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 StartFraction d squared y Over dx squared EndFraction
 at this point.
xequalstangent squared tminus1 ​, yequalscotangent t​; tequalsnegative StartFraction pi Over 4 EndFraction
Question content area bottom
Part 1
Write the equation of the tangent line.


We need to determine the tangent line equation and the second derivative \(\frac{d^2y}{dx^2}\) for the parametric curve given:

\[
x = \tan^2 t - 1, \quad y = \cot t, \quad t = -\frac{\pi}{4}.
\]

---

### Step 1: Find the point \((x, y)\) on the curve for \(t = -\frac{\pi}{4}\).

#### Compute \(x\):
\[
x = \tan^2\left(-\frac{\pi}{4}\right) - 1.
\]
\(\tan\left(-\frac{\pi}{4}\right) = -1\), so:
\[
x = (-1)^2 - 1 = 0.
\]

#### Compute \(y\):
\[
y = \cot\left(-\frac{\pi}{4}\right).
\]
\(\cot\left(-\frac{\pi}{4}\right) = -1\), so:
\[
y = -1.
\]

Thus, the point is \((0, -1)\).

---

### Step 2: Find the slope of the tangent line (\(\frac{dy}{dx}\)).

From parametric equations, the slope is:
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.
\]

#### Compute \(\frac{dx}{dt}\):
\[
x = \tan^2 t - 1 \implies \frac{dx}{dt} = 2\tan t \sec^2 t.
\]

#### Compute \(\frac{dy}{dt}\):
\[
y = \cot t \implies \frac{dy}{dt} = -\csc^2 t.
\]

#### Evaluate \(\frac{dy}{dx}\) at \(t = -\frac{\pi}{4}\):
At \(t = -\frac{\pi}{4}\):
\[
\tan\left(-\frac{\pi}{4}\right) = -1, \quad \sec^2\left(-\frac{\pi}{4}\right) = 2, \quad \csc^2\left(-\frac{\pi}{4}\right) = 2.
\]
\[
\frac{dx}{dt} = 2(-1)(2) = -4, \quad \frac{dy}{dt} = -2.
\]
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2}{-4} = \frac{1}{2}.
\]

The slope of the tangent line is \(\frac{1}{2}\).

---

### Step 3: Write the equation of the tangent line.

The equation of the tangent line is:
\[
y - y_1 = m(x - x_1),
\]
where \(m = \frac{1}{2}\), and \((x_1, y_1) = (0, -1)\).

\[
y - (-1) = \frac{1}{2}(x - 0) \implies y + 1 = \frac{1}{2}x.
\]

Final tangent line equation:
\[
y = \frac{1}{2}x - 1.
\]

---

### Part 2: Compute \(\frac{d^2y}{dx^2}\).

From parametric equations:
\[
\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \div \frac{dx}{dt}.
\]

#### Differentiate \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\csc^2 t}{2\tan t \sec^2 t}.
\]
Simplify:
\[
\frac{dy}{dx} = \frac{-\csc^2 t}{2\tan t (1 + \tan^2 t)}.
\]

Using the quotient rule, find \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\), then divide by \(\frac{dx}{dt} = -4\). The full derivative simplifies at \(t = -\frac{\pi}{4}\).

At \(t = -\frac{\pi}{4}\), \(\frac{d^2y}{dx^2}\) computes to \(-\frac{3}{4}\).

---

### Final Results:
1. Tangent line equation: \[
y = \frac{1}{2}x - 1.
\]
2. \(\frac{d^2y}{dx^2} = -\frac{3}{4}\).

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 StartFraction d squared y Over dx squared EndFraction at this point. x = tangent squared t - 1, y = cotangent t; t = negative StartFraction pi Over 4 EndFraction

Discover how to find the tangent line and second derivative of a parametric curve defined by x = tan²(t) - 1 and y = cot(t) at t = -π/4. This detailed guide covers calculus concepts, including derivatives, tangent line equations, and parametric differentiation. Suitable for Grades 11-12 or college calculus students.

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