Friday

August 1, 2014

August 1, 2014

Posted by **krishna** on Friday, October 14, 2011 at 1:54pm.

all i know is that its related to finding a derivative and i m basically completely lost....can please someone provide me some help?????

- calculus--please helpp -
**Steve**, Friday, October 14, 2011 at 2:12pmthe derivative gives the slope of the tangent line to the curve, at any point (x,y)

The line 3x-4y=0 has slope= 3/4

At any point on the curve (x,x^3 + k) the slope of the tangent line is 3x^2

So, you want to find x where 3x^2 = 3/4

x = 1/2

so, now you have y = 1/2^3 + k = 1/8 + k

What point on the line has y = 1/8 + k?

3* 1/2 - 4(1/8 + k) = 0

3/2 - 1/2 - 4k = 0

4k = 1**k**= 1/4

So, y = x^3 + 1/4

At x = 1/2, y = 1/8 + 1/4 = 3/8

Note that at the point (1/2 , 3/8) 3x-4y = 0

- calculus--please helpp -
**Steve**, Friday, October 14, 2011 at 3:06pmNote on missing answer.

x can also be -1/2 if 3x^2 = 3/4

In that case, k = -1/4

and the point of tangency is (-1/2 , -3/8)

Note also, that if the line had been 3x-4y=1, the same line would have touched the curve at both points, with k=0.

- calculus--please helpp -
**krishna**, Saturday, October 15, 2011 at 10:09amthanks so much!

**Related Questions**

Calculus - Tangent Line - Hi, im having problems with the following problem. The...

math - f(x) = 4e^-1x +13x +5 at (0,9) 1. All steps for finding the first ...

Calculus - If F(x)=x^3−7x+5, use the limit definition of the derivative to...

Calculus - Please help this is due tomorrow and I dont know how to Ive missed a ...

Calculus - If the line 3x-4y=0 is tangent in the first quadrant to the curve y=x...

Calculus - Consider the curve given by y^2 = 2+xy (a) show that dy/dx= y/(2y-x...

Calculus [finding slope of tangent line] - Find the slope of the tangent line to...

Calculus - If a tangent line is drawn to the parabola y = 3 - x^2 at any point ...

Calculus - Damon - Find the line which passes through the point (0, 1/4) and is ...

calculus - If a tangent line is drawn to the parabola y = 3 - x^2 at any point ...