What are the critical points for the plane curve defined by the equations x(t)=−sin(3t),y(t)=5t, and 0 \leq t<\pi ? Write your answer as a list of values of t, separated by commas. For example, if you found t=1 or t=2, you would enter 1,2 .
Q. What are the critical points for the plane curve defined by the equations x(t)=−sin(3t),y(t)=5t, and 0≤t<π ? Write your answer as a list of values of t, separated by commas. For example, if you found t=1 or t=2, you would enter 1,2 .
Find x′(t): To find the critical points of the plane curve, we need to find the values of t where the derivatives of x(t) and y(t) are both zero or do not exist. Let's start by finding the derivative of x(t) with respect to t. x(t)=−sin(3t) dtdx=−cos(3t)⋅3
Find y′(t): Now, let's find the derivative of y(t) with respect to t.y(t)=5tdtdy=5
No critical points for y(t): Since dtdy=5 is never zero and does not depend on t, there are no critical points for y(t) based on its derivative. Therefore, we only need to consider the critical points for x(t) based on dtdx.
Set x′(t) to zero: We set the derivative of x(t) equal to 0 to find the critical points for x(t).0=−cos(3t)×3This implies that cos(3t) must be equal to 0.
Solve for cos(3t)=0: The cosine function is zero at odd multiples of 2π. Therefore, we need to find the values of t such that 3t is an odd multiple of 2π within the given interval 0 \leq t < \pi.
Find values of t: Let's solve for t: 3t=2(2n+1)π, where n is an integer. t=6(2n+1)π
Solve for t: We need to find the values of n such that t is in the interval 0 \leq t < \pi. Let's start with n=0 and increase n to find all possible values of t.For n=0: t=6πFor n=1: n0For n1: n2
Find possible values of t: If we increase n to 3, we get t=67π, which is outside the interval 0 \leq t < \pi. Therefore, we only consider n=0, n=1, and n=2.
Consider n=0,1,2: The critical points for the plane curve defined by the equations x(t)=−sin(3t) and y(t)=5t are at t=6π, t=2π, and t=65π.
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