5.2E: Exercises for Section 5.2

In exercises 1 - 4, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.

1) \(x=3+t,\quad y=1−t\)

2) \( x=8+2t, \quad y=1\)

Answer

\(m=0\)

3) \( x=4−3t, \quad y=−2+6t\)

4) \( x=−5t+7, \quad y=3t−1\)

Answer

\(m= -\frac{3}{5}\)

In exercises 5 - 9, determine the slope of the tangent line, then find the equation of the tangent line at the given value of the parameter.

5) \( x=3\sin t,\quad y=3\cos t, \quad \text{for }t=\frac{π}{4}\)

6) \( x=\cos t, \quad y=8\sin t, \quad \text{for }t=\frac{π}{2}\)

Answer

Slope\(=0; y=8.\)

7) \( x=2t, \quad y=t^3, \quad \text{for } t=−1\)

8) \( x=t+\dfrac{1}{t}, \quad y=t−\dfrac{1}{t}, \quad \text{for }t=1\)

Answer

Slope is undefined; \( x=2\).

9) \( x=\sqrt{t}, \quad y=2t, \quad \text{for }t=4\)

In exercises 10 - 13, find all points on the curve that have the given slope.

10) \( x=4\cos t, \quad y=4\sin t,\) slope = \(0.5\)

Solution

\(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{4\cos t}{-4\sin t} = - \cot t.\)
Setting this derivative equal to \(0.5,\) we obtain the equation, \(\tan t = -2.\)
\( \tan t = -2 \implies \dfrac{y}{x} = -2 \implies y = -2x.\)
Note also that this pair of parametric equations represents the circle \(x^2 + y^2 = 16.\)
By substitution, we find that this curve has a slope of \(0.5\) at the points:
\(\left(\frac{4\sqrt{5}}{5},\frac{−8\sqrt{5}}{5}\right)\) and \(\left(\frac{-4\sqrt{5}}{5},\frac{8\sqrt{5}}{5}\right).\)

11) \( x=2\cos t, \quad y=8\sin t,\) slope= \(−1\)

12) \( x=t+\dfrac{1}{t}, \quad y=t−\dfrac{1}{t},\) slope= \(1\)

Answer

No points possible; undefined expression.

13) \( x=2+\sqrt{t}, \quad y=2−4t,\) slope= \(0\)

In exercises 14 - 16, write the equation of the tangent line in Cartesian coordinates for the given parameter \(t\).

14) \( x=e^{\sqrt{t}}, \quad y=1−\ln t^2, \quad \text{for }t=1\)

Answer

\( y=−\left(\frac{4}{e}\right)x+5\)

15) \( x=t\ln t, \quad y=\sin^2t, \quad \text{for }t=\frac{π}{4}\)

16) \( x=e^t, \quad y=(t−1)^2,\) at \((1,1)\)

Answer

\( y=-2x+3\)

17) For \( x=\sin(2t), \quad y=2\sin t\) where \( 0≤t<2π.\) Find all values of \(t\) at which a horizontal tangent line exists.

18) For \( x=\sin(2t), \quad y=2\sin t\) where \( 0≤t<2π\). Find all values of \(t\) at which a vertical tangent line exists.

Answer

A vertical tangent line exists at \(t = \frac{π}{4},\frac{5π}{4},\frac{3π}{4},\frac{7π}{4}\)

19) Find all points on the curve \( x=4\cos(t), \quad y=4\sin(t)\) that have the slope of \( \frac{1}{2}\).

20) Find \( \dfrac{dy}{dx}\) for \( x=\sin(t), \quad y=\cos(t)\).

Answer

\( \dfrac{dy}{dx}=−\tan(t)\)

21) Find the equation of the tangent line to \( x=\sin(t), \quad y=\cos(t)\) at \( t=\frac{π}{4}\).

22) For the curve \( x=4t, \quad y=3t−2,\) find the slope and concavity of the curve at \( t=3\).

Answer

\( \dfrac{dy}{dx}=\dfrac{3}{4}\) and \( \dfrac{d^2y}{dx^2}=0\), so the curve is neither concave up nor concave down at \( t=3\). Therefore the graph is linear and has a constant slope but no concavity.

23) For the parametric curve whose equation is \( x=4\cos θ, \quad y=4\sin θ\), find the slope and concavity of the curve at \( θ=\frac{π}{4}\).

24) Find the slope and concavity for the curve whose equation is \( x=2+\sec θ, \quad y=1+2\tan θ\) at \( θ=\frac{π}{6}\).

Answer

\( \dfrac{dy}{dx}=4, \quad \dfrac{d^2y}{dx^2}=−4\sqrt{3};\) the curve is concave down at \( θ=\frac{π}{6}\).

25) Find all points on the curve \( x=t+4, \quad y=t^3−3t\) at which there are vertical and horizontal tangents.

26) Find all points on the curve \( x=\sec θ, \quad y=\tan θ\) at which horizontal and vertical tangents exist.

Answer

No horizontal tangents. Vertical tangents at \( (1,0)\) and \((−1,0)\).

In exercises 27 - 29, find \( d^2y/dx^2\).

27) \( x=t^4−1, \quad y=t−t^2\)

28) \( x=\sin(πt), \quad y=\cos(πt)\)

Answer

\( d^2y/dx^2 = −\sec^3(πt)\)

29) \( x=e^{−t}, \quad y=te^{2t}\)

In exercises 30 - 31, find points on the curve at which tangent line is horizontal or vertical.

30) \( x=t(t^2−3), \quad y=3(t^2−3)\)

Answer

Horizontal \( (0,−9)\);
Vertical \( (±2,−6).\)

31) \( x=\dfrac{3t}{1+t^3}, \quad y=\dfrac{3t^2}{1+t^3}\)

In exercises 32 - 34, find \( dy/dx\) at the value of the parameter.

32) \( x=\cos t,y=\sin t, \quad \text{for }t=\frac{3π}{4}\)

Answer

\(dy/dx = 1\)

33) \( x=\sqrt{t}, \quad y=2t+4,t=9\)

34) \( x=4\cos(2πs), \quad y=3\sin(2πs), \quad \text{for }s=−\frac{1}{4}\)

Answer

\(dy/dx = 0\)

In exercises 35 - 36, find \( d^2y/dx^2\) at the given point without eliminating the parameter.

35) \( x=\frac{1}{2}t^2, \quad y=\frac{1}{3}t^3, \quad \text{for }t=2\)

36) \( x=\sqrt{t}, \quad y=2t+4, \quad \text{for }t=1\)

Answer

\(d^2y/dx^2 = 4\)

37) Find intervals for \(t\) on which the curve \( x=3t^2, \quad y=t^3−t\) is concave up as well as concave down.

38) Determine the concavity of the curve \( x=2t+\ln t, \quad y=2t−\ln t\).

Answer

Concave up on \( t>0\).

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