Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/221

 Given $$u = \sin \tfrac{x}{y}, x = e^t, y = t^2$$; find  $$\tfrac{du}{dt}.$$

Solution. $$\frac{\partial u}{\partial x} = \frac{1}{y} \cos \frac{x}{y}, \frac{\partial u}{\partial y} = -\frac{x}{y^2} \cos {\partial x}{\partial y}, \frac{dx}{dt} = e^t, \frac{dy}{dt} = 2t.$$

Substituting in (51), $$\frac{du}{dt} = (t - 2)\frac{e^t}{t^3} \cos \frac{e^t}{t^2}.$$ Ans.

Given $$u = e^{ax} (y - z), y = a \sin x, z = \cos x; \mbox{ find } \tfrac{du}{dx}.$$

Solution. $$\frac{\partial u}{\partial x} = ae^{ax} (y - z), \frac{\partial u}{\partial y} = e^{ax}, \frac{\partial u}{\partial z} = -e^{ax}; \frac{dy}{dx} = a \cos x, \frac{dz}{dx} = -\sin x.$$

Substituting in (54),

NOTE. In examples like the above, $$u$$ could, by substitution, be found explicitly in terms of the independent variable and then differentiated directly, but generally this process would be longer and in many cases could not be used at all.

Formulas (51) and (52) are very useful in all applications involving time rates of change of functions of two or more variables. The process is practically the same as that outlined in the rule given on §94 except that, instead of differentiating with respect to $$t$$ (Third Step), we find the partial derivatives and substitute in (51) or (52). Let us illustrate by an example.

The altitude of a circular cone is 100 inches, and decreasing at the rate of 10 inches per second; and the radius of the base is 50 inches, and increasing at the rate of 5 inches per second. At what rate is the volume changing?



Solution. Let $$x =$$radius of base, $$y$$ = altitude; then $$u = \tfrac{1}{3}\pi x^2 y$$ = volume,

Substitute in (51),

But $$x = 50$$, $$y = 100$$, $$\frac{dx}{dt}$$ = 5, $$\frac{dy}{dt}$$ = - 10.

Multiplying (51) and (52) through by dt, we get

and so on. Equations (55) and (56) define the quantity $$du$$, which is called a total differential of $$u$$ or a complete differential, and