Find the positive value of the parameter t corresponding to a point on the curve parametrized by
{x= t^2 +3 ; y=t^2+t
for which the tangent line passes through the origin.
I tried with 2t+3 and 2t+1
but it isn't successful
The slope at any point (x,y) is
dy/dx = (dy/dt)/(dx/dt) = (2t+1)/(2t)
We want to find t such that the tangent line is y=kx (that is, it passes through the origin). At that point
y = kx, so
k = (t^2+t)/(t^2+3)
That means that we need the slope = k, or
(2t+1)/(2t) = (t^2+t)/(t^2+3)
solve that and you get
t = 3+2√3
Check:
x = t^2+3 = 24+12√3
y = t^2+t = 24+14√3
dy/dx = 1/2 + 1/√3
y/x = 1/2 + 1/√3
so, y = kx, where k=1/2 + 1/√3
Well, it seems like you're having some trouble finding the right tangent line that passes through the origin. Let me help you out, but as a clown bot, I'll use some humor to make math a little more fun!
To find the positive value of the parameter t corresponding to a point on the curve where the tangent line passes through the origin, we need to find the point on the curve first.
Given that the curve is parametrized by {x = t^2 + 3; y = t^2 + t}, we can find the slope of the tangent line by taking the derivative of y with respect to x, which is dy/dx.
dy/dx = (dy/dt) / (dx/dt)
Taking the derivatives of x and y with respect to t, we get:
dx/dt = 2t
dy/dt = 2t + 1
Now, let's find the derivative dy/dx using the chain rule:
dy/dx = (dy/dt) / (dx/dt)
dy/dx = (2t + 1) / 2t
To find the point on the curve where the tangent line passes through the origin, the slope of the tangent line should be equal to the slope of a line passing through the origin, which is 0. Let's set dy/dx equal to 0 and solve for t:
(2t + 1) / 2t = 0
Uh-oh! We've got a problem here. Since the numerator (2t + 1) can never be equal to zero (because it is always positive), there is no positive value of t that makes the tangent line pass through the origin for this particular curve.
But don't worry, math isn't always about success. Sometimes, it's about exploring and learning from different attempts. So keep experimenting and don't be afraid to embrace a little failure along the way!
To find the positive value of the parameter t corresponding to a point on the curve for which the tangent line passes through the origin, we need to find the point of intersection between the curve and the line passing through the origin.
1. Equation of the tangent line passing through the origin (0,0): y = mx, where m is the slope of the tangent line.
2. To find the slope, we differentiate the equations x = t^2 + 3 and y = t^2 + t with respect to t.
dx/dt = 2t and dy/dt = 2t + 1.
3. The slope of the tangent line is given by the ratio dy/dt / dx/dt:
m = (2t + 1) / (2t).
4. Substitute the equation of the tangent line (y = mx) into the parametric equations:
t^2 + t = (2t + 1) / (2t) * (t^2 + 3).
5. Solve the above equation for t. Multiply both sides by 2t to eliminate the denominator:
2t^3 + 2t^2 = (2t + 1) * (t^2 + 3).
6. Expand and simplify:
2t^3 + 2t^2 = 2t^3 + 6t^2 + t^2 + 3.
7. Combine like terms:
2t^3 + 2t^2 = 3t^2 + t^2 + 3.
8. Rearrange the equation:
2t^3 + 2t^2 - 3t^2 - t^2 - 3 = 0.
9. Combine like terms:
2t^3 - 2t^2 - t^2 - 3 = 0.
10. Simplify the equation:
2t^3 - 3t^2 - 3 = 0.
11. Now, we can solve this equation for t. You can use numerical methods like Newton's method, the bisection method, or graph the equation and find the t-intercept to approximate the value of t.
To find the positive value of the parameter t for which the tangent line passes through the origin, we need to find the equation of the tangent line and then solve for t.
The equation of the tangent line can be found using the derivative of the parametric equations. We differentiate both x and y with respect to t to find dx/dt and dy/dt:
dx/dt = 2t
dy/dt = 2t + 1
The slope of the tangent line is given by dy/dx. We can calculate this by dividing dy/dt by dx/dt:
dy/dx = (dy/dt) / (dx/dt) = (2t + 1) / 2t
Now we have the slope of the tangent line, and we know that the line passes through the origin, which means the point (0, 0) lies on the line. So, we can substitute x = 0 and y = 0 into the equation of the line to find the value of t:
0 = (2t + 1) / 2t * 0
This expression is undefined since we can't divide by zero. Therefore, the tangent line does not pass through the origin.
It seems that your attempt to find the equation of the tangent line using 2t + 3 and 2t + 1 is not valid. Instead, we need to find dy/dx as explained above.
To summarize, there is no positive value of t for which the tangent line of the given curve passes through the origin.