Find an equation of the tangent(s) to the curve at a given point. Then graph the curve and the tangent(s).
x = 6sin(t), y = t^2 + t; (0,0)
How do I do this without the value of t? I don't see a way to eliminate the parameter.
Oh also if you can, how do you find d2y/d2x for parametrics?
The nice thing about parametric equations is that you don't have to eliminate the parameter.
dy/dx = dy/dt / dx/dt
dy/dx = (2t+1)/6cos(t)
This is zero at t = -1/2 and has a vertical asymptote whenever t = an odd multiple of pi/2.
The graph is a kind of double wave sine surve. It wanders back and forth in the x direction, but the escillations get longer in the y direction, since y uses t^2, instead of just t.
Mosey on over to wolfram dot com to see the big picture. Type in a command like
graph x=6*sin(t), y=t^2 + t, t=-5..5
try
graph x=6*sin(t), y=t^2 + t, t=-2..1.5 to see the horizontal tangent at t=-1/2
To find the equation of the tangent(s) to the curve without eliminating the parameter, you can proceed as follows:
1. Find the derivative of y with respect to x using the chain rule:
dy/dx = (dy/dt) / (dx/dt)
2. Substituting the given parametric equations:
x = 6sin(t), y = t^2 + t
we have:
dx/dt = 6cos(t)
dy/dt = 2t + 1
3. Calculate dy/dx:
dy/dx = (dy/dt) / (dx/dt) = (2t + 1) / (6cos(t))
4. To find the slope of the tangent at the point (0, 0) where t = 0, substitute t = 0 into dy/dx:
dy/dx = (2(0) + 1) / (6cos(0)) = 1 / 6
5. The slope of the tangent is 1/6, which means the equation of the tangent will have the form:
y = mx + b, where m is the slope of the tangent.
6. Plug in the point (0, 0) into the equation to find the y-intercept, b:
0 = (1/6)(0) + b
b = 0
7. Therefore, the equation of the tangent at the point (0, 0) is:
y = (1/6)x
8. To graph the curve x = 6sin(t), y = t^2 + t and the tangent y = (1/6)x, plot the points from t = 0 to t = 2π on the curve, and then draw a straight line through the point (0, 0) with a slope of 1/6.
To find the equation of the tangent line(s) to the curve without knowing the value of t, we can still proceed by finding the derivative of y with respect to x, which is dy/dx, and then evaluate it at the given point (0,0). Let's go step by step:
1. Start by differentiating both x and y with respect to t:
dx/dt = 6cos(t)
dy/dt = 2t + 1
2. Next, determine dy/dx by dividing dy/dt by dx/dt:
dy/dx = (dy/dt) / (dx/dt)
= (2t + 1) / (6cos(t))
3. Now, substitute t = 0 into dy/dx to find the slope of the tangent line at the point (0,0):
dy/dx = (2(0) + 1) / (6cos(0))
= 1 / 6
So, the slope of the tangent line at (0,0) is 1/6.
4. Using the point-slope form of the equation, we can define the equation of the tangent line as:
y - y1 = m(x - x1)
where (x1, y1) is the given point (0,0), and m is the slope we found:
y - 0 = (1/6)(x - 0)
y = (1/6)x
Now that we have the equation of the tangent line, we can proceed to graphing. Since the given curve is defined parametrically, we'll need to plot points based on different values of t to get a sense of the curve shape. Let's use a graphing tool to plot the curve and tangent line.