Calculus with Parametric Curves: Videos & Practice Problems

To find the derivative of parametric equations, express x and y as functions of t, then compute dy/dx by dividing dy/dt by dx/dt. For higher-order derivatives, differentiate the first derivative with respect to t and divide by dx/dt. To calculate the arc length of a parametric curve, use the formula $s=\int t_{\in }{\surd }^{({\frac{dx/dt}{}}^{)}+{\frac{dy/dt}{}}^{²})}dt$ over the interval [a, b].

Differentiation of Parametric Curves

In the study of parametric equations, we express the variables \(x\) and \(y\) as functions of a parameter \(t\). To find the derivative of these parametric equations, we utilize the relationship between the derivatives of \(y\) and \(x\) with respect to \(t\). Specifically, the derivative \(\frac{dy}{dx}\) can be calculated by dividing the derivative of \(y\) with respect to \(t\) by the derivative of \(x\) with respect to \(t\):

\[\frac{dy}{dx} = \frac{y'(t)}{x'(t)}\]

Here, \(y'(t)\) and \(x'(t)\) represent the derivatives of \(y\) and \(x\) with respect to \(t\), respectively. It is crucial to ensure that \(x'(t) \neq 0\) to avoid division by zero.

For example, consider the parametric equations \(x(t) = t^2 + 3t\) and \(y(t) = 2t^3 - 4\). To find the derivatives, we apply the power rule:

1. For \(y(t)\): \[ y'(t) = \frac{d}{dt}(2t^3 - 4) = 6t^2 \]2. For \(x(t)\): \[ x'(t) = \frac{d}{dt}(t^2 + 3t) = 2t + 3 \]

Substituting these derivatives into the formula for \(\frac{dy}{dx}\), we have:

\[\frac{dy}{dx} = \frac{6t^2}{2t + 3}\]

Next, if we want to find the equation of the tangent line at a specific point, we need to determine the coordinates of that point and the slope of the tangent line. For instance, if we want to find the tangent line when \(t = 1\), we first calculate:

1. \(x(1) = 1^2 + 3(1) = 4\) 2. \(y(1) = 2(1)^3 - 4 = -2\)

Thus, the point on the curve is \((4, -2)\).

Next, we find the slope at \(t = 1\) by substituting \(t = 1\) into the derivative:

\[\frac{dy}{dx} \bigg|_{t=1} = \frac{6(1)^2}{2(1) + 3} = \frac{6}{5}\]

With the slope and the point, we can use the point-slope form of the equation of a line:

\[y - y_1 = m(x - x_1)\]

Substituting in our values, we have:

\[y - (-2) = \frac{6}{5}(x - 4)\]

This equation represents the tangent line to the curve at the point \((4, -2)\). If desired, this can be rearranged into slope-intercept form by solving for \(y\).

Understanding how to derive and manipulate parametric equations is essential for analyzing curves and their properties, including tangent lines. Practice with various examples will solidify these concepts and enhance your problem-solving skills in calculus.

Differentiation of Parametric Curves Example 1

To find the derivative \(\frac{dy}{dx}\) for the parametric curve defined by \(x(t) = 3 \cos(t)\) and \(y(t) = 3 \sin(t)\), we start by applying the formula for the derivative of a parametric curve. This formula states that:

\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]

First, we need to compute the derivatives \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\).

For \(y(t) = 3 \sin(t)\), the derivative is:

\[\frac{dy}{dt} = 3 \cos(t)\]

For \(x(t) = 3 \cos(t)\), the derivative is:

\[\frac{dx}{dt} = -3 \sin(t)\]

Now, substituting these derivatives into the formula gives us:

\[\frac{dy}{dx} = \frac{3 \cos(t)}{-3 \sin(t)}\]

The \(3\) in the numerator and denominator cancels out, leading to:

\[\frac{dy}{dx} = -\frac{\cos(t)}{\sin(t)}\]

Recognizing that \(-\frac{\cos(t)}{\sin(t)}\) is the negative cotangent function, we can express this as:

\[\frac{dy}{dx} = -\cot(t)\]

This result provides the slope of the tangent line to the parametric curve at any point \(t\). Thus, the derivative of the given parametric equations is \(-\cot(t)\).

Write the equation of the tangent line in cartesian coordinates for the given parameter $t$.
$x=8\cos t$, $y=6\sin t$, $t=\frac{\pi }{4}$

• A. $y-3\sqrt{2}=-\frac{3}{4}(x-4\sqrt{2})$
• B. $y-3\sqrt{2}=\frac{3}{4}(x-4\sqrt{2})$
• C. $y-4\sqrt{2}=-\frac{3}{4}(x-3\sqrt{2})$
• D. $y-4\sqrt{2}=\frac{3}{4}(x-3\sqrt{2})$

Answer: (A)

Differentiation of Parametric Curves Example 2

To find points on the parametric curve defined by the equations \( x(t) = t - \frac{1}{t} \) and \( y(t) = t + \frac{1}{t} \) where the slope is zero, we need to determine where the derivative \( \frac{dy}{dx} \) equals zero. This requires calculating the derivatives \( y'(t) \) and \( x'(t) \) with respect to \( t \).

First, we find \( y'(t) \) by differentiating \( y(t) \):

Given \( y(t) = t + \frac{1}{t} \), we can rewrite \( \frac{1}{t} \) as \( t^{-1} \). Using the power rule:

\[y'(t) = \frac{d}{dt}(t) + \frac{d}{dt}(t^{-1}) = 1 - \frac{1}{t^2}\]

Next, we find \( x'(t) \) by differentiating \( x(t) \):

Given \( x(t) = t - \frac{1}{t} \), we similarly differentiate:

\[x'(t) = \frac{d}{dt}(t) - \frac{d}{dt}(t^{-1}) = 1 + \frac{1}{t^2}\]

Now, we can express the derivative \( \frac{dy}{dx} \) as:

\[\frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{1 - \frac{1}{t^2}}{1 + \frac{1}{t^2}}\]

To find where the slope is zero, we set \( \frac{dy}{dx} = 0 \). This occurs when the numerator is zero:

\[1 - \frac{1}{t^2} = 0\]

Solving for \( t^2 \), we add \( \frac{1}{t^2} \) to both sides:

\[1 = \frac{1}{t^2} \implies t^2 = 1\]

Taking the square root gives us \( t = \pm 1 \). Now, we need to find the corresponding \( x \) and \( y \) coordinates for these \( t \) values.

For \( t = 1 \):

\[x(1) = 1 - \frac{1}{1} = 0\]

\[y(1) = 1 + \frac{1}{1} = 2\]

Thus, one point is \( (0, 2) \).

For \( t = -1 \):

\[x(-1) = -1 - \frac{1}{-1} = 0\]

\[y(-1) = -1 + \frac{1}{-1} = -2\]

Thus, the second point is \( (0, -2) \).

In conclusion, the points on the curve where the slope is zero are \( (0, 2) \) and \( (0, -2) \).

Higher Order Derivatives of Parametric Curves

To find higher order derivatives of parametric curves, it's essential to understand the process for calculating the first and subsequent derivatives. Given parametric equations \( x(t) \) and \( y(t) \), the first derivative \( \frac{dy}{dx} \) can be found by differentiating \( y \) and \( x \) with respect to \( t \) and then dividing the results:

\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]

Once the first derivative is established, the second derivative \( \frac{d^2y}{dx^2} \) can be calculated by differentiating the first derivative with respect to \( t \) and then dividing by \( \frac{dx}{dt} \) again:

\[\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\]

For example, if we have \( x(t) = t^2 + 3t \) and \( y(t) = 2t^3 - 4 \), we first find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \):

\[\frac{dy}{dt} = 6t^2, \quad \frac{dx}{dt} = 2t + 3\]

Thus, the first derivative becomes:

\[\frac{dy}{dx} = \frac{6t^2}{2t + 3}\]

To find the second derivative, we apply the quotient rule to differentiate \( \frac{dy}{dx} \) with respect to \( t \):

\[\frac{d}{dt}\left(\frac{6t^2}{2t + 3}\right) = \frac{(2t + 3)(12t) - (6t^2)(2)}{(2t + 3)^2}\]

After simplifying, we arrive at:

\[\frac{d^2y}{dx^2} = \frac{12t^2 + 36t}{(2t + 3)^3}\]

For higher order derivatives, the same pattern applies: differentiate the most recent derivative with respect to \( t \) and divide by \( \frac{dx}{dt} \). This method can be repeated for third, fourth, or any higher order derivatives, ensuring a systematic approach to parametric curves.

Find $\frac{d^{2}y}{d{x}^2}$ for the parametric curve at the given point.
$x=8\cos t$, $y=6\sin t$, $t=\frac{\pi }{4}$

• A. $\frac{3}{2}$
• B. $-\frac{3}{2}$
• C. $-\frac{3\sqrt{2}}{16}$
• D. $\frac{3\sqrt{2}}{16}$

Answer: (C)

Higher Order Derivatives of Parametric Curves Example 3

To determine the concavity of the curve defined by the parametric equations \( x = t \) and \( y = t^3 - 3t \), we need to analyze the second derivative of \( y \) with respect to \( x \). A function is considered concave up when its second derivative is positive and concave down when it is negative.

First, we find the first derivative \( \frac{dy}{dx} \) using the chain rule. This involves calculating the derivatives of \( y \) and \( x \) with respect to \( t \). The derivative of \( y \) is:

\( y' = \frac{dy}{dt} = 3t^2 - 3 \)

For \( x \), since \( x = t \), we have:

\( x' = \frac{dx}{dt} = 1 \)

Thus, the first derivative \( \frac{dy}{dx} \) is given by:

\( \frac{dy}{dx} = \frac{y'}{x'} = \frac{3t^2 - 3}{1} = 3t^2 - 3 \)

Next, we need to find the second derivative \( \frac{d^2y}{dx^2} \). This requires taking the derivative of \( \frac{dy}{dx} \) with respect to \( t \) and then dividing by \( \frac{dx}{dt} \):

\( \frac{d^2y}{dx^2} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \div \frac{dx}{dt} \)

Calculating the derivative of \( \frac{dy}{dx} \):

\( \frac{d}{dt}(3t^2 - 3) = 6t \)

Since \( \frac{dx}{dt} = 1 \), we find:

\( \frac{d^2y}{dx^2} = \frac{6t}{1} = 6t \)

Now, to determine the concavity, we analyze the sign of \( 6t \). The function is concave up when \( 6t > 0 \) (i.e., \( t > 0 \)) and concave down when \( 6t < 0 \) (i.e., \( t < 0 \)). At \( t = 0 \), the second derivative equals zero, indicating a potential inflection point where the concavity changes.

In summary, the curve is concave up for \( t > 0 \) and concave down for \( t < 0 \). The point \( t = 0 \) serves as an inflection point where the concavity transitions.

Find $\frac{d^{3}y}{d{x}^3}$ for the parametric curve at the given point.
$x=2t^3$, $y=t^4$, $t=1$

• A. $-\frac{4}{9}$
• B. $-\frac{2}{9}$
• C. $\frac{2}{3}$
• D. $-\frac{1}{27}$

Answer: (D)

Arc Length of Parametric Curves

Parametric curves are represented by equations that define the x and y coordinates in terms of a parameter, typically denoted as t. To find the arc length of a parametric curve, we are essentially calculating the distance traveled along the curve between two points. This process can initially seem challenging, but it can be simplified by understanding the underlying principles.

The arc length, denoted as \( s \), can be derived from the concept of summing infinitely many small line segments along the curve. To calculate the arc length, we utilize the derivatives of the parametric equations \( x(t) \) and \( y(t) \). The formula for arc length is given by:

\[s = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]

Here, \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) represent the derivatives of the x and y coordinates with respect to the parameter t, and \( [a, b] \) is the interval over which we are calculating the arc length.

To illustrate this, consider the parametric equations \( x(t) = 2 \cos(t) \) and \( y(t) = 2 \sin(t) \) over the interval from \( 0 \) to \( \pi \). First, we compute the derivatives:

\[\frac{dx}{dt} = -2 \sin(t)\]

\[\frac{dy}{dt} = 2 \cos(t)\]

Substituting these derivatives into the arc length formula, we have:

\[s = \int_{0}^{\pi} \sqrt{(-2 \sin(t))^2 + (2 \cos(t))^2} \, dt\]

Squaring the terms inside the square root gives:

\[s = \int_{0}^{\pi} \sqrt{4 \sin^2(t) + 4 \cos^2(t)} \, dt\]

Factoring out the common term results in:

\[s = \int_{0}^{\pi} \sqrt{4(\sin^2(t) + \cos^2(t))} \, dt\]

Using the Pythagorean identity \( \sin^2(t) + \cos^2(t) = 1 \), we simplify this to:

\[s = \int_{0}^{\pi} \sqrt{4} \, dt = \int_{0}^{\pi} 2 \, dt\]

Integrating gives:

\[s = 2t \bigg|_{0}^{\pi} = 2\pi\]

Thus, the arc length of the curve defined by these parametric equations is \( 2\pi \). This example highlights the importance of understanding the relationship between the arc length formula and the distance formula, as well as the utility of derivatives in calculating distances along parametric curves.

Arc Length of Parametric Curves Example 4

To find the arc length of a parametric curve defined by the equations \( x(t) \) and \( y(t) \) from \( t = 0 \) to \( t = 1 \), we utilize the arc length formula. The arc length \( s \) can be expressed as:

\[s = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]

In this case, we need to compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For example, if \( x(t) = t^3 \), then:

\[\frac{dx}{dt} = 3t^2\]

And if \( y(t) = 2t^2 \), then:

\[\frac{dy}{dt} = 4t\]

Substituting these derivatives into the arc length formula gives us:

\[s = \int_{0}^{1} \sqrt{(3t^2)^2 + (4t)^2} \, dt\]

This simplifies to:

\[s = \int_{0}^{1} \sqrt{9t^4 + 16t^2} \, dt\]

Factoring out \( t^2 \) from the square root yields:

\[s = \int_{0}^{1} t \sqrt{9t^2 + 16} \, dt\]

Next, we can perform a substitution to simplify the integral. Let:

\[u = 9t^2 + 16 \quad \Rightarrow \quad du = 18t \, dt \quad \Rightarrow \quad dt = \frac{du}{18t}\]

Substituting \( dt \) into the integral, we have:

\[s = \int \frac{1}{18} \sqrt{u} \, du\]

Integrating \( u^{1/2} \) gives:

\[s = \frac{1}{18} \cdot \frac{2}{3} u^{3/2} + C\]

Substituting back for \( u \) results in:

\[s = \frac{1}{27} (9t^2 + 16)^{3/2}\]

Now, we evaluate this from \( t = 0 \) to \( t = 1 \):

\[s = \frac{1}{27} \left[ (9(1)^2 + 16)^{3/2} - (9(0)^2 + 16)^{3/2} \right]\]

This simplifies to:

\[s = \frac{1}{27} \left[ (9 + 16)^{3/2} - 16^{3/2} \right] = \frac{1}{27} \left[ 25^{3/2} - 16^{3/2} \right]\]

Calculating \( 25^{3/2} = 125 \) and \( 16^{3/2} = 64 \), we find:

\[s = \frac{1}{27} (125 - 64) = \frac{61}{27}\]

Thus, the arc length of the parametric curve from \( t = 0 \) to \( t = 1 \) is:

\[\boxed{\frac{61}{27}}\]

Find the length of the curve below on the interval $[0,4]$.
$x=\frac{1}{3}{(2t+3)}^{\frac{3}{2}}$, $y=\frac{1}{2}{t}^2+t$

• A. $16$
• B. $24$
• C. $12$
• D. $20$

Answer: (A)

Here’s what students ask on this topic:

How do you find the derivative of a parametric curve?

To find the derivative of a parametric curve, where x and y are expressed as functions of a parameter t (x(t) and y(t)), you calculate $\frac{d}{y}/\frac{d}{x}$ by dividing $\frac{d}{y}/\frac{d}{t}$ by $\frac{d}{x}/\frac{d}{t}$. Specifically, compute the derivatives of y(t) and x(t) with respect to t, then divide them: $\frac{d}{y}/\frac{d}{x}=\frac{\frac{d}{y}}{/}$. Ensure that $\frac{d}{x}/\frac{d}{t}$ is not zero to avoid division by zero. This process is essential for analyzing the slope of the curve at any given point.

How do you find the second derivative of a parametric curve?

To find the second derivative of a parametric curve, first compute the first derivative $\frac{d}{y}/\frac{d}{x}$. Then, differentiate this result with respect to t, and divide by $\frac{d}{x}/\frac{d}{t}$. The formula is: $\frac{\frac{d}{\frac{d}{y}}}{/}$. This process involves applying the chain rule and quotient rule, especially if the first derivative is a rational function. Simplify the result to express the second derivative in terms of t.

What is the formula for finding the arc length of a parametric curve?

The formula for finding the arc length of a parametric curve defined by x(t) and y(t) over the interval [a, b] is: $s_{=}{\int }_a^b\sqrt{{\frac{d}{x}}^{/}}dt$. Here, $\frac{d}{x}/\frac{d}{t}$ and $\frac{d}{y}/\frac{d}{t}$ are the derivatives of x(t) and y(t) with respect to t. This formula calculates the total distance traveled along the curve by summing infinitesimally small line segments.

How do you find the equation of a tangent line to a parametric curve?

To find the equation of a tangent line to a parametric curve at a specific t-value, follow these steps: 1) Compute the slope of the tangent line using $\frac{d}{y}/\frac{d}{x}=\frac{\frac{d}{y}}{/}$. 2) Find the point of tangency by substituting the t-value into x(t) and y(t). 3) Use the point-slope form of a line: $y-y_1=m(x-x_1)$, where m is the slope, and $x_1$ and $y_1$ are the coordinates of the point of tangency.

What are higher-order derivatives in parametric equations?

Higher-order derivatives in parametric equations involve differentiating the first derivative repeatedly. For example, the second derivative $\frac{\frac{d}{\frac{d}{y}}}{/}$ is found by differentiating the first derivative $\frac{d}{y}/\frac{d}{x}$ with respect to t and dividing by $\frac{d}{x}/\frac{d}{t}$. For the third derivative, repeat the process: differentiate the second derivative with respect to t and divide by $\frac{d}{x}/\frac{d}{t}$. This pattern continues for higher orders.