The point (7,2) lies on a curve. The tanent line to the curve at this point is y=(-2/3)(x)-(20/3). if dx/dt = 18.37, find dy/dt. Find the speed at this point

We are given that the point \((7, 2)\) lies on a curve, and the equation of the tangent line at this point is \( y = -\frac{2}{3}x - \frac{20}{3} \). Additionally, \( \frac{dx}{dt} = 18.37 \), and we are asked to find \( \frac{dy}{dt} \) and the speed at this point.

### Step 1: Understand the Slope of the Tangent Line
The equation of the tangent line is given as:
\[
y = -\frac{2}{3}x - \frac{20}{3}
\]
The slope of the tangent line, which represents \(\frac{dy}{dx}\) at the point \((7, 2)\), is the coefficient of \(x\). Hence,
\[
\frac{dy}{dx} = -\frac{2}{3}
\]

### Step 2: Use the Chain Rule to Relate \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)
We know that:
\[
\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}
\]
Substitute the given values:
\[
\frac{dy}{dt} = \left(-\frac{2}{3}\right) \cdot 18.37
\]
Now, calculate:
\[
\frac{dy}{dt} = -\frac{2}{3} \times 18.37 = -12.2467
\]
Thus, \( \frac{dy}{dt} = -12.25 \) (rounded to two decimal places).

### Step 3: Calculate the Speed at the Point
The speed at a point is given by the magnitude of the velocity vector, which is found using both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). The formula for speed is:
\[
\text{Speed} = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2}
\]
Substitute the known values:
\[
\text{Speed} = \sqrt{(18.37)^2 + (-12.25)^2}
\]
Now, calculate:
\[
\text{Speed} = \sqrt{337.0569 + 150.0625} = \sqrt{487.1194} = 22.07
\]
Thus, the speed at this point is approximately \( 22.07 \).

### Final Answers:
- \( \frac{dy}{dt} = -12.25 \)
- The speed at the point is approximately \( 22.07 \).

Do you want further details or have any questions?

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Here are some related questions:
1. How does the chain rule help in finding the rates of change?
2. What does the slope of the tangent line represent geometrically?
3. Why is the speed calculated using both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)?
4. How would the problem change if the tangent line had a different slope?
5. What is the physical interpretation of the speed in this context?

**Tip:** Speed is always the magnitude of velocity, while velocity itself can have direction (sign).

The point (7,2) lies on a curve. The tangent line to the curve at this point is y=(-2/3)(x)-(20/3). If dx/dt = 18.37, find dy/dt. Find the speed at this point.

Learn how to find dy/dt and the speed at a given point (7, 2) on a curve using the tangent line equation y = (-2/3)x - (20/3). This calculus problem covers differential calculus concepts, the chain rule, and speed calculation. Suitable for students in Grades 11-12.

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