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Chapter 8: Problem 39

Determine the \(t\) intervals on which the curve is concave downward or concave upward. $$ x=\sin t, \quad y=\cos t, \quad 0

Short Answer

Expert verified
The curve is concave downward when \( t \) ranges from \( 0 \) to \( \pi \).

Step by step solution

01

Calculate the derivative of \( x \) and \( y \)

First, calculate the derivative of \( x \) and \( y \) w.r.t. \( t \): \( dx/dt = \cos(t) \) and \( dy/dt = -\sin(t) \).
02

Derive the second derivatives

Next, calculate the second derivative of \( x \) and \( y \): \( d虏x/dt虏 = -\sin(t) \) and \( d虏y/dt虏 = -\cos(t) \).
03

Calculate the second order derivative of \( y \) w.r.t. \( x \)

Now, calculate the derivative of \( y \) w.r.t. \( x \) through the use of chain rule which gives \( dy/dx = (dy/dt)/(dx/dt) \). Therefore, \( dy/dx = -\sin(t)/\cos(t) = -\tan(t) \). Then, calculate the second order derivative of \( y \) w.r.t. \( x \), \( d虏y/dx虏 = d/dx(dy/dx) \) by applying the chain rule again. This gives \( d虏y/dx虏 = -sec虏(t)\). First derivative of \( y \) w.r.t. \( x \), \( dy/dx \), is negative for \( 0<t<\pi/2 \) and positive for \( \pi/2<t<\pi \). For the second derivative, since \( -sec虏(t) \) is always negative, we can say that the curve is concave downward for all \( t \) in the given interval \( 0<t<\pi \).
04

Interpret the result

To summarize: The curve is concave downward for all \( t \) in the given interval \( 0<t<\pi \).

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