How do you find the equation of the tangent and normal line to the curve #y=1+x^(2/3)# at (0,1)?

Answer 1

Any line through the point #(0,1)# will have the form:
#y = mx+1#
Substitute, the slope of the tangent line, #m_t = y'(0)# or the slope of the normal line, #m_n = -1/(y'(0))#

Usually we would use the form given in the above answer:

#y = mx+1#
Compute #y'(x)#:
#y'(x)= 2/3x^(-1/3)#

Evaluate at the x coordinate:

#m_t=y'(0) = 2/3(0)^(-1/3)#
But we have an exception to the procedure. Please observe that that the evaluation causes a division by 0, which implies that the tangent is the vertical passing the point #(0,1)#
Therefore, the tangent is the line, #x = 0#

The procedure for the normal line is similar.

#y = mx+1#
Compute #y'(x)#:
#y'(x)= 2/3x^(-1/3)#

Evaluate at the x coordinate:

#m_n=-1/(y'(0)) = -1/((2/3)0^(-1/3)) = 0#

Substitute 0 for the slope:

#y = 1#
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Answer 2

To find the equation of the tangent and normal lines to the curve (y = 1 + x^{2/3}) at the point (0,1), you can follow these steps:

  1. Find the derivative of the function (y = 1 + x^{2/3}) to get the slope of the tangent line at any point.
  2. Evaluate the derivative at the given point to find the slope of the tangent line at that point.
  3. Use the point-slope form of the equation of a line to write the equation of the tangent line.
  4. Use the negative reciprocal of the slope of the tangent line to find the slope of the normal line.
  5. Use the point-slope form of the equation of a line to write the equation of the normal line.

Let's go through the steps:

  1. The derivative of (y = 1 + x^{2/3}) is (dy/dx = (2/3)x^{-1/3}).
  2. Evaluate the derivative at (0,1): [dy/dx = (2/3)(0)^{-1/3} = \text{undefined}] Therefore, we need to use the limit definition to find the slope at (0,1): [\lim_{h \to 0} \frac{(1+(0+h)^{2/3}) - 1}{h} = \lim_{h \to 0} \frac{h^{2/3}}{h} = \lim_{h \to 0} h^{-1/3} = \infty] So, the slope of the tangent line at (0,1) is infinity.
  3. The equation of the tangent line at (0,1) is (x = 0).
  4. The slope of the normal line is the negative reciprocal of the slope of the tangent line, which is 0.
  5. Since the slope is 0, the equation of the normal line is (y = 1).

Therefore, the equations of the tangent and normal lines to the curve (y = 1 + x^{2/3}) at the point (0,1) are (x = 0) and (y = 1), respectively.

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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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