My Math Forum Difference between partial and implicit differentiation

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 June 6th, 2017, 04:15 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 Thanks: 2 Difference between partial and implicit differentiation Hi guys the question is: Find the equations of the tangent and normal lines to the curve 3x^2-14xy+8y^2-25=0 at the point P = (−1, 1). would i use partial or implicit differentiation here? Thanks !
 June 6th, 2017, 04:46 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 Use implicit differentiation to determine $\dfrac{dy}{dx}$, then evaluate at (-1,1) to find the tangent slope ... opposite reciprocal of that value to get the normal slope. Equations of both lines may be found using the point-slope form.
 June 6th, 2017, 09:50 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 "Partial differentiation" is used when you have a function, say, f, of two independent variables, x and y. "Implicit differentiation" is used when you have a formula in two variable, such as x and y, and one variable, y, is a function of the other. Here, you are not given a function of x and y but, rather, an expression involving x and y, and you are thinking of y as a function of x. $3x^2-14xy+8y^2-25=0$ so, differentiating with respect to x, $6x- 14y- 14xy'+ 16yy'= 0$. Then $(16y- 14x)y'= 14y- 6x$ so $y'= \frac{14y- 6x}{16y- 14x}= \frac{7y- 3x}{8y- 7x}$. You can use "partial differentiation" to get that result by thinking of "f" as a function of x and y- $f(x,y)= 3x^2- 14xy+ 8y^2= 25$. If, also, you think of x and y as functions of some third variable, u, using the chain rule, $\frac{df}{du}= \frac{\partial f}{\partial x}\frac{dx}{du}+ \frac{\partial f}{\partial y}\frac{dy}{du}= (6x- 14y)\frac{dx}{du}+ (-14x+ 16y}\frac{dy}{du}$. Since f(x, y) was a constant, its derivative with respect to any variable is 0: $(6x- 14y)\frac{dx}{du}+ (-14x+ 16y}\frac{dy}{du}= 0$. Now, take u= x so that dx/du= dx/dx= 1 and dy/du= dy/dx. That previous equation becomes $(6x- 14y)+ (-14x+ 16y}\frac{dy}{dx}= 0$. From that $\frac{dy}{dx}= \frac{6x- 14y}{16y- 14x}= \frac{7- 3x}{8y- 7x}$ as before.

### find the slope of tangent and normal to the curve x^2 y^2=25 at point -3,4

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