Determine the equation of the tangent line to a parametrically defined curve.

Find all points on the curve defined by the equations [tex]x(t)=2 \sin (8 t)[/tex] and [tex]y(t)=-9 \cos (8 t)[/tex] with a slope of [tex]-\frac{9 \sqrt{3}}{2}[/tex] when [tex]0 \leq t\ \textless \ \frac{\pi}{8}[/tex].

Find the derivatives d t d x ​ and d t d y ​ of the parametric equations.
Calculate the slope d x d y ​ = d x / d t d y / d t ​ = 2 9 ​ tan ( 8 t ) .
Set the slope equal to − 2 9 3 ​ ​ and solve for t , obtaining t = 12 π ​ .
Substitute t = 12 π ​ into the parametric equations to find the point ( 3 ​ , 2 9 ​ ) .

The point is ( 3 ​ , 2 9 ​ ) ​ .
Explanation

Problem Setup We are given the parametric equations x ( t ) = 2 sin ( 8 t ) and y ( t ) = − 9 cos ( 8 t ) , and we want to find the points on the curve where the slope of the tangent line is − 2 9 3 ​ ​ for 0 ≤ t < 8 π ​ .

Compute Derivatives First, we need to find d t d x ​ and d t d y ​ .
d t d x ​ = d t d ​ ( 2 sin ( 8 t )) = 2 ⋅ 8 cos ( 8 t ) = 16 cos ( 8 t ) d t d y ​ = d t d ​ ( − 9 cos ( 8 t )) = − 9 ⋅ ( − 8 ) sin ( 8 t ) = 72 sin ( 8 t )

Calculate dy/dx Next, we find d x d y ​ using the chain rule: d x d y ​ = d x / d t d y / d t ​ = 16 cos ( 8 t ) 72 sin ( 8 t ) ​ = 2 9 ​ tan ( 8 t )

Set dy/dx to Given Slope We are given that d x d y ​ = − 2 9 3 ​ ​ . So we set the expression for d x d y ​ equal to this value and solve for t :
2 9 ​ tan ( 8 t ) = − 2 9 3 ​ ​ tan ( 8 t ) = − 3 ​

Solve for t The general solution for tan ( 8 t ) = − 3 ​ is: 8 t = arctan ( − 3 ​ ) + nπ = − 3 π ​ + nπ t = − 24 π ​ + 8 nπ ​ where n is an integer.

Find Valid t Values We want to find the values of t in the interval 0 ≤ t < 8 π ​ .
For n = 1 , t = − 24 π ​ + 8 π ​ = 24 − π + 3 π ​ = 24 2 π ​ = 12 π ​ .
For n = 2 , t = − 24 π ​ + 8 2 π ​ = 24 − π + 6 π ​ = 24 5 π ​ .
For n = 3 , t = − 24 π ​ + 8 3 π ​ = 24 − π + 9 π ​ = 24 8 π ​ = 3 π ​ , which is greater than 8 π ​ , so it's not in the interval. Thus, the only value of t in the interval 0 ≤ t < 8 π ​ is t = 12 π ​ .

Find the Point (x, y) Now we substitute t = 12 π ​ into the parametric equations to find the corresponding point ( x , y ) :
x = 2 sin ( 8 t ) = 2 sin ( 8 ⋅ 12 π ​ ) = 2 sin ( 3 2 π ​ ) = 2 ⋅ 2 3 ​ ​ = 3 ​ y = − 9 cos ( 8 t ) = − 9 cos ( 8 ⋅ 12 π ​ ) = − 9 cos ( 3 2 π ​ ) = − 9 ⋅ ( − 2 1 ​ ) = 2 9 ​ So the point is ( 3 ​ , 2 9 ​ ) .

Final Answer Therefore, the point on the curve with a slope of − 2 9 3 ​ ​ is ( 3 ​ , 2 9 ​ ) .


Examples
Parametric equations are incredibly useful in physics, especially when analyzing projectile motion. Imagine you're tracking a ball thrown through the air. The horizontal and vertical positions of the ball can be described by parametric equations where time is the parameter. By finding the slope of the trajectory at a specific time, you can determine the ball's direction of motion, which is crucial for predicting where it will land or how it will interact with other objects in its path. This blend of math and physics provides a powerful tool for understanding and predicting real-world phenomena.

Answered by GinnyAnswer | 2025-07-03

The slope of the tangent line to the curve defined by x ( t ) = 2 sin ( 8 t ) and y ( t ) = − 9 cos ( 8 t ) at t = 12 π ​ yields the point ( 3 ​ , 2 9 ​ ) , and the equation of the tangent line can be derived from this using the point-slope form. The tangent line's slope matches the desired slope of − 2 9 3 ​ ​ .
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Answered by Anonymous | 2025-07-04