Menu Top
Complete Course of Mathematics
Topic 1: Numbers & Numerical Applications Topic 2: Algebra Topic 3: Quantitative Aptitude
Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
Topic 7: Mensuration Topic 8: Trigonometry Topic 9: Sets, Relations & Functions
Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
Topic 13: Linear Programming Topic 14: Index Numbers & Time-Based Data Topic 15: Financial Mathematics
Topic 16: Statistics & Probability


Content On This Page
Tangents and Normals to a Curve: Equations Approximations using Differentials Errors and Differentials (Absolute, Relative, Percentage Error)


Applications of Derivatives: Tangents, Normals, Approximations, Errors



Tangents and Normals to a Curve: Equations

The derivative of a function at a point provides the instantaneous rate of change and the slope of the curve at that specific point. This geometric interpretation is directly used to find the equations of the tangent line and the normal line to the curve at a given point.

Tangent Line

The tangent line is a straight line that passes through a point on a curve and has the same instantaneous slope as the curve at that point. It represents the best linear approximation of the curve near that point.


Normal Line

The normal line to a curve at a point is the line perpendicular to the tangent line at that same point.

Tangent and Normal lines to a curve at point P

Special Cases for Tangents and Normals

The formulas for the normal line need careful consideration when the tangent line is horizontal or vertical.


Example 1. Find the equations of the tangent and normal lines to the curve $y = x^3 - 2x + 1$ at the point $(2, 5)$.

Answer:

The given curve is $y = f(x) = x^3 - 2x + 1$. The given point is $P(x_1, y_1) = (2, 5)$.

First, verify that the point $(2, 5)$ lies on the curve by substituting $x=2$ into the function:

$f(2) = (2)^3 - 2(2) + 1 = 8 - 4 + 1 = 5$. Since $y=5$, the point $(2, 5)$ is indeed on the curve.

Step 1: Find the derivative $f'(x)$.

The derivative of the function gives the general slope of the tangent line at any point $x$.

"$f'(x) = \frac{d}{dx}(x^3 - 2x + 1)$"

Using the Power Rule, Constant Multiple Rule, and Sum/Difference Rule:

"$f'(x) = 3x^{3-1} - 2x^{1-1} + 0 = 3x^2 - 2$"

So, $f'(x) = 3x^2 - 2$.

Step 2: Find the slope of the tangent line at $x_1 = 2$.

The slope of the tangent at the point $(2, 5)$ is $f'(x)$ evaluated at $x=2$.

"$m_{\text{tangent}} = f'(2) = 3(2)^2 - 2$"

[Evaluate $f'(x)$ at $x=2$]

$= 3(4) - 2 = 12 - 2 = 10$"

The slope of the tangent line at $(2, 5)$ is 10.

Step 3: Find the equation of the tangent line.

Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, with the point $(x_1, y_1) = (2, 5)$ and slope $m = 10$:

"$y - 5 = 10(x - 2)$"

[Point-slope form]

Expand the right side:

"$y - 5 = 10x - 20$"

Solve for $y$ to get the slope-intercept form (or rearrange to general form):

"$y = 10x - 20 + 5$"

"$y = 10x - 15$"

Alternatively, in general form: $10x - y - 15 = 0$.

The equation of the tangent line is $y = 10x - 15$ (or $10x - y - 15 = 0$).

Step 4: Find the slope of the normal line.

The normal line is perpendicular to the tangent line. Since the slope of the tangent is $m_{\text{tangent}} = 10$, which is not zero, the slope of the normal line is the negative reciprocal:

"$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{10}$"

[Negative reciprocal of tangent slope]

Step 5: Find the equation of the normal line.

Using the point-slope form $y - y_1 = m(x - x_1)$, with the point $(x_1, y_1) = (2, 5)$ and slope $m = -\frac{1}{10}$:

"$y - 5 = -\frac{1}{10}(x - 2)$"

To clear the fraction, multiply both sides by 10:

"$10(y - 5) = -1(x - 2)$"

"$10y - 50 = -x + 2$"

Rearrange into the general form $Ax + By + C = 0$:

"$x + 10y - 50 - 2 = 0$"

"$x + 10y - 52 = 0$"

The equation of the normal line is $x + 10y - 52 = 0$.



Approximations using Differentials

Derivatives can be used to estimate how a function's value changes in response to a small change in its input. This application is based on the idea that, for a small interval, the function's graph is closely approximated by its tangent line. The change in the function's value ($\Delta y$) is approximated by the change in the y-value along the tangent line, which is given by the differential $dy$.

Differentials

Let $y = f(x)$ be a differentiable function. Consider a point $(x, f(x))$ on the graph.

Geometrically, if we move along the tangent line at $(x, f(x))$ by an amount $dx$ horizontally, the corresponding vertical change along the tangent line is $dy$. Recall that the slope of the tangent is $f'(x)$, so $f'(x) = \frac{\text{vertical change}}{\text{horizontal change}} = \frac{dy}{dx}$. This definition of $dy$ formalizes this relationship.

Differentials dy and dx vs actual change Delta y

Now consider the actual change in the function value when $x$ changes by a small amount $\Delta x = dx$. The actual change in $y$ is $\Delta y = f(x + \Delta x) - f(x)$.

From the limit definition of the derivative, $f'(x) = \lim\limits_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$. For small $\Delta x$, the difference quotient is a good approximation of the derivative:

$\frac{f(x + \Delta x) - f(x)}{\Delta x} \approx f'(x)$

Multiplying by $\Delta x$:

$f(x + \Delta x) - f(x) \approx f'(x) \Delta x$

Since $\Delta y = f(x + \Delta x) - f(x)$ and $dy = f'(x) dx$, setting $dx = \Delta x$, we get the approximation:

$\Delta y \approx dy = f'(x) \Delta x$ (when $\Delta x$ is small)

This means that for a small change in $x$, the change in $y$ along the tangent line ($dy$) is approximately equal to the actual change in $y$ along the curve ($\Delta y$).


Approximation Formula using Differentials

The approximation $\Delta y \approx dy$ can be used to approximate the value of a function at a point near a known point. Since $\Delta y = f(x + \Delta x) - f(x)$, we have:

$f(x + \Delta x) - f(x) \approx dy$

$f(x + \Delta x) - f(x) \approx f'(x) \Delta x$

Adding $f(x)$ to both sides, we get the approximation formula:

$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$

This formula is used to approximate the value of $f$ at a point $(x+\Delta x)$ that is close to a point $x$ where both $f(x)$ and $f'(x)$ are easy to calculate. The term $f(x) + f'(x) \Delta x$ is the value of the $y$-coordinate on the tangent line at $x + \Delta x$. Hence, this is also known as the linear approximation or tangent line approximation of $f(x + \Delta x)$.


Example 1. Use differentials to approximate the value of $\sqrt{25.3}$.

Answer:

We want to approximate $\sqrt{25.3}$. This can be viewed as evaluating a function $f(x) = \sqrt{x}$ at a point near a value where the function and its derivative are easy to calculate. We will use the approximation formula $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.

Step 1: Identify $f(x)$, a suitable $x$, and $\Delta x$.

The function is $f(x) = \sqrt{x}$.

We need to choose a value of $x$ close to 25.3 for which $f(x)$ (i.e., $\sqrt{x}$) is easy to calculate. $x=25$ is a good choice since $\sqrt{25}=5$ is known.

Then $x + \Delta x = 25.3$. So, $\Delta x = (x + \Delta x) - x = 25.3 - 25 = 0.3$. This is a small change, suitable for approximation using differentials.

Step 2: Calculate $f(x)$.

Evaluate the function at the chosen value of $x=25$:

"$f(25) = \sqrt{25} = 5$"

Step 3: Find the derivative $f'(x)$.

Find the derivative of $f(x) = \sqrt{x}$. Rewrite $\sqrt{x}$ using the power notation: $f(x) = x^{1/2}$.

Using the Power Rule ($\frac{d}{dx}(x^n) = nx^{n-1}$ with $n=1/2$):

"$f'(x) = \frac{1}{2}x^{(1/2)-1} = \frac{1}{2}x^{-1/2}$"

Rewrite with a positive exponent: $f'(x) = \frac{1}{2\sqrt{x}}$.

Step 4: Evaluate $f'(x)$ at the chosen $x$.

Evaluate the derivative at $x=25$:

"$f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{2(5)} = \frac{1}{10} = 0.1$"

Step 5: Apply the approximation formula.

Substitute the values of $f(25)$, $f'(25)$, and $\Delta x = 0.3$ into the formula $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$:

"$f(25.3) \approx f(25) + f'(25) \cdot \Delta x$"

"$\sqrt{25.3} \approx 5 + (0.1)(0.3)$"

Calculate the product $(0.1)(0.3) = 0.03$:

"$\sqrt{25.3} \approx 5 + 0.03$"

"$\sqrt{25.3} \approx 5.03$"

The approximate value of $\sqrt{25.3}$ using differentials is 5.03.

(A calculator gives $\sqrt{25.3} \approx 5.0299103...$. The approximation using differentials is very close to the actual value for a small $\Delta x$).



Errors and Differentials (Absolute, Relative, Percentage Error)

Differentials are not only useful for approximating function values but also for estimating the error that propagates through a calculation when there is a small error in the input measurement. If we measure a quantity $x$ with some error, and then calculate another quantity $y = f(x)$ based on that measurement, the error in $x$ will cause an error in $y$. Derivatives and differentials help us estimate this resulting error in $y$.

Definitions of Errors

Let $x$ be the measured value of a variable, and let $\Delta x$ be the error in the measurement of $x$. $\Delta x$ is the difference between the measured value and the true value. $|\Delta x|$ is the magnitude of the error.

Let $y = f(x)$ be the quantity calculated based on the measured value $x$. The actual change in $y$ due to the error $\Delta x$ is $\Delta y = f(x + \Delta x) - f(x)$. We want to estimate this error $\Delta y$.

These formulas allow us to use the derivative to estimate the magnitude of the error in a computed quantity $y$ based on a known error in the measured quantity $x$.


Example 1. The radius of a sphere is measured to be 7 cm with a possible error in measurement of $\pm 0.02$ cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in calculating the volume of the sphere.

Answer:

The quantity being calculated is the volume of a sphere, $V$, based on the measured radius, $r$. So, $V = f(r) = \frac{4}{3}\pi r^3$.

Given information:

  • Measured value of the independent variable: $r = 7$ cm.
  • Possible error in the measurement of $r$: $\Delta r = \pm 0.02$ cm. For calculating the maximum error, we use the magnitude of this error, $dr = \Delta r = 0.02$ cm.

We need to estimate the resulting errors in the calculated volume $V$.

Step 1: Find the derivative of $V$ with respect to $r$.

The derivative $\frac{dV}{dr}$ gives the rate of change of volume with respect to radius.

"$\frac{dV}{dr} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)$"

Using the Constant Multiple Rule and the Power Rule:

$= \frac{4}{3}\pi \cdot \frac{d}{dr}(r^3) = \frac{4}{3}\pi (3r^2) = 4\pi r^2$"

So, $\frac{dV}{dr} = 4\pi r^2$.

Step 2: Estimate the maximum propagated Absolute Error ($\Delta V \approx dV$).

The differential $dV$ is given by $dV = \frac{dV}{dr} dr$. We use the measured value $r=7$ and the maximum error $dr = 0.02$.

"$dV = (4\pi r^2) dr$"

Substitute $r=7$ and $dr=0.02$:

"$dV = 4\pi (7)^2 (0.02)$"

$= 4\pi (49) (0.02)$"

$= 196\pi (0.02)$"

$= 3.92\pi$"

The estimated maximum possible absolute error in the volume is $\Delta V \approx dV = 3.92\pi$. The units are cm³ (units of $V$).

Maximum Possible Absolute Error $\approx$ $3.92\pi$ cm³.

Step 3: Estimate the maximum propagated Relative Error ($\frac{\Delta V}{V} \approx \frac{dV}{V}$).

The relative error in $V$ is $\frac{dV}{V}$. We need the calculated value of $V$ at $r=7$.

"$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (7)^3$"

[Volume at $r=7$]

$= \frac{4}{3}\pi (343) = \frac{1372\pi}{3}$"

Now calculate the ratio $\frac{dV}{V}$ using $dV = 3.92\pi$ and $V = \frac{1372\pi}{3}$:

Relative Error $\approx \frac{dV}{V} = \frac{3.92\pi}{1372\pi / 3}$

$= 3.92 \cdot \frac{3}{1372} = \frac{11.76}{1372}$"

Alternatively, and often more simply, using the relationship $dV = f'(r) dr$ and $V=f(r)$: Relative Error $\approx \frac{f'(r) dr}{f(r)} = \frac{4\pi r^2 dr}{(4/3)\pi r^3} = \frac{3 dr}{r}$.

Substitute $dr = 0.02$ and $r=7$:

Relative Error $\approx \frac{3(0.02)}{7} = \frac{0.06}{7}$"

The estimated maximum possible relative error in the volume is $\frac{0.06}{7}$. This is a dimensionless quantity.

Maximum Possible Relative Error $\approx$ $\frac{0.06}{7}$.

Step 4: Estimate the maximum propagated Percentage Error.

The percentage error is the relative error multiplied by 100%.

Percentage Error $\approx \left(\frac{0.06}{7}\right) \times 100\%$"

$= \frac{6}{7}\% \approx 0.857\%$"

The estimated maximum possible percentage error in the volume is approximately $0.857\%$.

Maximum Possible Percentage Error $\approx$ $\frac{6}{7}\% \approx 0.857\%$.