A variable `Delta ABC` in the xy plane has its orthocentre at vertex B , a fixed vertex 'A' at the ongin and the third vertex restricted to lie on the parabola `y = 1+ (7x^2)/36` The point B starts at the point (0, 1) at time t=0 and moves upward along the y axis at a constant velocity of 2 cm/sec. How fast is the area of the triangle increasing when t=7/2 sec

Answer

To solve the problem step by step, we will follow the given information and derive the necessary equations to find out how fast the area of triangle ABC is increasing when \( t = \frac{7}{2} \) seconds.

### Step 1: Understand the Triangle Configuration
- Vertex A is at the origin \( (0, 0) \).
- Vertex B starts at \( (0, 1) \) and moves upward along the y-axis at a velocity of 2 cm/sec.
- Vertex C lies on the parabola \( y = 1 + \frac{7x^2}{36} \).

### Step 2: Determine the Position of Point B
At time \( t \), the coordinates of point B can be expressed as:
\[ B(0, 1 + 2t) \]

### Step 3: Find the Coordinates of Point C
Since point C lies on the parabola, we have:
\[ C(x, 1 + \frac{7x^2}{36}) \]
At the point where the y-coordinates of B and C are equal:
\[ 1 + 2t = 1 + \frac{7x^2}{36} \]
This simplifies to:
\[ 2t = \frac{7x^2}{36} \]
From this, we can solve for \( x^2 \):
\[ x^2 = \frac{72t}{7} \]
Thus, the x-coordinate of point C is:
\[ x = \sqrt{\frac{72t}{7}} \]

### Step 4: Calculate the Area of Triangle ABC
The area \( A \) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( BC \) is the distance between points B and C, which is:
\[ BC = x \]
And the height from point A to line BC is the y-coordinate of point B:
\[ \text{height} = 1 + 2t \]
Thus, the area becomes:
\[ A = \frac{1}{2} \times x \times (1 + 2t) \]
Substituting \( x = \sqrt{\frac{72t}{7}} \):
\[ A = \frac{1}{2} \times \sqrt{\frac{72t}{7}} \times (1 + 2t) \]

### Step 5: Differentiate the Area with Respect to Time
To find how fast the area is increasing, we differentiate \( A \) with respect to \( t \):
\[ \frac{dA}{dt} = \frac{1}{2} \left( \frac{d}{dt} \left( \sqrt{\frac{72t}{7}} \right) (1 + 2t) + \sqrt{\frac{72t}{7}} \frac{d}{dt}(1 + 2t) \right) \]

Calculating the derivatives:
1. For \( \sqrt{\frac{72t}{7}} \):
\[ \frac{d}{dt} \left( \sqrt{\frac{72t}{7}} \right) = \frac{1}{2} \left( \frac{72}{7} \right)^{1/2} \cdot \frac{1}{t^{1/2}} = \frac{36}{7\sqrt{72t}} \]
2. For \( (1 + 2t) \):
\[ \frac{d}{dt}(1 + 2t) = 2 \]

Substituting these into the derivative of area:
\[ \frac{dA}{dt} = \frac{1}{2} \left( \frac{36}{7\sqrt{72t}} (1 + 2t) + \sqrt{\frac{72t}{7}} \cdot 2 \right) \]

### Step 6: Evaluate at \( t = \frac{7}{2} \)
Substituting \( t = \frac{7}{2} \):
1. Calculate \( x \):
\[ x = \sqrt{\frac{72 \cdot \frac{7}{2}}{7}} = \sqrt{36} = 6 \]
2. Calculate \( 1 + 2t \):
\[ 1 + 2 \cdot \frac{7}{2} = 8 \]
3. Substitute into \( \frac{dA}{dt} \):
\[ \frac{dA}{dt} = \frac{1}{2} \left( \frac{36}{7 \cdot 6} \cdot 8 + 6 \cdot 2 \right) \]
\[ = \frac{1}{2} \left( \frac{288}{42} + 12 \right) = \frac{1}{2} \left( \frac{288 + 504}{42} \right) = \frac{1}{2} \cdot \frac{792}{42} = \frac{396}{42} = 9.42857 \text{ cm}^2/\text{s} \]

Thus, the area of triangle ABC is increasing at a rate of approximately \( 9.43 \text{ cm}^2/\text{s} \) when \( t = \frac{7}{2} \) seconds.