Question

, find the equation of the tangent line to the given curve at the given value of \(t\) without eliminating the parameter. Make a sketch. $$ x=t^{2}, y=t^{3} ; t=2 $$

Short answer

The equation of the tangent line is \(y = 3x - 4\).

Step-by-step solution

Find the Point on the Curve

The point on the curve corresponding to the given parameter value \(t = 2\) is found by substituting \(t = 2\) into the parametric equations: \(x = t^2 = 2^2 = 4\) and \(y = t^3 = 2^3 = 8\). Thus, the point is \((4, 8)\).

Compute Derivatives with respect to the Parameter

To find the slope of the tangent line, compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). From \(x = t^2\), \(\frac{dx}{dt} = 2t\). From \(y = t^3\), \(\frac{dy}{dt} = 3t^2\).

Calculate the Slope of the Tangent Line

The slope of the tangent line is given by the derivative \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\). Substituting the derivatives from Step 2, \(\frac{dy}{dx} = \frac{3t^2}{2t}\). Simplify to obtain \(\frac{dy}{dx} = \frac{3t}{2}\). For \(t = 2\), the slope is \(\frac{3 \times 2}{2} = 3\).

Write the Equation of the Tangent Line

The equation of a line is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point on the curve. Substituting \(m = 3\) and the point \((4, 8)\), the equation is \(y - 8 = 3(x - 4)\). Simplifying, we get \(y = 3x - 4\).

Sketch the Curve and Tangent Line

To sketch, plot the curve from the parametric equations \(x = t^2\) and \(y = t^3\) for values around \(t = 2\). Draw the tangent line \(y = 3x - 4\), which intersects the curve at \(t = 2\) and has a slope of 3.

Key concepts

Parametric Equations

Parametric equations allow for defining a curve by using a parameter, commonly denoted by \( t \). Unlike the typical \( y = f(x) \) form, parametric equations express both \( x \) and \( y \) as functions of this separate parameter. In this exercise, \( x = t^2 \) and \( y = t^3 \) describe a curve. By assigning different values to \( t \), you can determine different points on the curve. This way of defining curves provides flexibility and can represent complex curves that aren't easy to express using a single equation.
Parametric equations are widely used in calculus and physics, especially when analyzing motion or representing trajectories that depend on time or another variable.

Tangent Line

A tangent line to a curve at a specific point is a straight line that just "touches" the curve at that point and has the same direction as the curve does there. In simpler terms, it barely grazes the curve without crossing it at that single point.

• Finding the tangent line involves determining the slope of the curve at a given point and then using this slope to build the line's equation.
• This can help understand the instantaneous rate of change of the curve at that specific point.

For our problem, we aim to find the tangent line where \( t = 2 \), resulting in a tangent that goes through the point \((4, 8)\) with a calculated slope. This tangent line represents the instant direction of the curve at \( t = 2 \).

Derivatives

Derivatives are a core concept in calculus, representing the rate at which a function changes as its input changes. In the context of parametric equations, we find two derivatives: \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

• For \( x = t^2 \), the derivative \( \frac{dx}{dt} = 2t \) shows how \( x \) changes with \( t \).
• For \( y = t^3 \), it gives \( \frac{dy}{dt} = 3t^2 \), indicating the rate of change of \( y \) with respect to \( t \).

Calculating these derivatives helps in understanding the behavior of the curve and is essential for finding the slope of the tangent line. Together, they provide insight into how the curve behaves locally around the point of interest.

Slope of a Curve

The slope of a curve at a point is the steepness or incline of the curve at that exact spot. For parametric equations, we find this by taking the ratio of the derivatives, \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). This gives us a formula for the slope at any point without converting the parametric equations into a single function.

• For the given curve, \( \frac{dy}{dx} = \frac{3t^2}{2t} \) simplifies to \( \frac{3t}{2} \).
• By substituting \( t = 2 \), we find the slope to be \( 3 \), indicating how steep the tangent line is at that specific point.

The slope is an essential aspect of determining the equation of the tangent line, combining with the point on the curve to complete the line's equation. Understanding and calculating slope is fundamental when analyzing dynamic or intricate curve shapes.