Find the normal to the curve x=tt+1x=tt+1 and y=tt−1y=tt−1 at the point T where t = 2.
I tried converting this into one equation (y=x(t+1)t−1y=x(t+1)t−1) but don't see how to get the answer which says:
dydx=−(t+1)2(t−1)2dydx=−(t+1)2(t−1)2
and that T=(23,2)T=(23,2), 3x−27y+52=03x−27y+52=0
I'd appreciate any pointers.
all that duplicated text is impossible to parse. Try typing it in, rather than copy/pasting.
Use t^2 for t squared
Find the normal to the curve
X=t÷[t+1] and Y=t÷[t-1]at point T where t=2.
I tried converting this into one equation Y=x(t+1)÷(t-1) but don't see how to get the answer which says:
dy÷dx=-(t+1)^2÷(t-1)^2
and that T=(2÷3,2), 3x−27y+52=0
Is this ok
Those problems i got are for an pdf i got from my sir
All the steps in your work are correct.
To find the normal to the curve at a given point, we need to find the derivative of the curve with respect to the parameter t and evaluate it at the point of interest.
Given the parametric equations x = t(t+1) and y = t(t-1), we want to find the normal at the point where t = 2.
Step 1: Find the derivatives of x and y with respect to t.
- For x: differentiate x = t(t+1) with respect to t:
dx/dt = d/dt(t² + t) = 2t + 1.
- For y: differentiate y = t(t-1) with respect to t:
dy/dt = d/dt(t² - t) = 2t - 1.
Step 2: Evaluate the derivatives at t = 2.
- For x: dx/dt = 2(2) + 1 = 5.
- For y: dy/dt = 2(2) - 1 = 3.
Step 3: Find dy/dx by dividing dy/dt by dx/dt.
- dy/dx = (dy/dt)/(dx/dt) = 3/5 = 0.6
So, dy/dx = 0.6 at the point T (when t = 2).
Step 4: Find the equation of the normal line.
A normal line to a curve at a given point has a slope that is the negative reciprocal of the slope of the curve at that point.
Since the slope of the curve at T is 0.6, the slope of the normal line will be -1/0.6 = -5/3.
We can use the point-slope form of a line (y - y₁) = m(x - x₁), where (x₁, y₁) is the point of interest and m is the slope of the line.
Substituting the given point T (where t = 2) into the equation, we have:
- x₁ = 2(2 + 1) = 6
- y₁ = 2(2 - 1) = 2
Therefore, the equation of the normal line is:
- (y - 2) = (-5/3)(x - 6)
- Rearranging, we get: y = (-5/3)x + 20/3
So, the normal to the curve at the point T (where t = 2) is given by the equation y = (-5/3)x + 20/3.
To check if this is the correct answer, we can convert the parametric equations x = t(t+1) and y = t(t-1) into a single equation:
- Substituting t = 2, we get x = 2(2+1) = 6 and y = 2(2-1) = 2. So the point T is (6, 2).
The equation of the normal line is: y = (-5/3)x + 20/3.
- 3x + 5y - 20 = 0 (by multiplying through by 3 to eliminate fractions)
Comparing this equation with the given answer 3x - 27y + 52 = 0, it seems there might be a mistake in the answer provided. Please double-check the given answer or refer to alternative resources for confirmation.