Question
Topic-(a)- 4.6: Approximating Values of a Function Using Local Linearity and Linearization
Topic-(b)- 3.2: Implicit Differentiation
Topic-(c)- 5.12: Exploring Behaviors of Implicit Relations
Topic-(d)- 4.4: Introduction to Related Rates
5. Consider the curve defined by the equation \(x^{2}+3y+2y^{2}=48\) . It can be shown that \(\frac{dy}{dx}=\frac{-2x}{3+4y}\)
(a) There is a point on the curve near (2, 4) with x-coordinate 3. Use the line tangent to the curve at (2, 4) to approximate the y-coordinate of this point.
(b) Is the horizontal line y = 1 tangent to the curve? Give a reason for your answer.
(c) The curve intersects the positive x-axis at the point \(\left ( \sqrt{48},0 \right )\). Is the line tangent to the curve at this point vertical? Give a reason for your answer.
(d) For time t ≥ 0, a particle is moving along another curve defined by the equation \(y^{3}+2xy=24\) . At the instant the particle is at the point (4, 2), the y-coordinate of the particle’s position is decreasing at a rate of 2 units per second. At that instant, what is the rate of change of the x-coordinate of the particle’s position with respect to time?
▶️Answer/Explanation
5(a) There is a point on the curve near (2, 4) with x-coordinate 3. Use the line tangent to the curve at (2, 4) to approximate the y -coordinate of this point.
\(\frac{dy}{dx}| _{(x,y) = (2,4)} = \frac{-2(2)}{3+4(4)} = \frac{4}{19}\)
\(y = 4 -\frac{4}{19} (3-2) = \frac{72}{19}\)
5(b) Is the horizontal line y =1 tangent to the curve? Give a reason for your answer.
\(\frac{dy}{dx}=\frac{-2x}{3+4y}=0\Rightarrow x=0\)
And so, if the horizontal line y =1 is tangent to the curve, the point of tangency must be (0, 1). However, the point (0,1) is not on the curve, because \(0^{2}+3.1+2.1^{2}=5\neq 48\),
Therefore, the horizontal line y =1 is not tangent to the curve.
5(c) The curve intersects the positive x -axis at the point \(\left ( \sqrt{48},0 \right )\). Is the line tangent to the curve at this point vertical? Give a reason for your answer.
At the point \(\left ( \sqrt{48},0 \right )\), the slope of the line tangent to the
curve is \(\frac{dy}{dx}=\frac{-2\sqrt{48}}{3+4(0)}\) The denominator of \(\frac{dy}{dx}\\\ is\\\ 3 +4 (0)\), which does not equal 0.
Therefore, the line tangent to the curve at this point is not vertical.
5(d) For time \( t\geq 0\), a particle is moving along another curve defined by the equation \(y^{3}+2xy =24\). At the instant the particle is at the point (4, 2), the y-coordinate of the particle’s position is decreasing at
arate of 2 units per second. At that instant, what is the rate of change of the x -coordinate of the particle’s position with respect to time?
\(3y^{2}\frac{dy}{dt}+2x\frac{dy}{dt}+2y\frac{dx}{dt}=0\)
\(\frac{dy}{dt}=-2\)
\(3(2)^{2}(-2)+2(4)(-2)+2(2)\frac{dx}{dt}=0\Rightarrow \frac{dx}{dt}=\frac{40}{4}=10\)
The rate of change with respect to time in the x-coordinate is 10 units per second.