2.5 Derivative As Tangent Lineap Calculus

Question 2 From rst principles, nd the derivative f0(t) of the function f(t) = p t. Hence nd the point on the graph y = p t, where the tangent line has slope 1 4. Also nd the equation of that tangent line and sketch the tangent line. 2.5 Derivative as Tangent Line​ Notes 2.5 Video Approximate value of function using tangent line Notes 2.5 Video Given graph of g (x), find when g'0, g'2.5 Video Given table of h’ (x), find when h (x) is increasing /decreasing & tangent approximation. When a problem asks you to find the equation of the tangent line, you’ll always be asked to evaluate at the point where the tangent line intersects the graph. You’ll need to find the derivative, and evaluate at the given point.

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Section 2-1 : Tangent Lines And Rates Of Change

1. For the function (fleft( x right) = 3{left( {x + 2} right)^2}) and the point (P) given by (x = - 3) answer each of the following questions.

  1. For the points (Q) given by the following values of (x) compute (accurate to at least 8 decimal places) the slope, ({m_{PQ}}), of the secant line through points (P) and (Q).
    1. -3.5
    2. -3.1
    3. -3.01
    4. -3.001
    5. -3.0001
    1. -2.5
    2. -2.9
    3. -2.99
    4. -2.999
    5. -2.9999
  2. Use the information from (a) to estimate the slope of the tangent line to (fleft( x right)) at (x = - 3) and write down the equation of the tangent line.

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a For the points (Q) given by the following values of (x) compute (accurate to at least 8 decimal places) the slope, ({m_{PQ}}), of the secant line through points (P) and (Q). Show SolutionLineup
  1. -3.5
  2. -3.1
  3. -3.01
  4. -3.001
  5. -3.0001
  1. -2.5
  2. -2.9
  3. -2.99
  4. -2.999
  5. -2.9999

The first thing that we need to do is set up the formula for the slope of the secant lines. As discussed in this section this is given by,

[{m_{PQ}} = frac{{fleft( x right) - fleft( { - 3} right)}}{{x - left( { - 3} right)}} = frac{{3{{left( {x + 2} right)}^2} - 3}}{{x + 3}}]

Now, all we need to do is construct a table of the value of ({m_{PQ}}) for the given values of (x). All of the values in the table below are accurate to 8 decimal places, but in this case the values terminated prior to 8 decimal places and so the “trailing” zeros are not shown.

(x)({m_{PQ}})(x)({m_{PQ}})
-3.5-7.5-2.5-4.5
-3.1-6.3-2.9-5.7
-3.01-6.03-2.99-5.97
-3.001-6.003-2.999-5.997
-3.0001-6.0003-2.9999-5.9997


b Use the information from (a) to estimate the slope of the tangent line to (fleft( x right)) at (x = - 3) and write down the equation of the tangent line. Show Solution

From the table of values above we can see that the slope of the secant lines appears to be moving towards a value of -6 from both sides of (x = - 3) and so we can estimate that the slope of the tangent line is : (require{bbox} bbox[2pt,border:1px solid black]{{m = - 6}}).

The equation of the tangent line is then,

[y = fleft( { - 3} right) + mleft( {x - left( { - 3} right)} right) = 3 - 6left( {x + 3} right)hspace{0.5in} Rightarrow hspace{0.5in},require{bbox} bbox[2pt,border:1px solid black]{{y = - 6x - 15}}]

Here is a graph of the function and the tangent line.


For a function (z = f (x, y)), we learned that the partial derivatives (dfrac{∂f}{∂x}text{ and} dfrac{∂f}{∂y}) represent the (instantaneous) rate of change of (f) in the positive (x) and (y) directions, respectively. What about other directions? It turns out that we can find the rate of change in any direction using a more general type of derivative called a directional derivative.

Definition 2.5: directional derivative

Let (f (x, y)) be a real-valued function with domain (D) in (mathbb{R}^2), and let ((a,b)) be a point in (D). Let (textbf{v}) be a unit vector in (mathbb{R}^2). Then the directional derivative of (textbf{f}) at (mathbf{(a,b)}) in the direction of (mathbf{v}), denoted by (D_v f(a,b)), is defined as

[D_v f(a,b)=lim limits_{h to 0}dfrac{f((a,b)+htextbf{v})-f(a,b)}{h}label{Eq2.8}]

Notice in the definition that we seem to be treating the point ((a,b)) as a vector, since we are adding the vector (htextbf{v}) to it. But this is just the usual idea of identifying vectors with their terminal points, which the reader should be used to by now. If we were to write the vector (textbf{v}) as (textbf{v} = (v_1 ,v_2)), then

[D_v f (a,b)=lim limits_{h to 0}dfrac{f (a+ hv_1 ,b + hv_2)− f (a,b)}{h}label{Eq2.9}]

From this we can immediately recognize that the partial derivatives (dfrac{∂f}{∂x}text{ and} dfrac{∂f}{∂y}) are special cases of the directional derivative with (textbf{v} = textbf{i} = (1,0)text{ and } textbf{v} = textbf{j} = (0,1)), respectively. That is, (dfrac{∂f}{∂x} = D_i ftext{ and } dfrac{∂f}{∂y} = D_j f). Since there are many vectors with the same direction, we use a unit vector in the definition, as that represents a “standard” vector for a given direction.

If (f (x, y)) has continuous partial derivatives (dfrac{∂f}{∂x}text{ and }dfrac{∂f}{∂y}) (which will always be the case in this text), then there is a simple formula for the directional derivative:

Theorem 2.2

Let (f (x, y)) be a real-valued function with domain (D) in (mathbb{R}^2) such that the partial derivatives (dfrac{∂f}{∂x}text{ and }dfrac{∂f}{∂y}) exist and are continuous in (D). Let ((a,b)) be a point in (D), and let (textbf{v} = (v_1 ,v_2)) be a unit vector in (mathbb{R}^2). Then

[D_v f (a,b) = v_1 dfrac{∂f}{∂x} (a,b)+ v_2 dfrac{∂f}{∂y} (a,b)label{Eq2.10}]

Proof: Note that if (textbf{v} = textbf{i}) = (1,0) then the above formula reduces to (D_v f (a,b) = dfrac{∂f}{∂x} (a,b)), which we know is true since (D_i f = dfrac{∂f}{∂x}), as we noted earlier. Similarly, for (textbf{v} = textbf{j} = (0,1)) the formula reduces to (D_v f (a,b) = dfrac{∂f}{∂y} (a,b)), which is true since (D_j f = dfrac{∂f}{∂y}). So since (textbf{i} = (1,0)text{ and }textbf{j} = (0,1)) are the only unit vectors in (mathbb{R}^2) with a zero component, then we need only show the formula holds for unit vectors (textbf{v} = (v_1 ,v_2)text{ with }v_1 neq 0 text{ and }v_2 neq 0). So fix such a vector (textbf{v}) and fix a number (h neq 0).

Then

[ f (a+ hv_1 ,b + hv_2)− f (a,b) = f (a+ hv_1 ,b + hv_2)− f (a+ hv_1 ,b)+ f (a+ hv_1 ,b)− f (a,b)label{Eq2.11}]

Since (h neq 0 text{ and }v_2 neq 0), then (hv_2 neq 0) and thus any number (c) between (b text{ and }b + hv_2) can be written as (c = b+alpha hv_2) for some number (0 < alpha < 1). So since the function (f (a+hv_1 , y)) is a realvalued function of (y) (since (a + hv_1) is a fixed number), then the Mean Value Theorem from single-variable calculus can be applied to the function (g(y) = f (a + hv_1 , y)) on the interval ([b,b + hv_2]) (or ([b + hv_2 ,b]) if one of (h text{ or }v_2) is negative) to find a number (0 < alpha < 1) such that

[nonumber dfrac{∂f}{∂y} (a+ hv_1 ,b +alpha hv_2) = g ′ (b +alpha hv_2)=dfrac{g(b + hv_2)− g(b)}{b + hv_2 − b}=dfrac{f (a+ hv_1 ,b + hv_2)− f (a+ hv_1 ,b)}{hv_2}]

and so

[nonumber f (a+ hv_1 ,b + hv_2)− f (a+ hv_1 ,b) = hv_2 dfrac{∂f}{∂y} (a+ hv_1 ,b +alpha hv_2) .]

By a similar argument, there exists a number (0 < beta < 1) such that

[nonumber f (a+ hv_1 ,b)− f (a,b) = hv_1 dfrac{∂f}{∂x} (a+beta hv_1 ,b) .]

Thus, by Equation ref{Eq2.11}, we have

[nonumber begin{align} dfrac{f (a+ hv_1 ,b + hv_2)− f (a,b)}{h}&=dfrac{hv_2 dfrac{∂f}{∂y} (a+ hv_1 ,b +alpha hv_2)+ hv_1 dfrac{∂f}{∂x} (a+beta hv_1 ,b)}{h} [4pt] nonumber &=v_2 dfrac{∂f}{∂y} (a+ hv_1 ,b +alpha hv_2)+ v_1 dfrac{∂f}{∂x} (a+beta hv_1 ,b)end{align}]

so by Equation ref{Eq2.9} we have

[ nonumber begin{align} D_v f (a,b)&=lim limits_{h to 0}dfrac{f (a+ hv_1 ,b + hv_2)− f (a,b)}{h} [4pt] nonumber &=lim limits_{h to 0}left [v_2 dfrac{∂f}{∂y} (a+ hv_1 ,b +alpha hv_2)+ v_1 dfrac{∂f}{∂x} (a+beta hv_1 ,b)right ] [4pt] nonumber &= v_2 dfrac{∂f}{∂y} (a,b)+ v_1 dfrac{∂f}{∂x} (a,b) text{ by the continuity of } dfrac{∂f}{∂x} text{ and } dfrac{∂f}{∂y}text{, so} [4pt] nonumber D_v f (a,b) &= v_1 dfrac{∂f}{∂x} (a,b)+ v_2 dfrac{∂f}{∂y} (a,b) end{align}]

[nonumber text{after reversing the order of summation.}tag{(textbf{QED})}]

Note that (D_v f (a,b) = v cdot left (dfrac{∂f}{∂x} (a,b), dfrac{∂f}{∂y} (a,b) right )). The second vector has a special name:

2.5 Derivative As Tangent Lineup Calculus Formula

Definition 2.6

For a real-valued function (f (x, y)), the gradient of (f), denoted by (nabla f), is the vector

[nabla f =left ( dfrac{∂f}{∂x} , dfrac{∂f}{∂y} right ) label{Eq2.12}]

in (mathbb{R}^2). For a real-valued function (f (x, y, z)), the gradient is the vector

[nabla f = left ( dfrac{∂f}{∂x} , dfrac{∂f}{∂y} , dfrac{∂f}{∂z}right ) label{Eq2.13}]

in (mathbb{R}^ 3). The symbol (nabla) is pronounced “del”.

Corollary 2.3

[nonumber D_v f = textbf{v} cdot nabla f]

2.5 Derivative As Tangent Lineup Calculus Calculator

Example 2.15

Find the directional derivative of (f (x, y) = x y^2 + x^3 y) at the point (1,2) in the direction of (textbf{v} = left ( dfrac{1}{sqrt{2}},dfrac{1}{sqrt{2}}right )).

Solution

We see that (nabla f = (y^2 +3x^2 y,2x y+ x^3 )), so

2.5 Derivative As Tangent Lineup Calculus Equation

[nonumber D_v f (1,2) = textbf{v}cdot nabla f (1,2) = left ( dfrac{1}{sqrt{2}},dfrac{1}{sqrt{2}}right ) cdot (2^2 +3(1)^2 (2),2(1)(2)+1^3 ) = dfrac{15}{sqrt{2}}]

A real-valued function (z = f (x, y)) whose partial derivatives (dfrac{∂f}{∂x}text{ and }dfrac{∂f}{∂y}) exist and are continuous is called continuously differentiable. Assume that (f (x, y)) is such a function and that (nabla f neq textbf{0}). Let (c) be a real number in the range of (f) and let (textbf{v}) be a unit vector in (mathbb{R}^2) which is tangent to the level curve (f (x, y) = c) (see Figure 2.4.1).

The value of (f (x, y)) is constant along a level curve, so since (textbf{v}) is a tangent vector to this curve, then the rate of change of (f) in the direction of (textbf{v}) is 0, i.e. (D_v f = 0). But we know that (D_v f = textbf{v} cdot nabla f = leftlVert textbf{v} rightrVert leftlVert nabla f rightrVert cos theta ), where (theta) is the angle between (textbf{v} text{ and }nabla f). So since (leftlVert textbf{v} rightrVert = 1 text{ then }D_v f = leftlVert nabla f rightrVert cos theta ). So since (nabla f neq textbf{0}text{ then }D_v f = 0 ⇒ cos theta = 0 ⇒ theta = 90^circ). In other words, (nabla f perp textbf{v}), which means that (nabla f) is normal to the level curve.

In general, for any unit vector (textbf{v}) in (mathbb{R}^2), we still have (D_v f = leftlVert nabla f rightrVert cos theta text{, where }theta) is the angle between (textbf{v} text{ and }nabla f). At a fixed point ((x, y)) the length (leftlVert nabla f rightrVert) is fixed, and the value of (D_v f) then varies as (theta) varies. The largest value that (D_v f) can take is when (cos theta = 1 (theta = 0^circ )), while the smallest value occurs when (cos theta = −1 (theta = 180^circ )). In other words, the value of the function (f) increases the fastest in the direction of (nabla f) (since (theta = 0^circ) in that case), and the value of (f) decreases the fastest in the direction of (−nabla f) (since (theta = 180^circ) in that case). We have thus proved the following theorem:

Theorem 2.4

Lineap

Let (f (x, y)) be a continuously differentiable real-valued function, with (nabla f neq 0). Then:

  1. The gradient (nabla f) is normal to any level curve (f (x, y) = c).
  2. The value of (f (x, y)) increases the fastest in the direction of (nabla f).
  3. The value of (f (x, y)) decreases the fastest in the direction of (−nabla f).

Example 2.16

In which direction does the function (f (x, y) = x y^2 + x^3 y) increase the fastest from the point (1,2)? In which direction does it decrease the fastest?

2.5 Derivative As Tangent Lineup Calculus Solver

Solution

Since (nabla f = (y^2 + 3x^2 y,2xy + x^3 )), then (nabla f (1,2) = (10,5) neq textbf{0}). A unit vector in that direction is (textbf{v} = dfrac{nabla f}{leftlVert nabla f rightrVert} = left (dfrac{2}{sqrt{5}} ,dfrac{1}{sqrt{5}}right )). Thus, (f) increases the fastest in the direction of (left (dfrac{2}{sqrt{5}} ,dfrac{1}{sqrt{5}}right )) and decreases the fastest in the direction of (left (dfrac{−2}{sqrt{5}},dfrac{−1}{sqrt{5}}right )).

Though we proved Theorem 2.4 for functions of two variables, a similar argument can be used to show that it also applies to functions of three or more variables. Likewise, the directional derivative in the three-dimensional case can also be defined by the formula (D_v f = textbf{v}cdot nabla f).

Example 2.17

The temperature (T) of a solid is given by the function (T(x, y, z) = e^−x + e^{−2y} + e^{4z}text{, where }x, y, z) are space coordinates relative to the center of the solid. In which direction from the point (1,1,1) will the temperature decrease the fastest?

Solution

Since (nabla f = (−e^{−x} ,−2e^{−2y} ,4e^{4z} )), then the temperature will decrease the fastest in the direction of (−nabla f (1,1,1) = (e^{−1} ,2e^{−2} ,−4e^4 )).

Contributors and Attributions

  • Michael Corral (Schoolcraft College). The content of this page is distributed under the terms of the GNU Free Documentation License, Version 1.2.