Alex is a civil engineer working on the design of a curved road. The slope of the road at certain points is crucial for safety reasons, so Alex is tasked with calculating the slope of the road at a specific point along the curve. The road follows the curve described by the function:
R\left(x\right)=\frac{1}{2\sqrt{3x+1}}
Alex needs to find the slope of the road at the point Q(0,1 over 2)
 to ensure that vehicles can safely navigate this section.  Find an equation of the line tangent to the curve y=R(x)
 at Q
.

To find the slope of the road at point \( Q(0, \frac{1}{2}) \), we need to find the derivative of the function \( R(x) = \frac{1}{2\sqrt{3x + 1}} \) at \( x = 0 \). This derivative represents the slope of the tangent line to the curve at that point.

### Step 1: Differentiate \( R(x) \)
The function \( R(x) = \frac{1}{2\sqrt{3x + 1}} \) can be rewritten for convenience as:
\[
R(x) = \frac{1}{2} (3x + 1)^{-\frac{1}{2}}
\]

Now, let’s differentiate \( R(x) \) with respect to \( x \):
1. Apply the power rule and chain rule to differentiate \( (3x + 1)^{-\frac{1}{2}} \).
   
\[
R'(x) = \frac{1}{2} \cdot -\frac{1}{2} (3x + 1)^{-\frac{3}{2}} \cdot 3
\]

2. Simplify this expression:
\[
R'(x) = -\frac{3}{4} (3x + 1)^{-\frac{3}{2}}
\]

### Step 2: Calculate \( R'(0) \)
To find the slope of the tangent line at \( x = 0 \), substitute \( x = 0 \) into \( R'(x) \):

\[
R'(0) = -\frac{3}{4} (3 \cdot 0 + 1)^{-\frac{3}{2}} = -\frac{3}{4} \cdot 1 = -\frac{3}{4}
\]

So, the slope of the tangent line at \( Q \) is \( -\frac{3}{4} \).

### Step 3: Write the Equation of the Tangent Line
The equation of a line with slope \( m \) passing through a point \( (x_1, y_1) \) is:
\[
y - y_1 = m(x - x_1)
\]

Here, \( (x_1, y_1) = (0, \frac{1}{2}) \) and \( m = -\frac{3}{4} \), so:
\[
y - \frac{1}{2} = -\frac{3}{4}(x - 0)
\]

Simplifying:
\[
y = -\frac{3}{4}x + \frac{1}{2}
\]

### Answer
The equation of the tangent line to the curve \( y = R(x) \) at the point \( Q(0, \frac{1}{2}) \) is:
\[
y = -\frac{3}{4}x + \frac{1}{2}
\]

Would you like more details on each differentiation step or have any questions?

Here are some follow-up questions to expand understanding:
1. What is the significance of the tangent slope in terms of road safety for vehicles?
2. How does the derivative's sign (positive or negative) impact the design considerations for the curve?
3. What additional factors should Alex consider for safe road design on curves?
4. How would the equation of the tangent change if the road’s curve function changed slightly?
5. How can curvature (second derivative) provide further insights into the road design at this point?

**Tip:** When finding a tangent line's equation, always confirm the derivative and the point are correctly applied in the slope formula to avoid errors.

Alex is a civil engineer working on the design of a curved road. The slope of the road at certain points is crucial for safety reasons, so Alex is tasked with calculating the slope of the road at a specific point along the curve. The road follows the curve described by the function: R(x) = 1/(2√(3x + 1)). Alex needs to find the slope of the road at the point Q(0, 1/2) to ensure that vehicles can safely navigate this section. Find an equation of the line tangent to the curve y = R(x) at Q.

Learn how to calculate the slope of a road curve at a specific point using derivatives. This problem involves differentiating the function R(x) = 1/(2√(3x + 1)) and finding the equation of the tangent line at Q(0, 1/2), with step-by-step solutions. Ideal for students studying calculus in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation