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Part 1

Find an equation for the line tangent to the curve at the point defined by the given value of t.

x equals 5 sine t comma y equals 5 cosine t comma t equals StartFraction 3 pi Over 4 EndFractionx=5sint, y=5cost, t=3π4

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Part 1

A.

yequals=55StartRoot 2 EndRoot2xplus+1

B.

yequals=minus−xplus+55StartRoot 2 EndRoot2

C.

yequals=55xplus+55StartRoot 2 EndRoot2

D.

yequals=xminus−55StartRoot 2 EndRoot

To find the equation of the tangent line to the parametric curve defined by \( x = 5\sin(t) \) and \( y = 5\cos(t) \) at the point where \( t = \frac{3\pi}{4} \), follow these steps:

### 1. **Find the coordinates of the point on the curve:**

Substitute \( t = \frac{3\pi}{4} \) into the parametric equations for \( x \) and \( y \):

\[
x = 5\sin\left(\frac{3\pi}{4}\right)
\]
\[
y = 5\cos\left(\frac{3\pi}{4}\right)
\]

Evaluate these:

\[
\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}
\]
\[
\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}
\]

So the coordinates of the point are:

\[
x = 5 \cdot \frac{\sqrt{2}}{2} = \frac{5\sqrt{2}}{2}
\]
\[
y = 5 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\frac{5\sqrt{2}}{2}
\]

### 2. **Find the slope of the tangent line:**

The slope of the tangent line to the parametric curve is given by \( \frac{dy}{dx} \). We find this by differentiating \( y \) and \( x \) with respect to \( t \):

\[
\frac{dx}{dt} = \frac{d}{dt}(5\sin t) = 5\cos t
\]
\[
\frac{dy}{dt} = \frac{d}{dt}(5\cos t) = -5\sin t
\]

So the slope \( \frac{dy}{dx} \) is:

\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-5\sin t}{5\cos t} = -\tan t
\]

At \( t = \frac{3\pi}{4} \):

\[
\tan\left(\frac{3\pi}{4}\right) = -1
\]

Thus, the slope of the tangent line is \( 1 \).

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

The equation of the tangent line can be written in point-slope form:

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

Substituting the point \(\left(\frac{5\sqrt{2}}{2}, -\frac{5\sqrt{2}}{2}\right)\) and the slope \( m = 1 \):

\[
y + \frac{5\sqrt{2}}{2} = 1\left(x - \frac{5\sqrt{2}}{2}\right)
\]

Simplifying, we get:

\[
y = x - 5\sqrt{2}
\]

### 4. **Compare with the given options:**

The correct answer is:

**D. \( y = x - \frac{5\sqrt{2}}{2} \)**

Would you like further details or have any questions?

Here are some additional related questions:
1. What is the general process for finding the slope of a tangent line in parametric form?
2. How does the point-slope form of a line equation work?
3. Why is \( \tan\left(\frac{3\pi}{4}\right) \) equal to \(-1\)?
4. How can you verify the slope of a curve at a specific point using derivatives?
5. What is the significance of the parametric form in describing curves?

**Tip:** Always ensure the angle is correctly evaluated when dealing with trigonometric functions, as they significantly affect the slope and the final equation of the tangent line.

Learn how to find the equation of the tangent line to the parametric curve x = 5sin(t), y = 5cos(t) at the point where t = 3π/4. This problem involves parametric equations, trigonometry, and the slope of a tangent line formula. Suitable for advanced high school students.

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