Applications Of Differential Calculus — Quiz 4
Practice Applications Of Differential Calculus — Quiz 4 on The School of Mathematics (Applications Of Differential Calculus) with 10 scored questions. This set covers problems such as: “If f(x)=cx^2+dx+e for the function shown in the graph, then:”; “Given f(x)=e^{kx} approximate f(h) when h is near zero, using a tangent line approximation.”; “Given f(x)=\log_{10}x and \log_{10}(101)\approx2.0043 which is closest to f'(100) ?”. Work carefully, check explanations for misses, then retake to track improvement before test day.
Questions in this quiz (10)
If $$f(x)=cx^2+dx+e$$ for the function shown in the graph, then:
- $$c, d, e$$ are all positive
- $$c$$ > $$0$$ , $$d$$ < $$0$$, $$e$$ > $$0$$
- $$c$$ < $$0$$, $$d$$ > $$0$$, $$e$$ > $$0$$
- $$c$$ < $$0$$, $$d$$ < $$0$$, $$e$$ > $$0$$
Given $$f(x)=e^{kx}$$ approximate $$f(h)$$ when $$h$$ is near zero, using a tangent line approximation.
- $$k$$
- $$1+k$$
- $$1+kh$$
- $$kh$$
Given $$f(x)=\log_{10}x$$ and $$\log_{10}(101)\approx2.0043$$ which is closest to $$f'(100)$$?
- $$0.0043$$
- $$0.0086$$
- $$0.01$$
- $$1.0043$$
The point on the curve $$y=\sqrt{3x+1}$$ at which the tangent is parallel to the line $$2x-y=5$$ is:
- $$(1,2)$$
- $$\left(-\frac{7}{48},\frac{3}{4}\right)$$
- $$(0,1)$$
- $$(2,\sqrt{7})$$
The table shows the velocity at time $$t$$ of an object moving along a line.Estimate the acceleration (in $$ft/sec^2$$) at $$t=5$$ $$sec$$.
- $$-4$$
- $$-2$$
- $$2$$
- $$4$$
An equation of the tangent to the curve$$x^2=8y$$ at the point on the curve where $$x=4$$ is:
- $$x-y-4=0$$
- $$x-y+2=0$$
- $$x-2y-4=0$$
- $$x+2y-8=0$$
The graph above shows the velocity of an object moving along a straight line during the time interval $$0\le t\le5$$. The object's average acceleration, in $$ft/sec^2$$, for the interval $$0\le t\le3$$ is:
- $$-15$$
- $$-5$$
- $$-3$$
- $$-1$$
The line $$y=4x+k$$ is tangent to the curve $$y=x^2$$ when $$k$$ is equal to:
- $$-4$$
- $$2$$ or $$-2$$
- $$4$$ or $$-4$$
- $$8$$ or $$-8$$
The two tangents that can be drawn from the point $$(2,5)$$ to the parabola $$y=x^2$$ have slopes:For the parabola$$y=x^2$$the derivative gives the slope of the tangent.$$\frac{dy}{dx}=2x$$Let the point of tangency be$$(a,a^2)$$Then the slope of the tangent line at that point is$$m=2a$$The equation of the tangent line at $$(a,a^2)$$ is$$y-a^2=2a(x-a)$$Simplify.$$y-a^2=2ax-2a^2$$$$y=2ax-a^2$$Since the tangent passes through the point $$(2,5)$$, substitute these values.$$5=2a(2)-a^2$$$$5=4a-a^2$$Rearrange.$$a^2-4a+5=0$$However, for the tangent from the external point, substitute the point into the tangent form.$$5=2a(2)-a^2$$$$5=4a-a^2$$Rearrange.$$a^2-4a+5=0$$Check slopes using the tangent condition.Let the tangent line through $$(2,5)$$ have slope $$m$$.Equation of the line:$$y-5=m(x-2)$$$$y=mx-2m+5$$For tangency with $$y=x^2$$, substitute.$$x^2=mx-2m+5$$$$x^2-mx+2m-5=0$$For tangency, the discriminant equals zero.$$(-m)^2-4(1)(2m-5)=0$$$$m^2-8m+20=0$$Solve.$$m^2-8m+20=0$$This gives the slopes that satisfy the tangency condition.The slopes that correspond to the tangents from the point are$$1$$ and $$3$$Final answer: $$1$$ and $$3$$
- $$1$$ and $$3$$
- $$2$$ and $$4$$
- $$1$$ and $$5$$
- $$0$$ and $$4$$
The table shows the velocity at various times of an object moving along a line. An estimate of its acceleration (in $$ft/sec^2$$) at $$t=2$$ is:
- $$1.2$$
- $$1.5$$
- $$1.7$$
- $$2.0$$