How do you find the equation of the tangent line for the curve given by x = 2t and #y = t^2 + 5# at the point where t = 1?

Answer 1

#x-y+4=0#.

#t=1, x=2t, y=t^2+5 rArr x=2, y=6,# so the pt. of contact is #(x,y)=(2,6)#.
#x=2t, y=t^2+5 rArr dx/dt=2, dy/dt=2t#
By the Rule of Parametric Diff., #dy/dx=(dy/dt)/(dx/dt)=2t/2=t#.
#:." The Slope of tgt. at "(t=1) = [dy/dx]_(t=1)=1#
To sum up, the tgt. passes thro. the pt.#(2,6)#, and, has slope=1#.
#:." The eqn. of tgt. is ": y-6=1(x-2), i.e., x-y=4+0#.
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Answer 2

To find the equation of the tangent line at ( t = 1 ), we first find the corresponding values of ( x ) and ( y ) at ( t = 1 ), which are ( x = 2(1) = 2 ) and ( y = (1)^2 + 5 = 6 ).

Next, we find the derivative of ( y ) with respect to ( t ), which is ( \frac{dy}{dt} = 2t ).

At ( t = 1 ), the slope of the tangent line is ( m = \frac{dy}{dt}\bigg|_{t=1} = 2(1) = 2 ).

Using the point-slope form of a line equation ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is a point on the line, we substitute ( x_1 = 2 ), ( y_1 = 6 ), and ( m = 2 ) to get the equation of the tangent line:

( y - 6 = 2(x - 2) )

Simplify to get the equation of the tangent line:

( y = 2x + 2 )

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Answer 3

To find the equation of the tangent line for the curve given by (x = 2t) and (y = t^2 + 5) at the point where (t = 1), we first need to find the coordinates of the point where (t = 1).

Substitute (t = 1) into the equations (x = 2t) and (y = t^2 + 5) to find the coordinates:

(x = 2(1) = 2) and (y = (1)^2 + 5 = 6)

So, at (t = 1), the coordinates are ((2, 6)).

Next, we find the slope of the tangent line at this point. The slope of the tangent line to a curve at a given point is given by the derivative of the curve's equation with respect to (t), evaluated at that point.

The derivative of (y = t^2 + 5) with respect to (t) is (dy/dt = 2t).

At (t = 1), the slope of the tangent line is (2(1) = 2).

So, the slope of the tangent line at the point ((2, 6)) is (m = 2).

Now that we have the slope of the tangent line and a point on the line, we can use the point-slope form of the equation of a line to find the equation of the tangent line:

(y - y_1 = m(x - x_1))

Substituting (x_1 = 2), (y_1 = 6), and (m = 2) into the equation:

(y - 6 = 2(x - 2))

Expanding and simplifying:

(y - 6 = 2x - 4)

(y = 2x + 2)

Therefore, the equation of the tangent line for the curve (x = 2t) and (y = t^2 + 5) at the point where (t = 1) is (y = 2x + 2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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