Problem 62 - Calculus Early Transcendentals

Question

Consider the following parametric curves. a. Determine \(d y / d x\) in terms of \(t\) and evaluate it at the given value of \(t.\) b. Make a sketch of the curve showing the tangent line at the point corresponding to the given value of \(t.\) $$x=\cos t, y=8 \sin t ; t=\pi / 2$$

Step-by-Step Solution

Answer

Question: Determine the slope of the tangent line to the curve defined parametrically by \(x = \cos(t)\) and \(y = 8\sin(t)\) at the point corresponding to \(t = \pi/2\), and sketch the curve with the tangent line at this point. Answer: The slope of the tangent line to the curve is 0 at the point corresponding to \(t = \pi/2\). The tangent line is horizontal, passing through point P(0, 8). The curve resembles an ellipse and oscillates between the cosine and sine functions, with the sine function scaled by 8.

Step 1: Derivatives of x and y in terms of t

Let's start by finding the derivatives of x and y with respect to t. $$x = \cos(t) \Rightarrow \frac{d x}{d t} = -\sin(t)$$ $$y = 8\sin(t) \Rightarrow \frac{d y}{d t} = 8\cos(t)$$

Step 2: Computing \(\frac{d y}{d x}\) from the chain rule

Now, we use the chain rule to find the expression for \(\frac{dy}{dx}\). $$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{8\cos(t)}{-\sin(t)} = -8\frac{\cos(t)}{\sin(t)}$$

Step 3: Evaluating \(\frac{d y}{d x}\) at \(t = \frac{\pi}{2}\)

Now, let's evaluate the expression for \(\frac{dy}{dx}\) at \(t=\pi/2\): $$\frac{d y}{d x}\Bigg|_{t = \pi/2} = -8\frac{\cos(\pi/2)}{\sin(\pi/2)} = -8\frac{0}{1} = 0$$ So, \(\frac{dy}{dx} = 0\) when \(t = \pi/2\).

Step 4: Sketching the curve and tangent line

Now, let's sketch the curve and the tangent line at the point corresponding to \(t = \pi/2\). First, we need the coordinates of this point: $$x(\pi/2) = \cos(\pi/2) = 0$$ $$y(\pi/2) = 8 \sin(\pi/2) = 8$$ So, the point is P(0, 8). Since \(\frac{dy}{dx} = 0\) at this point, the tangent line is horizontal and passes through P(0, 8). To sketch the curve, note that x and y are simply trigonometric functions, with x being a cosine wave and y being a sine wave scaled by 8. The curve oscillates between the two functions, resembling an ellipse. Place the point P(0, 8) and draw a horizontal line tangent to the curve.