{"id":231,"date":"2023-09-20T22:48:26","date_gmt":"2023-09-20T22:48:26","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/velocities-and-rates-of-change\/"},"modified":"2024-08-05T01:48:23","modified_gmt":"2024-08-05T01:48:23","slug":"defining-the-derivative-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/defining-the-derivative-learn-it-3\/","title":{"raw":"Defining the Derivative: Learn It 3","rendered":"Defining the Derivative: Learn It 3"},"content":{"raw":"

Velocities and Rates of Change<\/h2>\r\n

Now that we can evaluate a derivative, we can apply it to velocity problems. Recall that if [latex]s(t)[\/latex] is the position of an object moving along a coordinate axis, the average velocity<\/span> of the object over a time interval [latex][a,t][\/latex] if [latex]t>a[\/latex] or [latex][t,a][\/latex] if [latex]t<a[\/latex] is:<\/p>\r\n

[latex]v_{\\text{avg}}=\\dfrac{s(t)-s(a)}{t-a}[\/latex]<\/div>\r\n

As the values of [latex]t[\/latex] approach [latex]a[\/latex], the values of [latex]v_{\\text{avg}}[\/latex] approach the value we call the instantaneous velocity<\/span> at [latex]a[\/latex]. That is, instantaneous velocity at [latex]a[\/latex], denoted [latex]v(a)[\/latex], is given by<\/p>\r\n

[latex]v(a)=s^{\\prime}(a)=\\underset{t\\to a}{\\lim}\\dfrac{s(t)-s(a)}{t-a}[\/latex]<\/div>\r\n

To better understand the relationship between average velocity and instantaneous velocity, see Figure 7. In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time [latex]t=a[\/latex]. The slope of the secant line (shown in green) is the average velocity of the object over the time interval [latex][a,t][\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"This Figure 7. The slope of the secant line is the average velocity over the interval [latex][a,t][\/latex]. The slope of the tangent line is the instantaneous velocity.[\/caption]\r\n\r\n

We can use the definitions to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using the difference quotient.<\/p>\r\n

\r\n

A lead weight on a spring is oscillating up and down. Its position at time [latex]t[\/latex] with respect to a fixed horizontal line is given by [latex]s(t)= \\sin t[\/latex]. Use a table of values to estimate [latex]v(0)[\/latex]. Check the estimate by using the definition of a derivative.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]\"A Figure 8. A lead weight suspended from a spring in vertical oscillatory motion.[\/caption]\r\n\r\n

[reveal-answer q=\"fs-id1169738960040\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1169738960040\"]<\/p>\r\n

We can estimate the instantaneous velocity at [latex]t=0[\/latex] by computing a table of average velocities using values of [latex]t[\/latex] approaching [latex]0[\/latex], as shown in the table below.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Average velocities using values of [latex]t[\/latex] approaching [latex]0[\/latex]<\/caption>\r\n
[latex]t[\/latex]<\/th>\r\n[latex]\\frac{\\sin t - \\sin 0}{t-0}=\\frac{\\sin t}{t}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
[latex]-0.1[\/latex]<\/td>\r\n[latex]0.998334166[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]-0.01[\/latex]<\/td>\r\n[latex]0.9999833333[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]-0.001[\/latex]<\/td>\r\n[latex]0.999999833[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]0.001[\/latex]<\/td>\r\n[latex]0.999999833[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]0.01[\/latex]<\/td>\r\n[latex]0.9999833333[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]0.1[\/latex]<\/td>\r\n[latex]0.998334166[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

From the table we see that the average velocity over the time interval [latex][-0.1,0][\/latex] is [latex]0.998334166[\/latex], the average velocity over the time interval [latex][-0.01,0][\/latex] is [latex]0.9999833333[\/latex], and so forth. Using this table of values, it appears that a good estimate is [latex]v(0)=1[\/latex].<\/p>\r\n

By using the definition of a derivative, we can see that<\/p>\r\n

[latex]v(0)=s^{\\prime}(0)=\\underset{t\\to 0}{\\lim}\\dfrac{\\sin t- \\sin 0}{t-0}=\\underset{t\\to 0}{\\lim}\\dfrac{\\sin t}{t}=1[\/latex].<\/div>\r\n

Thus, in fact, [latex]v(0)=1[\/latex].<\/p>\r\n

[\/hidden-answer]<\/p>\r\n<\/section>\r\n

As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change<\/strong> of a function, or the rate of change of a function at any point along the function.<\/p>\r\n

\r\n

instantaneous rate of change<\/h3>\r\n

The instantaneous rate of change<\/strong> of a function [latex]f(x)[\/latex] at a value [latex]a[\/latex] is its derivative [latex]f^{\\prime}(a)[\/latex].<\/p>\r\n<\/section>\r\n

\r\n

A homeowner sets the thermostat so that the temperature in the house begins to drop from [latex]70^{\\circ}\\text{F}[\/latex] at 9 p.m., reaches a low of [latex]60^{\\circ}[\/latex] during the night, and rises back to [latex]70^{\\circ}[\/latex] by 7 a.m. the next morning.<\/p>\r\n

Suppose that the temperature in the house is given by [latex]T(t)=0.4t^2-4t+70[\/latex] for [latex]0\\le t\\le 10[\/latex], where [latex]t[\/latex] is the number of hours past 9 p.m. Find the instantaneous rate of change of the temperature at midnight.<\/p>\r\n

[reveal-answer q=\"fs-id1169739297924\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1169739297924\"]<\/p>\r\n

Since midnight is [latex]3[\/latex] hours past 9 p.m., we want to compute [latex]T^{\\prime }(3)[\/latex]. Refer to the definition of a derivative.<\/p>\r\n

[latex]\\begin{array}{lllll}T^{\\prime}(3) & =\\underset{t\\to 3}{\\lim}\\frac{T(t)-T(3)}{t-3} & & & \\text{Apply the definition.} \\\\ & =\\underset{t\\to 3}{\\lim}\\frac{0.4t^2-4t+70-61.6}{t-3} & & & \\begin{array}{l}\\text{Substitute }T(t)=0.4t^2-4t+70 \\, \\text{and} \\\\ T(3)=61.6. \\end{array} \\\\ & =\\underset{t\\to 3}{\\lim}\\frac{0.4t^2-4t+8.4}{t-3} & & & \\text{Simplify.} \\\\ & =\\underset{t\\to 3}{\\lim}\\frac{0.4(t-3)(t-7)}{t-3} & & & =\\underset{t\\to 3}{\\lim}\\frac{0.4(t-3)(t-7)}{t-3} \\\\ & =\\underset{t\\to 3}{\\lim}0.4(t-7) & & & \\text{Cancel.} \\\\ & =-1.6 & & & \\text{Evaluate the limit.} \\end{array}[\/latex]<\/div>\r\n

The instantaneous rate of change of the temperature at midnight is [latex]-1.6^{\\circ}\\text{F}[\/latex] per hour.<\/p>\r\n

[\/hidden-answer]<\/p>\r\n<\/section>\r\n

\r\n

[ohm_question]162461[\/ohm_question]<\/p>\r\n<\/section>","rendered":"

Velocities and Rates of Change<\/h2>\n

Now that we can evaluate a derivative, we can apply it to velocity problems. Recall that if [latex]s(t)[\/latex] is the position of an object moving along a coordinate axis, the average velocity<\/span> of the object over a time interval [latex][a,t][\/latex] if [latex]t>a[\/latex] or [latex][t,a][\/latex] if [latex]t\n

[latex]v_{\\text{avg}}=\\dfrac{s(t)-s(a)}{t-a}[\/latex]<\/div>\n

As the values of [latex]t[\/latex] approach [latex]a[\/latex], the values of [latex]v_{\\text{avg}}[\/latex] approach the value we call the instantaneous velocity<\/span> at [latex]a[\/latex]. That is, instantaneous velocity at [latex]a[\/latex], denoted [latex]v(a)[\/latex], is given by<\/p>\n

[latex]v(a)=s^{\\prime}(a)=\\underset{t\\to a}{\\lim}\\dfrac{s(t)-s(a)}{t-a}[\/latex]<\/div>\n

To better understand the relationship between average velocity and instantaneous velocity, see Figure 7. In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time [latex]t=a[\/latex]. The slope of the secant line (shown in green) is the average velocity of the object over the time interval [latex][a,t][\/latex].<\/p>\n

\"This
Figure 7. The slope of the secant line is the average velocity over the interval [latex][a,t][\/latex]. The slope of the tangent line is the instantaneous velocity.<\/figcaption><\/figure>\n

We can use the definitions to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using the difference quotient.<\/p>\n

\n

A lead weight on a spring is oscillating up and down. Its position at time [latex]t[\/latex] with respect to a fixed horizontal line is given by [latex]s(t)= \\sin t[\/latex]. Use a table of values to estimate [latex]v(0)[\/latex]. Check the estimate by using the definition of a derivative.<\/p>\n

\"A
Figure 8. A lead weight suspended from a spring in vertical oscillatory motion.<\/figcaption><\/figure>\n