What is the slope of the tangent to the curve `x =t^(2) + 3t-8, y =2t^(2) - 2t -5` at t=2?

Options

A. `7//6`
B. `6//7`
C. 1
D. `5//6`

Answer

To find the slope of the tangent to the curve defined by the parametric equations \( x = t^2 + 3t - 8 \) and \( y = 2t^2 - 2t - 5 \) at \( t = 2 \), we will follow these steps:

### Step 1: Differentiate \( x \) and \( y \) with respect to \( t \)

1. Differentiate \( x = t^2 + 3t - 8 \):
\[ \frac{dx}{dt} = \frac{d}{dt}(t^2 + 3t - 8) = 2t + 3 \]

2. Differentiate \( y = 2t^2 - 2t - 5 \):
\[ \frac{dy}{dt} = \frac{d}{dt}(2t^2 - 2t - 5) = 4t - 2 \]

### Step 2: Find the slope \( \frac{dy}{dx} \)

The slope of the tangent line is given by the formula:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{4t - 2}{2t + 3} \]

### Step 3: Evaluate the slope at \( t = 2 \)

Now we substitute \( t = 2 \) into the slope formula:
\[ \frac{dy}{dx} \bigg|_{t=2} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7} \]

### Conclusion

The slope of the tangent to the curve at \( t = 2 \) is:
\[ \frac{6}{7} \]

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