Question

Let $B\equiv(0,3)$ and $C\equiv(4,0)$. The point A is moving on the line $y=2x$ at the rate of 2 units/second. The area of $\triangle ABC$ is increasing at the rate of

• A. $\frac{11}{\sqrt{5}}(units)^{2}/sec$
• B. $\frac{11}{5}(units)^{2}/sec$
• C. $\frac{13}{\sqrt{5}}(units)^{2}/sec$
• D. $\frac{13}{5}(units)^{2}/sec$

Hint:

Calculus Tip: When a point moves along a specific curve or line at a given speed, always express its position parametrically using a single variable (like $h$) and use the distance formula from a fixed point (usually the origin) to link the speed to the derivative of that parameter ($\frac{dh}{dt}$).

Answer 1

The Correct Option is A

Concept: Calculus - Application of Derivatives (Rates of Change) and Coordinate Geometry.

Step 1: Define the coordinates of the moving point A. Point A lies on the line $y=2x$. Let the x-coordinate of point A be $h$. Then, its y-coordinate is $2h$. Thus, $A \equiv (h, 2h)$.

Step 2: Determine the rate of change of coordinate $h$. We are given that point A is moving along the line $y=2x$ at a speed of 2 units/second. This speed is the rate of change of the distance from the origin $O(0,0)$ to $A(h, 2h)$. Distance $OA = \sqrt{h^2 + (2h)^2} = \sqrt{5h^2} = \sqrt{5}h$. The rate of change is $\frac{d(OA)}{dt} = 2$. Differentiating $OA$ with respect to time $t$: $\sqrt{5} \frac{dh}{dt} = 2$. Therefore, $\frac{dh}{dt} = \frac{2}{\sqrt{5}}$. Let's call this Equation (i).

Step 3: Set up the formula for the area of $\triangle ABC$. The area ($\alpha$) of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Substitute the vertices $A(h, 2h)$, $B(0, 3)$, and $C(4, 0)$: $\alpha = \frac{1}{2}|h(3 - 0) + 0(0 - 2h) + 4(2h - 3)|$.

Step 4: Simplify the area expression. Expand the terms inside the absolute value: $\alpha = \frac{1}{2}|3h + 0 + 8h - 12| = \frac{1}{2}|11h - 12|$. Since A is moving along the line from the origin upwards (indicated by positive speed), $h$ increases, making $11h - 12$ positive for sufficiently large $h$. For the rate of increase, we differentiate the expression: $\alpha = \frac{11h - 12}{2}$.

Step 5: Differentiate the area with respect to time to find the final rate. Differentiate $\alpha$ with respect to time $t$: $\frac{d\alpha}{dt} = \frac{d}{dt} \left( \frac{11}{2}h - 6 \right) = \frac{11}{2} \frac{dh}{dt}$. Substitute the value of $\frac{dh}{dt}$ from Equation (i): $\frac{d\alpha}{dt} = \frac{11}{2} \left( \frac{2}{\sqrt{5}} \right) = \frac{11}{\sqrt{5}}$. $$ \therefore \text{The area is increasing at the rate of } \frac{11}{\sqrt{5}} \text{ sq units/sec.} $$