001

f) As the second point gets closer and closer to x=0x=0x=0, the slope of the secant line gets smaller and smaller and approaches 0. Therefore, the slope of the tangent line at the point x=0x=0x=0 is 0.

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Example: Drawing Tangent Lines
For each of the following curves, sketch the tangent line at the indicated points, then answer the questions stated

a)
i) How many tangent lines to this curve have a slope of 0?

2 points (the points where the curve has a turning point

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b)

i) How many tangent lines to this curve have a slope of 0?

2 points (at the top and bottom of the circle

ii) How many tangent lines to this curve have an infinite slope?

2 points (at the left and right-most points of the circle

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c)
i) What can we conclude about the slope of the tangent line of a straight line?

The slope of the tangent line on a point that is on a straight line is the same as the slope of the straight line

ii) What is the slope of the tangent line at the turning point (a.k.a. cusp) of the absolute value function?

The slope is not defined because we can draw many different tangent lines at that point.

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Example: Calculating Slopes of Tangents
Determine the slope of the tangent to the following curves at the indicated point.
a) f(x)=x2−xf(x)=x^2-xf(x)=x2−x at the point x=0x=0x=0

m=lim⁡h→0 f(0+h)−f(0)h\displaystyle m=\lim_{h\to0}\ \frac{f\left(0+h\right)-f\left(0\right)}{h}m=h→0lim​ hf(0+h)−f(0)​
m=lim⁡h→0 ((0+h)2−(0+h))−(02−0)h\displaystyle m=\lim_{h\to0}\ \frac{\left(\left(0+h\right)^2-\left(0+h\right)\right)-\left(0^2-0\right)}{h}m=h→0lim​ h((0+h)2−(0+h))−(02−0)​
m=lim⁡h→0 (h2−h)−0h\displaystyle m=\lim_{h\to0}\ \frac{\left(h^2-h\right)-0}{h}m=h→0lim​ h(h2−h)−0​
m=lim⁡h→0 h2−hh\displaystyle m=\lim_{h\to0}\ \frac{h^2-h}{h}m=h→0lim​ hh2−h​

If we "substittue" the value h→0h\to0h→0, we get 00\frac{0}{0}
00​, which doesn't make sense. Let's simplify this expression further:
m=lim⁡h→0 h(h−1)h\displaystyle m=\lim_{h\to0}\ \frac{h\left(h-1\right)}{h}m=h→0lim​ hh(h−1)​
m=lim⁡h→0 h−1\displaystyle m=\lim_{h\to0}\ h-1m=h→0lim​ h−1
m=0−1m=0-1m=0−1
m=−1m=-1m=−1

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b) f(x)=x−1f\left(x\right)=\sqrt{x-1}f(x)=x−1​ at the point x=2x=2x=2

m=lim⁡h→0 f(2+h)−f(2)h\displaystyle m=\lim_{h\to0}\ \frac{f\left(2+h\right)-f\left(2\right)}{h}m=h→0lim​ hf(2+h)−f(2)​
m=lim⁡h→0 (2+h)−1−2−1h\displaystyle m=\lim_{h\to0}\ \frac{\sqrt{\left(2+h\right)-1}-\sqrt{2-1}}{h}m=h→0lim​ h(2+h)−1​−2−1​​
m=lim⁡h→0 1+h−1h\displaystyle m=\lim_{h\to0}\ \frac{\sqrt{1+h}-1}{h}m=h→0lim​ h1+h​−1​

If we "substitute" the value h→0h\to0h→0, we get 00\frac{0}{0}00​, which doesn't make sense. Let's simplify this expression further:
m=lim⁡h→0 1+h−1h×1+h+11+h+1\displaystyle m=\lim_{h\to0}\ \frac{\sqrt{1+h}-1}{h}\times\frac{\sqrt{1+h}+1}{\sqrt{1+h}+1}m=h→0lim​ h1+h​−1​×1+h​+11+h​+1​
m=lim⁡h→0 1+h−1h(1+h+1)\displaystyle m=\lim_{h\to0}\ \frac{1+h-1}{h\left(\sqrt{1+h}+1\right)}m=h→0lim​ h(1+h​+1)1+h−1​
m=lim⁡h→0 hh(1+h+1)\displaystyle m=\lim_{h\to0}\ \frac{h}{h\left(\sqrt{1+h}+1\right)}m=h→0lim​ h(1+h​+1)h​
m=lim⁡h→0 11+h+1\displaystyle m=\lim_{h\to0}\ \frac{1}{\sqrt{1+h}+1}m=h→0lim​ 1+h​+11​
m=11+0+1\displaystyle m=\frac{1}{\sqrt{1+0}+1}m=1+0​+11​
m=12\displaystyle m=\frac{1}{2}m=21​

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c) f(x)=2xf\left(x\right)=\frac{2}{x}f(x)=x2​ at the point x=1x=1x=1

m=lim⁡h→0 f(1+h)−f(1)h\displaystyle m=\lim_{h\to0}\ \frac{f\left(1+h\right)-f\left(1\right)}{h}m=h→0lim​ hf(1+h)−f(1)​
m=lim⁡h→0 21+h−21h\displaystyle m=\lim_{h\to0}\ \frac{\frac{2}{1+h}-\frac{2}{1}}{h}m=h→0lim​ h1+h2​−12​​
m=lim⁡h→0 21+h−2h\displaystyle m=\lim_{h\to0}\ \frac{\frac{2}{1+h}-2}{h}m=h→0lim​ h1+h2​−2​

If we "substitute" the value h→0h\to 0h→0, we get 00\frac{0}{0}00​ which doesn't make sense. Let's simplify this expression further:
m=lim⁡h→0 [2−2(1+h)1+h]h\displaystyle m=\lim_{h\to0}\ \frac{\left[\frac{2-2\left(1+h\right)}{1+h}\right]}{h}m=h→0lim​ h[1+h2−2(1+h)​]​
m=lim⁡h→0 [2−2−2h1+h]h\displaystyle m=\lim_{h\to0}\ \frac{\left[\frac{2-2-2h}{1+h}\right]}{h}m=h→0lim​ h[1+h2−2−2h​]​
m=lim⁡h→0 [−2h1+h]h\displaystyle m=\lim_{h\to0}\ \frac{\left[\frac{-2h}{1+h}\right]}{h}m=h→0lim​ h[1+h−2h​]​
m=lim⁡h→0 −21+h\displaystyle m=\lim_{h\to0}\ \frac{-2}{1+h}m=h→0lim​ 1+h−2​
m=−21+0\displaystyle m=-\frac{2}{1+0}m=−1+02​
m=−2m=-2m=−2

Practice: Calculating Slopes of Tangents
Caluclate the slope of the tangent line to the following curves at the indicated point.

y=4x−1y=\sqrt{4x-1}y=4x−1​ at x=1x=1x=1

Enter the exact value (simplify but do not round)

I don't know

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Example: Equation of the Tangent Line
Find the equation of the tangent line to y=5+x−x2y=5+x-x^2y=5+x−x2 at the point where x=1x=1x=1.

At x=1x=1x=1, y=5+1−12=5y=5+1-1^2=5y=5+1−12=5.
So we are looking for the equation of the tangent line at the point (x0, y0)=(1, 5)\left(x_0,\ y_0\right)=\left(1,\ 5\right)(x0​, y0​)=(1, 5).

The slope of the line is
m=lim⁡h→0 f(1+h)−f(1)h\displaystyle m=\lim_{h\to0}\ \frac{f\left(1+h\right)-f\left(1\right)}{h}m=h→0lim​ hf(1+h)−f(1)​
m=lim⁡h→0 [5+(1+h)−(1+h)2]−[5+1−12]h\displaystyle m=\lim_{h\to0}\ \frac{\left[5+\left(1+h\right)-\left(1+h\right)^2\right]-\left[5+1-1^2\right]}{h}m=h→0lim​ h[5+(1+h)−(1+h)2]−[5+1−12]​
m=lim⁡h→0 [5−h−h2]−[5]h\displaystyle m=\lim_{h\to0}\ \frac{\left[5-h-h^2\right]-\left[5\right]}{h}m=h→0lim​ h[5−h−h2]−[5]​
m=lim⁡h→0 −h−h2h\displaystyle m=\lim_{h\to0}\ \frac{-h-h^2}{h}m=h→0lim​ h−h−h2​
m=lim⁡h→0 h(−1−h)h\displaystyle m=\lim_{h\to0}\ \frac{h\left(-1-h\right)}{h}m=h→0lim​ hh(−1−h)​
m=lim⁡h→0−1−h\displaystyle m=\lim_{h\to0}-1-hm=h→0lim​−1−h
m=−1m=-1m=−1

Therefore, the equation of the tangent line at this particular point is:
y−5=−1(x−1)y-5=-1(x-1)y−5=−1(x−1)
y=−x+6y=-x+6y=−x+6

Practice: Equation of the Tangent Lines
Find the equation of the tangent line to the curve y=2−3x+1y=2-\sqrt{3x+1} y=2−3x+1​ at the point x=0x=0x=0.

Enter the equation of the tangent line in the form

y=mx+b

I don't know

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Example: Slope of Tangent Line
Find the point(s) (if any) on the curve f(x)=1+3xf\left(x\right)=1+\frac{3}{x}f(x)=1+x3​ where the tangent line is perpendicular to the line 3y−x=33y-x=33y−x=3.

1. Find the slope of the given line
Rearranging the given line, we get y=13x+1y=\frac{1}{3}x+1y=31​x+1, so the slope is 13\frac{1}{3}31​

2. Take the negative reciprocal of this slope --> this is the slope of the line that is perpendicular to the given line
The tangent line perpendicular to this line will have slope −31 or −3-\frac{3}{1}\ or\ -3−13​ or −3 

3. Use the slope of tangent formula to find the slope of the tangent line to the curve at the point x=ax=ax=a
Using the formula, we can calculate the slope of the tangent line at the point x=ax=ax=a:
m=lim⁡h→0 f(a+h)−f(a)h\displaystyle m=\lim_{h\to0}\ \frac{f\left(a+h\right)-f\left(a\right)}{h}m=h→0lim​ hf(a+h)−f(a)​
m=lim⁡h→0 [1+3a+h]−[1+3a]h\displaystyle m=\lim_{h\to0}\ \frac{\left[1+\frac{3}{a+h}\right]-\left[1+\frac{3}{a}\right]}{h}m=h→0lim​ h[1+a+h3​]−[1+a3​]​

4. Simplify the limit expression as much as possible, then evaluate the limit
m=lim⁡h→0 [a+h+3a+h]−[a+3a]h\displaystyle m=\lim_{h\to0}\ \frac{\left[\frac{a+h+3}{a+h}\right]-\left[\frac{a+3}{a}\right]}{h}m=h→0lim​ h[a+ha+h+3​]−[aa+3​]​
m=lim⁡h→0 [(a+h+3)a−(a+3)(a+h)(a+h)a]h\displaystyle m=\lim_{h\to0}\ \frac{\left[\frac{\left(a+h+3\right)a-\left(a+3\right)\left(a+h\right)}{\left(a+h\right)a}\right]}{h}m=h→0lim​ h[(a+h)a(a+h+3)a−(a+3)(a+h)​]​
m=lim⁡h→0 [a2+ah+3a−a2−ah−3a−3h(a+h)a]h\displaystyle m=\lim_{h\to0}\ \frac{\left[\frac{a^2+ah+3a-a^2-ah-3a-3h}{\left(a+h\right)a}\right]}{h}m=h→0lim​ h[(a+h)aa2+ah+3a−a2−ah−3a−3h​]​
m=lim⁡h→0 [−3h(a+h)a]h\displaystyle m=\lim_{h\to0}\ \frac{\left[\frac{-3h}{\left(a+h\right)a}\right]}{h}m=h→0lim​ h[(a+h)a−3h​]​
m=lim⁡h→0 −3(a+h)a\displaystyle m=\lim_{h\to0}\ \frac{-3}{\left(a+h\right)a}m=h→0lim​ (a+h)a−3​
m=−3(a+0)a\displaystyle m=-\frac{3}{\left(a+0\right)a}m=−(a+0)a3​
m=−3a2\displaystyle m=-\frac{3}{a^2}m=−a23​

5. Set the limit found in step 4 to the slope value you found in step 2
We need this slope to equal −3-3−3:
−3=−3a2\displaystyle -3=-\frac{3}{a^2}−3=−a23​

6. Solve for the value aaa
a2=1a^2=1a2=1
a=±1a=\pm1a=±1

7. Find the corresponding yyy values
x=1 → y=1+31=4x=1\ \to\ y=1+\frac{3}{1}=4x=1 → y=1+13​=4 the point is (1, 4)\left(1,\ 4\right)(1, 4)
x=−1 → y=1+3−1=−2x=-1\ \to\ y=1+\frac{3}{-1}=-2x=−1 → y=1+−13​=−2 the point is (−1, −2)\left(-1,\ -2\right)(−1, −2)
Therefore, the points on the curve where the tangent line is perpendicular to the line 3y−x=33y-x=33y−x=3 are (1, 4)\left(1,\ 4\right)(1, 4) and (−1, −2)\left(-1,\ -2\right)(−1, −2)

Practice: Slope of Tangent Line
Find the point on the curve y=2x2−xy=2x^2-xy=2x2−x where the tangent line is horizontal.

Enter the x-coordinate of the point in which the tangent is horizontal

Enter the y-coordinate of the point in which the tangent is horizontal

I don't know