Tangent Lines to Parametrized Curves

In Exercises 1−14, find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of d²y/dx² at this point.

x = sec² t − 1, y = tan t, t = −π/4

검증된 단계별 안내

First, find the derivatives of the parametric equations with respect to the parameter \(t\). Compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) from the given functions \(x = \sec^{2} t - 1\) and \(y = \tan t\).

Next, find the slope of the tangent line \(\frac{dy}{dx}\) at the given value of \(t = -\frac{\pi}{4}\) by using the chain rule for parametric curves: \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

Evaluate \(x\) and \(y\) at \(t = -\frac{\pi}{4}\) to find the specific point on the curve where the tangent line touches.

Use the point-slope form of a line, \(y - y_0 = m(x - x_0)\), where \(m\) is the slope found in step 2 and \((x_0, y_0)\) is the point found in step 3, to write the equation of the tangent line.

To find \(\frac{d^{2}y}{dx^{2}}\) at \(t = -\frac{\pi}{4}\), first compute \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\), then divide by \(\frac{dx}{dt}\) using the formula \(\frac{d^{2}y}{dx^{2}} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\), and finally evaluate at \(t = -\frac{\pi}{4}\).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

11m

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Parametric Equations and Curves

Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted t. Understanding how x and y depend on t allows us to analyze the curve's shape and behavior by varying t, rather than expressing y directly as a function of x.

Finding the Tangent Line to a Parametric Curve

The tangent line at a point on a parametric curve is found using derivatives dx/dt and dy/dt. The slope of the tangent line is dy/dx = (dy/dt) / (dx/dt), evaluated at the given t. Using this slope and the point coordinates, the tangent line equation can be written in point-slope form.

Second Derivative d²y/dx² for Parametric Curves

The second derivative d²y/dx² measures the curvature of the parametric curve and is found by differentiating dy/dx with respect to x. Using the chain rule, d²y/dx² = (d/dt(dy/dx)) / (dx/dt). Evaluating this at the given t provides the curve's concavity at that point.

Tangent Lines to Parametrized CurvesIn Exercises 1−14, find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of d²y/dx² at this point.x = sec² t − 1, y = tan t, t = −π/4