The equation of the curve is y = 15x^3/2 - x^5/2. The point P (1,14) lies on the curve. Show that the equation of the tangent to the curve at P is y = 20x - 6.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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Given the curve :

`y= 15x^(3/2) - x^(5/2)`

We need to find the equation of the tangent line at the point P(1,14).

==> The equation of the line is:

y-y1 = m (x-x1)

==> y- 14 = m (x-1)

==> y= m(x-1) + 14............(1)

Now we need to find the slope m.

We know that the slope of the tangent line is the derivative at the tangent point which is x= 1.

==> m = y'(1)

`==gt y' = 15*3/2 x^(1/2) - 5/2 x^(3/2) `

`==gt y'= 45/2 x^(1/2) - (5/2) x^(3/2) `

`==gt y'(1)= 45/2 - 5/2 = 40/2 = 20`

==> Then, the slope m = 20.

Now we will substitute into (1).

==> y= m(x-1) + 14

==> y= 20(x-1) + 14

==> y= 20x - 20 + 14

==> y= 20x - 6

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