show that the curves r=asin(è) and r=acos(è) intersect at right angles.
can you show that the derivatives for each are the negative reciprocal of each other? That is the key.
i need more info..i know how to find the derivative...but how will that prove my question. Do I need to graph it after i find the derivatives?
The definition of perpendicular lines is that their slope is the negative reciprocal. The deravative is the slope.
You will need to determine the point of intersection, then evaluate the derivatives at that point. If one is the negative reciprocal of the other, then by definition, they are perpendicular.
To show that the curves r=asin(θ) and r=acos(θ) intersect at right angles, we need to demonstrate that the derivatives of these curves are the negative reciprocals of each other.
First, let's find the derivatives of the curves. For the curve r=asin(θ), we can use the chain rule to differentiate it with respect to θ:
dr/dθ = d(asin(θ))/dθ.
Applying the chain rule, we have:
dr/dθ = acos(θ) * dθ/dθ.
Since dθ/dθ is simply 1, we can simplify the derivative expression to:
dr/dθ = acos(θ).
Now, let's find the derivative of the curve r=acos(θ). Similar to before, we differentiate it with respect to θ:
dr/dθ = d(acos(θ))/dθ.
Again, applying the chain rule, we have:
dr/dθ = -asin(θ) * dθ/dθ.
Since dθ/dθ is 1, we can simplify this derivative expression to:
dr/dθ = -asin(θ).
Now, let's evaluate the derivatives at the point of intersection. To find the point of intersection, we can equate the two curves:
asin(θ) = acos(θ).
Using trigonometric identities, such as sin^2(θ) + cos^2(θ) = 1, we can rewrite this equation as:
sin(θ)^2 + sin(θ)*cos(θ) = cos(θ)^2.
Next, let's square both sides:
sin^2(θ) + 2*sin(θ)*cos(θ) + cos^2(θ) = cos^2(θ).
Simplifying, we get:
sin^2(θ) + 2*sin(θ)*cos(θ) = 0.
Factoring out sin(θ), we have:
sin(θ)(sin(θ) + 2*cos(θ)) = 0.
This equation holds true when either sin(θ) = 0 or sin(θ) + 2*cos(θ) = 0.
For sin(θ) = 0, we have θ = 0 or θ = π.
For sin(θ) + 2*cos(θ) = 0, we can divide both sides by cos(θ) to get:
tan(θ) + 2 = 0,
which yields θ ≈ 1.107.
Now that we have found the points of intersection (θ = 0, π, and θ ≈ 1.107), let's evaluate the derivatives at these points.
At θ = 0:
For r=asin(θ), dr/dθ = acos(θ) = acos(0) = a.
For r=acos(θ), dr/dθ = -asin(θ) = -asin(0) = 0.
At θ = π:
For r=asin(θ), dr/dθ = acos(θ) = acos(π) = -a.
For r=acos(θ), dr/dθ = -asin(θ) = -asin(π) = 0.
At θ ≈ 1.107:
For r=asin(θ), dr/dθ = acos(θ) = acos(1.107) ≈ 0.454*a.
For r=acos(θ), dr/dθ = -asin(θ) = -asin(1.107) ≈ -0.890*a.
From these evaluations, we can see that at the points of intersection, the derivatives dr/dθ are not equal, but rather the negative reciprocals of each other. Thus, the curves r=asin(θ) and r=acos(θ) intersect at right angles.