Find an equation of the tangent to the curve at the given point by two methods:?

Question

(a) without eliminating the parameter and (b) by first eliminating the parameter.

x= tanT y=secT; (1, radical 2)

Please help, I don't understand this problem!

Answer 1

Point-Slope formula y - y0 = m (x-x0)

Parametric form dy/dx = dy/dt / dx/dt
dy/dt = secTtanT
dx/dt = sec^2(T)

dy/dx = tanT / secT
we know tanT = 1 and secT=radical2 (from the point given)
dy/dx= slope = 1 /radical2
y - y0 = m (x - x0)
y - radical2 = 1/radical2 * (x-1)

Eliminate parameter:
we know tan^2(T) + 1 = sec^2(T) by Pyth ID
x^2 + 1 = y^2
now find slope with implicit differentiation
2x = 2y y'
y' = x/y
y' = 1 / radical2
(same slope as above)

Answer 2

The parameter we are talking about is the T. To find the tangent line of a normal function, you would take the derivative, right? So, if we take the derivative of x and y with respect to T, that is dx/dT and dy/dT, we get dx/dT=sec^2(T) and dy/dT=secT*tanT. But the tangent line doesn't use dx/dT or dy/dT, it uses dy/dx, but (dy/dT)/(dx/dT)=dy/dx (the dT's cancel eachother).

So secT*tanT/sec^2(T)=tanT/secT. Since tan=sin/cos and sec=1/cos, this reduces to simply sinT. Now plug in T and you'll have dy/dx. To find T, plug in the point into the original equation. You'll find T=pi/4. Now that you have the slope and a point, you can find the tangent line.


Now, that was without eliminating the parameter. Eliminating the parameter means getting rid of T before doing anything. You can do this by squaring x and tanT to get x^2=tan^2(T) Using trig identites, x^2=sec^2(T)-1, and solve for secant and get sec(T)=sqrt(x^2+1)=y. Now with the equation y=sqrt(x^2+1), you can find the tangent line easily.

Answer 3

the factor (a million,-a million) would not lie on the curve y=x^2+3x+4. assume the factor on the curve is (a,a^2+3a+4) then the gradient m= 2a+3 and the equation is y=(2a+3)x+c and sub x=a, y=a^2+3a+4 to grant a^2+3a+4=(2a+3)a+c giving c=4-a^2 so the equation of the tangent is y=(2a+3)x+4-a^2. Now make this bypass by using (a million, -a million) which provides 2 feasible values for a and subsequently the equations of the mandatory tangents. you will desire to get y=-x and y=11x-12

Answer 4

The slope is just (dy/dt)/(dx/dt) the dt's cancel and you end up with just dy/dx. Just use the point slope to come up with the line.