Applications Of Differential Calculus — Quiz 1
Practice Applications Of Differential Calculus — Quiz 1 on The School of Mathematics (Applications Of Differential Calculus) with 10 scored questions. This set covers problems such as: “An equation of the tangent line to the hyperbola x^2-y^2=9 at the point (5,4) on the curve is:”; “The slope of the curve y^3+xy^2=10 at the point where y=1 is:”; “Differentiate implicitly. y^2-2xy+4=0 2y\frac{dy}{dx}-2\left(x\frac{dy}{dx}+y\right)=0 Distribute. 2y\frac{…”. Work carefully, check explanations for misses, then retake to track improvement before test day.
Questions in this quiz (10)
An equation of the tangent line to the hyperbola$$x^2-y^2=9$$at the point $$(5,4)$$ on the curve is:
- $$y=\frac{5}{4}x-\frac{9}{4}$$$$y=\frac{5}{4}x-\frac{9}{2}$$
- $$y=\frac{5}{4}x-\frac{9}{4}$$
- $$y=\frac{4}{5}x$$
- $$y=\frac{5}{4}x$$
The slope of the curve$$y^3+xy^2=10$$at the point where $$y=1$$ is:
- $$-1$$
- $$-\frac{1}{2}$$
- $$-\frac{1}{21}$$
- $$1$$
Differentiate implicitly.$$y^2-2xy+4=0$$$$2y\frac{dy}{dx}-2\left(x\frac{dy}{dx}+y\right)=0$$Distribute.$$2y\frac{dy}{dx}-2x\frac{dy}{dx}-2y=0$$Group the terms with $$\frac{dy}{dx}$$.$$\left(2y-2x\right)\frac{dy}{dx}=2y$$Solve for $$\frac{dy}{dx}$$.$$\frac{dy}{dx}=\frac{2y}{2y-2x}$$$$\frac{dy}{dx}=\frac{y}{y-x}$$A tangent is vertical when the denominator is $$0$$ and the numerator is not $$0$$.So set$$y-x=0$$$$x=y$$Now substitute $$x=y$$ into the original equation.$$y^2-2y^2+4=0$$$$-y^2+4=0$$$$y^2=4$$$$y=\pm2$$Therefore, the tangent is vertical when$$y=\pm2$$Final answer: $$y=\pm2$$
- $$y=0$$
- $$y=\pm2$$
- $$y=1$$
- $$y=\pm\sqrt{2}$$
The slope of the curve$$y^2-2xy+x=3$$at the point $$(1,2)$$ is:
- $$-2$$
- $$-1$$
- $$1$$
- $$\frac{3}{2}$$
The volume of a sphere is given by$$V=\frac{4}{3}\pi r^3$$Use a tangent line linear approximation to approximate the increase in volume, in cubic inches, when the radius of a sphere is increased from $$4$$ inches to $$4.05$$ inches.
- $$0.2\pi$$
- $$0.8\pi$$
- $$1.6\pi$$
- $$3.2\pi$$
An equation of the tangent line to the curve$$y=x\cos x$$ at the point $$\left(\frac{\pi}{2},0\right)$$ is:
- $$y=-x+\frac{\pi}{2}$$
- $$y=x-\frac{\pi}{2}$$
- $$y=\frac{\pi}{2}-x$$
- $$y=-\frac{\pi}{2}x+\frac{\pi^2}{4}$$
When $$x=2$$, the equation$$x^2+y^3=12$$has the solution $$y=2$$. Using a tangent line to the graph of the curve, approximate $$y$$ when $$x=2.03$$.
- $$1.94$$
- $$1.99$$
- $$2.02$$
- $$2.06$$
The tangent to the curve$$y=xe^{-2x}$$ is horizontal when $$x$$ is equal to:
- $$0$$
- $$\frac{1}{2}$$
- $$1$$
- $$2$$
The minimum value of the slope of the curve$$y=x^4+2x^2-3x$$ is:
- $$-3$$
- $$-1$$
- $$0$$
- $$1$$
The total number of local maximum and minimum points of the function whose derivative, for all $$x$$, is given by$$f'(x)=(x-2)^2(x+1)(x-4)^3$$ is:
- $$0$$
- $$1$$
- $$2$$
- $$3$$